01 circular motion - osbincbse.com

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Chapter 01: Circular Motion Formulae Section 1: Angular Displacement, Relation

Between Linear Velocity and Angular Velocity

1. Angular velocity:

i. = r

v

where, v = linear velocity r = radius of the circle along which

particle performs circular motion.

ii. = t

where, = angular displacement of the particle in circular motion during time interval t.

iii. = 2n where, n = frequency of revolution of particle

in circular motion.

iv. = T

2

where, T = period of revolution of particle performing circular motion.

2. Angular displacement: = t 3. Time period:

i. T = 2 r

v

ii. T =

2

4. Frequency of revolution:

n = 1

T 2

5. Linear velocity: i. v = r ii. v = 2nr Section 2: Angular Acceleration 1. Angular acceleration:

i. = t

where, = change in the angular velocity of a particle in circular motion during a time interval t.

ii. = 2 n

t

where, n = change in frequency of the particle in circular motion during a time interval t.

Section 3: Centripetal and Tangential Acceleration 1. Centripetal (or radial) acceleration:

ar = r

v2

= v = r2

2. Tangential acceleration: aT = r 3. Resultant or total acceleration:

a = 2 2t r t ra a 2a a cos

where, = angle made by ar with at

a = 2 2t ra a when = 90.

4. For U.C.M.:

a = ar = 2v

r= v

at = 0 Section 4: Centripetal and Centrifugal Forces 1. Centripetal force:

i. Fc = 2mv

r ii. Fc = mv

iii. Fc = mr2 iv. Fc = 42mrn2

v. Fc = 2

2

4 mr

T

where, m = mass of particle performing circular motion

Section 5: Motion of a Vehicle along a Curved

Unbanked Road 1. The necessary centripetal force:

Fc = mg = 2mv

r

where, m = mass of vehicle v = velocity of the vehicle r = radius of the curve road

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Std. XII Sci.: Physics Numericals

= coefficient of friction between the tyres of the vehicle and the surface of the road.

2. The maximum velocity with which a vehicle

can take a turn safely without skidding:

v = rg 3. The maximum angular velocity with which

a vehicle can take a turn safely without skidding:

= g

r

Section 6: Banking of Roads For motion of vehicles along a banked curve road: 1. The proper velocity or optimum velocity:

v = rg tan

where, = angle of banking 2. The maximum velocity without skidding:

vmax = s

s

tanrg

1 tan

where, s = coefficient of friction between the tyres

of the vehicle and surface of the road 3. Angle of banking:

= tan12v

rg

or tan =2v

rg

4. Height of inclined road:

h = d sin where, d = distance between the two front or rear

wheels. 5. The maximum velocity with which a vehicle

can go on a banked curved road without toppling:

v = d

rg2H

where, H = height of centre of gravity (C.G.) of the vehicle from the road.

Section 7: Conical Pendulum 1. Linear speed of bob:

v = rg tan

2. Angular velocity:

= g

cosl =

g tan

r

3. Periodic time:

T = 2

= 2cos

g

l

= 2h

g

where, l = length of conical pendulum h = the height of the fixed support from the

centre of the circle or axial height of the cone

= semivertical angle of the cone. 4. Tension in the string:

T = mg

cos

Section 8: Vertical Circular Motion 1. Velocity at any point in vertical circular

motion:

i. vP = 2Lv 2gr (1 cos)

ii. vL = 5rg

iii. vH = rg

iv. vM = 3rg

where, vP = velocity of the particle at any

point P along the circle. vL = Minimum velocity at the lowest

point on the circle so that it can safely travel along the vertical circle (looping the loop).

vH = Minimum velocity of the particle at the highest point on the circle so that the string will not be slackened.

vM = Minimum velocity of the particle at a mid-way point so that it can travel along the circle.

r = radius of the vertical circle. = angle between the position vectors

at the given position of particle and that of the lowest point on the vertical circle.

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Chapter 01: Circular Motion

2. Relation between velocities at different points in vertical circular motion:

i. 2Lv = 2

Hv + 4gr

ii. 2Mv = 2

Hv + 2gr 3. Tension at: i. Any point P,

TP = 2mv

r+ mg cos

ii. Lowest point L,

TL = 2Lmv

r + mg =

2Hmv

r + 5mg

iii. Highest point H,

TH = 2Hmv

r mg =

2Lmv

r 5 mg

iv. Midway point M,

TM = 2Mmv

r=

2Lmv

r 2 mg

4. Total energy at any point:

E = 1

2 mv2 + mgr (1 cos )

= 5

2 mgr

Section 9: Kinematical Equations Analogy between translatory motion and circular motion

No. Translatory Motion Circular Motion

1. Linear velocity v =

d r

dt

Angular velocity =

d

dt

2. Linear acceleration

a =

2

d v d r

dt dt

Angular acceleration

=

2

d d

dt dt

3. Linear momentum p

= m v

Angular momentum L = I

4. Linear impulse = F

(t) = p Angular impulse =

(t) = L

5. Force F = m

a Torque

= I

6. Work W = F

. r Work W =

.

7.

Kinetic energy of translation

Et = 1

2mv2

Kinetic energy of rotation

Er = 1

2I2

8. Equations of linear motion Equations of rotational motion

i. v = u + at i. 2 = 1 + t

ii. s = ut + 1

2at2 ii. = 1t +

2

1t2

iii. v2 u2 = 2 as iii. 22

21 = 2

iv. sn = u + a

2(2n 1) iv. n = 1 +

2

(2n 1)

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Std. XII Sci.: Physics Numericals Shortcuts 1. Angular acceleration (): If number of rotations or revolutions is given,

then

= 2

t

(n2 n1) rad/s2

2. If linear velocity is given, apply ar = r

v2

and

if angular velocity is given, apply ar = 2r to find ar.

3. If number of revolutions in a particular time is given, apply ar = 42n2r.

4. If some mass is placed on a rotating body, then angular motion changes due to frictional force between the mass and the body.

Coefficient of friction can be calculated by applying the formula,

2mv

r= mg or

= 2v

rg =

2r

g

=

2 24 n r

g

5. Breaking tension is the maximum centripetal force which is given by the relation,

F = 2mv

r = m2r

6. Speed of a vehicle on a banked road or

circular turn depends upon the curvature of the road.

tan = 2v

rg

7. To avoid skidding, v rg where is the

coefficient of friction between the tyres and road.

= h2

d where d is the distance between two

wheels. 8. In case of conical pendulum if l be the length

and r be the radius of the horizontal circle then height of the rigid point of suspension is calculated by the formula,

h = 2 2rl

9. Use the expressions, Lv = 5gr and Tv = gr

for the following cases: i. Bucket full of water whirled in a vertical

circle. ii. Motor cyclist riding in a vertical circle

in hollow sphere.

10. For looping the loop of radius r, the minimum height from which the body should be released is given by,

h = 5r

2

11. i. For a vehicle moving over a convex

bridge which is in the shape of an arc of a circle,

N = mg – 2mv

r

ii. For the motor cyclist at the upper most point in the globe of death in a circus,

N = 2mv

r – mg

where, N = normal reaction acting on the

vehicle or the motorcycle.

Solved Examples Section 1: Angular Displacement, Relation


Example 1.1 A wheel of radius 2 metre is making 60 revolutions per minute. Calculate the angular velocity of any point on the rim. Solution: Given: n = 60 r.p.m. = 60/60 = 1 rps, r = 2 m To find: Angular velocity () Formula: = 2n Calculation: From formula, = 2 1 = 2 rad/s Ans: The angular velocity of any point on the rim

of the wheel is 2 rad/s. Example 1.2 Find the angular displacement of a particle moving on a circle with angular velocity (2/3) rad/s in 15 s. Solution: Given: t = 15 s, = (2/3) rad/s To find: Angular displacement () Formula: = t Calculation: From formula,

= 2

153

= 10 π rad Ans: The angular displacement of the particle is

10 rad.

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Example 1.3 What is the angular speed of the second hand of a clock? If the second hand is 10 cm long, then find the linear speed of its tip. Solution: Given: r = 10 cm = 101 m, T = 60 s To find: i. Angular speed () ii. Linear speed (v)

Formulae: i. = 2

T

ii. v = r

Calculation: From formula (i),

= 2

T

=

60

14.32

= 1.047 101 rad/s From formula (ii), v = r v = 1.047 101 0.1 v = 1.047 102 m/s. Ans: i. The angular speed of the second hand of

a clock is 1.047 101 rad/s. ii. The linear speed of the tip of the second

hand is 1.047 102 m/s. Example 1.4 If a body rotates in a horizontal circle of radius 15 cm with an angular velocity of 0.8 rad/s, then what is its linear velocity? Solution: Given: r = 15 cm = 0.15 m, = 0.8 rad/s To find: Linear velocity (v) Formula: v = r Calculation: From formula, v = 0.15 0.8 v = 0.12 m/s. Ans: The linear velocity of the rotating body is

0.12 m/s. Example 1.5 The linear velocity of a body is 0.2 m/s. If it rotates in a horizontal circle having radius 0.5 m, what is its angular velocity? Solution: Given: r = 0.5 m, v = 0.2 m/s To find: Angular velocity () Formula: v = r Calculation: From formula,

= v

r =

0.2

0.5

= 0.4 rad/s Ans: The angular velocity of the rotating body is

0.4 rad/s.

Example 1.6 A particle is revolving in a circle of radius 10 cm with linear speed of 20 m/s. Find (i) its period of revolution and (ii) frequency. Solution: Given: r = 10 cm = 10 102 m, v = 20 m/s To find: i. Period of revolution (T) ii. Frequency (n)

Formulae: i. T = 2 r

v

ii. n = 1

T


T = 22 3.14 10 10

20

= 3.14 102 T = 0.0314 s From formula (ii),

n = 1

T=

1

0.0314

n = 31.85 s–1 Ans: i. The period of revolving particle is

0.0314 s ii. The frequency of the particle is

n = 31.85 s–1. Example 1.7 What is the angular velocity of the minute hand of a clock? If the minute hand is 5 cm long, what is the linear velocity of its tip? [Oct 04] Solution: Given: R = 5 cm = 5 102 m, T = 1h = 3600 s To find: i. Angular velocity () ii. Linear velocity of the tip (v)

Formulae: i. = 2

T

ii. v = r Calculation: From formula (i),

= 2

T

=

2 3.14

3600

= 1.74 103 rad/s From formula (ii), v = 5 102 1.74 103 v = 8.7 105 m/s Ans: For the minute hand of the clock, i. The angular velocity is 1.74 10–3 rad/s. ii. The linear velocity of the tip is

8.7 10–5 m/s.

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Example 1.8 An aircraft takes a turn along a circular path of radius 2000 metre. If the linear speed of the aircraft is 500 m/s, find the angular speed and the time required by it to complete half the circular path. Solution: Given: r = 2000 m, v = 500 m/s To find: i. Angular speed () ii. Time required to complete half

the circular path (t)

Formulae: i. = v

r ii. v =

2 r / 2

t


= 500

2000

= 0.25 rad/s From formula (ii),

t = 2 3.14 2000

2 500

t = 12.56 s Ans: i. The angular speed of the aircraft is

0.25 rad/s. ii. The time required by the aircraft to

complete half the circular path is 12.56 s.

Section 2: Angular Acceleration Example 2.1 A particle performing UCM changes its angular velocity from 70 r.p.m to 130 r.p.m in 18 s. Find the angular acceleration of the particle. Solution:

Given: n1 = 70/min = 70

60/s,

n2 = 130/min = 60

130/s, t = 18 s

To find: Angular acceleration ()

Formula: = 2 1

t

=

2 12 n n

t

Calculation: From formula,

= 2 12 n 2 n

t

=

130 702

60 6018

=

130 702

6018

= 2 60

60 18

3.142

9 9

= 0.349 rad/s2 Ans: The angular acceleration of the particle is

0.349 rad/s2. Example 2.2 The frequency of a particle performing circular motion changes from 60 r.p.m to 180 r.p.m in 20 second. Calculate the angular acceleration. [Oct 98] Solution:

Given: n1 = 60 r.p.m = 60

60 = 1 rev/s,

n2 = 180 r.p.m = 180

60 = 3 rev/s,

t = 20 s To find: Angular acceleration ()

Formula: = 2 1

t


= 2 12 n 2 n

t

=

2 3 1

20

= 2 3.142 2

20

= 3.142

5

= 0.6284 rad/s2 Ans: Angular acceleration of the particle is

0.6284 rad/s2. Example 2.3 A fly wheel rotating at 420 r.p.m. slows down at a constant rate of 2 rad/s2. What time is required to stop the fly wheel ? Solution:

Given: n1 = 420 r.p.m. =420

60 r.p.s.

= 7 r.p.s = 2 rad/s2, 2 = 0 To find: Time required (t) Formulae: i. = 2n ii. 2 = 1 + t Calculation: From formula (i), 1 = 2n1 = 2 7 = 14 rad/sec

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From formula (ii), 0 = 14 + (2) t 0 = 14 2t 2t = 14

t = 14

2

=

14

2

22

7

t = 22 s Ans: The time required to stop the flywheel is 22 s. Example 2.4 A particle performs circular motion with a constant angular acceleration of 4 rad/s2. If the radius of the circular path is 20 cm and the initial angular speed of the particle is 2 rad/s, find the angular speed of the particle after 0.5 sec. Solution: Given: = 4 rad/s2, r = 20 cm = 0.2 m, 1 = 2 rad/s, t = 0.5 s To find: Angular speed (2) Formula: 2 = 1 + t Calculation: From formula, 2 = 2 + 4 0.5 = 2 + 2 2 = 4 rad/s Ans: The angular speed of the particle after 0.5 s is

4 rad/s. Example 2.5 A stone tied to one end of a string is whirled in a horizontal circle of radius 40 cm with a frequency of 30 r.p.m. Find the angular velocity and linear velocity of the stone. Solution: Given: r = 40 cm = 40 102 m

n = 30 rpm = 30

60 rps = 0.5 rps

To find: i. Angular velocity () ii. Linear velocity (v) Formulae: i. = 2n ii. v = r Calculation: From formula (i), = 2 3.14 0.5 = 1.0 3.14 = 3.14 rad/s From formula (ii), v = 40 102 3.14 = 4 3.14 102 = 125.6 102 v = 1.256 m/s Ans: For the stone: i. the angular velocity is 3.14 rad/s ii. the linear velocity is 1.256 m/s.

Example 2.6 A satellite revolves around the earth in a circular orbit of radius 7000 km. If its period of revolution is 2 h, calculate its angular speed, linear speed and its centripetal acceleration. Solution: Given: r = 7000 km = 7000 103 m, T = 2h = 2 3600 = 7200 s To find: i. Angular speed () ii. Linear speed (v) iii. Centripetal acceleration (ar)

Formulae: i. = 2

T

ii. v = r

iii. ar = r2 Calculation: From formula (i),

= 2 3.14

7200

= 8.72 104 rad/s From formula (ii), v = 7000 103 8.72 104 v = 61.04 102 v = 6.104 103 m/s. v = 6.104 km/s. From formula (iii), ar = 7000 103 (8.72 104)2 = 7 (8.72)2 10–5 ar = 5.32 m/s2 Ans: For the revolving satellite: i. the angular speed is 8.72 104 rad/s, ii. the linear speed is 6.104 km/s iii. the centripetal acceleration is 5.32 m/s2 Example 2.7 The earth moves round the sun in an almost circular orbit of radius 1.5 1011 m with constant angular speed. Calculate its i. angular velocity ii. linear velocity iii. centripetal acceleration [Given: 1 year = 3.156 107 sec, Mass of earth = 5.98 1024 kg] Solution: Given: r = 1.5 1011 m, m = 5.98 1024 kg, T = 1 yr = 3.156 107 s To find: i. Angular velocity () ii. Linear velocity (v) iii. Centripetal acceleration (ar)

Formulae: i. = 2

T

ii. v = r

iii. ar = r2

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= 7

2 3.14

3.156 10

1.99 10–7 rad/s. From formula (ii), v = 1.5 1011 1.99 107 v = 2.985 104 m/s From formula (iii), ar = 1.5 1011 (1.99 107)2

ar = 2antilog log 1.5 log 1.99

1011 10–14

= antilog 0.1716 2 0.2989 10–3

= antilog 0.1761 0.5978 10–3

= antilog 0.7739 10–3

ar = 5.941 10–3 m/s2

Ans: For the earth moving round the sun: i. the angular velocity is 1.99 10–7 rad/s, ii. the linear velocity is 2.985 104 m/s and iii. the centripetal acceleration is

5.941 10–3 m/s2. Section 3: Centripetal and Tangential Acceleration Example 3.1 A 0.5 kg mass is rotated in a horizontal circle of radius 20 cm. Calculate the centripetal acceleration if the angular velocity is 0.6 rad/s Solution: Given: m = 0.5 kg, r = 20 cm = 0.2 m, = 0.6 rad/s To find: Centripetal acceleration (ar)

Formulae: i. v = r ii. ar = r

v2

Calculation: From formula (i), v = r = 0.2 0.6 m/s From formula (ii),

ar = 2(0.2 0.6)

0.2

ar = 0.072 m/s2 Ans: The centripetal acceleration of the mass is

0.072 m/s2. Example 3.2 A stone tied to the end of a string which is 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude of centripetal acceleration?

Solution: Given: r = 80 cm = 0.80 m, n = 14/25 s1 To find: Centripetal acceleration (ar) Formula: Centripetal acceleration, ar = r2 Calculation: From formula,

ar = 42n2r ….[ = 2n]

= 4 (3.142)2 2

14

25

0.8

=

2 2

2

4 3.142 14 0.8

25

= 9.908 ar 9.91 m/s2 Ans: The magnitude of centripetal acceleration is

9.91 m/s2. Example 3.3 A racing car completes 5 rounds of a circular track in 2 minutes. Find the radius of the track if the car has uniform centripetal acceleration of 2 m/s2. [Oct 13] Solution: Given: 5 rounds = 2r(5), t = 2 minutes = 120 s To find: Radius (R) Formulae: acp = 2r Calculation: From formula, acp = 2r

2 = 2v

r

But v = 2 r (5)

t

=

10 r

t

2 = 2 2

2

100 r

rt

r = 120 120

100

= 144 m

Ans: The radius of the track is 144 m. Example 3.4 The tangential acceleration of the tip of a blade is 47.13 m/s2 and its centripetal acceleration is 473.9 m/s2. Find the value of linear acceleration of the tip of the blade. Solution: Given: at = 47.13 m/s2, ar = 473.9 m/s2 To find: Linear acceleration (a)

Formula: a = 2 2r ta a

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a = 2 2(473.9) (47.13)

Now, (473.9)2 = antilog [2 log(473.9)] = antilog [2 2.6757] = antilog [5.3514] = 2.246 105 (47.13)2 = antilog [2 log(47.13)] = antilog [2 1.6733] = antilog [3.3466] = 2221

ar = 2 2473.9 47.13

= 52.246 10 2221

= 226821

= antilog 1log 226821

2

= antilog1

5.35562

= antilog [2.6778] ar = 476.2 = 4.762 102 m/s2 Ans: The value of linear acceleration of the tip of

the blade is 4.762 102 m/s2. Example 3.5 A particle is revolving in a circle. Its angular speed increases from 2 rad/s to 40 rad/s in 19 sec. The radius of the circle is 20 cm. Compare the ratio of centripetal acceleration to tangential acceleration. Solution: Given: 1 = 2 rad /s 2 = 40 rad /s t = 19 s r = 20 cm = 20 102 m = 0.2 m To find: Ratio of centripetal to tangential

acceleration (ar : at) Formulae: i. ar = 2r ii. at = r Calculation: From formula (i), ar = r 2 = 0.2 (40)2 = 0.2 1600 ar = 320 m /s2

From formula (ii) at = r

= r 2 1

t

= 0.2 40 2

19

= 0.2 38

19

= 0.2 2 = 0.4 m /s2

r

t

a

a =

320

0.4

= 3200

4

ar : at = 800 : 1 Ans: The ratio of centripetal acceleration to

tangential acceleration of the particle is 800 : 1. Example 3.6 The tangential acceleration of a body is 29.48 m/s2 and its linear acceleration is 52.3 m/s2. Find its radial acceleration. Solution: Given : a = 52.3 m/s2, at = 29.48 m/s2 To find: Radial acceleration (ar)

Formula: a = 2 2t ra a or a2 = 2

ta + 2ra

or ar = 2 2ta a


ar = 2 2(52.3) (29.48)

Now, (52.3)2 = antilog [2 log (52.3)] = antilog [2 1.7185] = antilog [3.4370] = 2735 (29.48)2 = antilog [2 log (29.48)] = antilog [2 1.4695] = antilog [2.9390] = 869

ar = 2 252.3 29.48

= 2735 869

= 1866

= antilog 1log 1866

2

= antilog1

3.27092

= antilog [1.63545] ar = 43.1 m/s2 Ans: The radial acceleration of the body is

43.1 m/s2.

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Section 4: Centripetal and Centrifugal Forces Example 4.1 A 0.2 kg mass is rotated in a horizontal circle of radius 10 cm. If its angular speed is 0.4 rad/s, find the centripetal force acting on it. Given: m = 0.2 kg, r = 10 cm = 10 102 m, = 0.4 rad /s To find: Centripetal force (Fc) Formula: Fc = mr2 Calculation: From formula, Fc = 0.2 10 102 (0.4)2 = 0.32 102 N Fc = 0.0032 N Ans: The centripetal force acting on the rotating

mass is 0.0032 N. Example 4.2 A car of mass 1500 kg rounds a curve of radius 250m at 90 km/hour. Calculate the centripetal force acting on it. [Feb 13] Solution: Given: m = 1500 kg, r = 250 m,

v = 90 km/h = 5

9018

= 25 m/s

To find: Centripetal force (FCP)

Formula: FCP = 2mv

r


FCP = 21500 25

250

FCP = 3750 N Ans: The centripetal force acting on the car is

3750 N. Example 4.3 A stone of mass 1 kg is whirled in horizontal circle attached at the end of a 1 m long string. If the string makes an angle of 30 with vertical, calculate the centripetal force acting on the stone. (g = 9.8 m/s2). [Mar 14] Solution: Given: m = 1 kg, l = 1 m, = 30, g = 9.8 m/s2 To find: Centripetal force (FCP)

Formulae: i. FCP = 2mv

r

ii. v = rg tan

Calculation: Substituting formula (ii) in (i),

FCP = 2

m rg tan

r

= mg tan = 1 9.8 tan 30

= 9.8 1

3=

9.8

1.732

= 5.658 N Ans: The centripetal force acting on stone is 5.658 N. Example 4.4 A coin is placed on a revolving disc which revolves at 60 r.p.m. It does not slip-off when it is at 15 cm from the axis of rotation. What should be the distance of the coin from the axis of rotation so that it does not slip-off, when the speed of the revolving disc is changed to 75 r.p.m? Solution: Given: r1 = 15 cm = 0.15 m

n1 = 60 r.p.m = 60

60 = 1 r.p.s.

1 = 2n1 = 2 rad/s = 2 3.14 = 6.28 rad/s

n2 = 75 r.p.m. = 75

60 r.p.s

= 1.25 r.p.s 2 = 1.25 2 3.14 = 7.85 rad/sec To find: Distance of the coin from the axis (r2) Formula: mr1

21 = mr2

22


r2 = 2

1 122

r

=

2

2

0.15 6.28

7.85

r2 = antilog [log (0.15) + log(6.28)2 – log(7.85)2] = antilog [log(0.15) + 2 log(6.28)

– 2 log (7.85)] = antilog [ 1.1761+ 2 0.7980 – 2 0.8949] = antilog [ 1 .1761 + 1.5960 – 1.7898]

= antilog [0.7726 + 2 .2102]

= antilog [ 2 .9828] = 0.09612 m 0.096 m r2 = 9.6 cm Ans: The distance of the coin from the axis so that

it does not slip-off is 9.6 cm.

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Example 4.5 A string breaks under tension of 10 kgwt. If the string is used to revolve a body of mass 1.2 kg in a horizontal circle of radius 50 cm, what is the maximum speed with which the body can be revolved? What is its period then? Solution: Given: Tension, Tmax = 10 9.8 = 98N, m = 1.2 kg, r = 0.5 m To find: i. Maximum speed (vmax) ii. Time Period (T)

Formulae: i. Tmax = 2maxmv

r

ii. v = 2 r

T


2maxv =

m

rTmax

= 2.1

5.098 = 40.833

vmax = 40.83 = 6.39 m/s. From formula (ii),

T = 39.6

5.014.32 =

3.14

6.39

T = 0.4914 s Ans: i. The maximum speed with which the

body can be revolved is 6.39 m/s. ii. The period of revolution of the body is

0.4914 s. Example 4.6 A stone of mass 0.3 kg is tied to the end of a string. It is whirled in a circle of radius 1m with a speed of 40 rev min1. If the string can withstand a maximum tension of 200 N, then find the tension and maximum speed with which the stone can be whirled. Solution: Given: m = 0.3 kg, r = 1 m,

n = 40 rpm = 40

60rps =

2

3rps,

= 2n = 2 2

3 =

4

3

rad s1,

Tmax = 200 N To find: i. Tension (T) ii. Maximum speed (vmax) Formulae: i. Tension, T = mr2

ii. Tmax = 2maxmv

r


T = 0.3 1 2

4

3

T = 5.258 N Now, for maximum velocity, tension is

maximum. From formula (ii),

vmax = maxT r

m

=

200 1

0.3

=

2000

3

vmax = antilog 1log 2000 log 3

2

= antilog 13.3010 0.4771

2

= antilog 12.8239

2

= antilog 1.41195

= antilog 1.4120

vmax = 25.82 ms1 Ans: For the stone whirled in a circle: i. the tension in the string is 5.258 N. ii. the maximum speed is 25.82 m/s. Example 4.7 A certain string breaks under a tension of 45 kg-wt. A mass of 100 g is attached to this string of length 500 cm and whirled in a horizontal circle. Find the maximum number of revolutions per second without breaking the string. [Mar 92] Solution: Given: T = 45 kg-wt = 45 9.8 N, m = 100 g = 100 103 g = 0.1 kg, r = 500 cm = 5 m To find: Maximum number of revolution per

second (n)

Formulae: i. F = T = 2mv

r

ii. v = r = r (2n) Calculation: From formula (i) and (ii),

T = 2m(2 r n)

r

= m 42 n2r

n2 = 2

T

4 r m

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n = 2

45 9.8

4 (3.14) 5 0.1

= 2

45 4.9

3.14

=45 4.9

3.14

= antilog 1log 45 log 4.9 log 3.14

2

= antilog 11.6532 0.6902 0.4969

2

= antilog 12.3434 0.4969

2

= antilog {1.1717 – 0.4969} = antilog (0.6748) n = 4.73 Hz Ans: The maximum number of revolutions of the

mass without breaking the string are 4.729. Example 4.8 An electron of mass 9 1031 kg moves in a circular orbit of radius 5.3 1011 m around a proton in a hydrogen atom. If the charges are 1.6 1019 C each, find the frequency of revolution. Solution: Given: me = 9 1031 kg, r = 5.3 1011 m, qe = qp = 1.6 1019 C (magnitudes only) To find: Frequency of revolution (n)

Formula: F = 0

1

4. e p

2

q q

r= 42 mrn2


n = e p

2 30

q q 1.

4 mr 4 = e e

2 20

q q 1

4 mr r 4

= e2

0

q 1 1

r 4 mr 4

=

19 9

11 2 31 11

1.6 10 9 10

5.3 10 4 3.14 9 10 5.3 10

n = 1.6

5.3 10–8

1

29

2 30 11

10

4 3.14 0.53 10 10

=1.6

5.3 10–8

1

250

2

10

4 3.14 0.53

=1.6

5.3

1

2

2

1

4 3.14 0.53

10+17

= antilog log 1.6 log 5.3

1 1log 4 log 3.14 log 0.53

2 2

10+17

=1

antilog 0.2041 0.7243 0.60212

10.4969 1.7243

2

10+17

= antilog 1.4798 0.3011 0.4969

12 1.7243

2

10+17

= antilog 1.1787 0.4969 1.8621 10+17

= 17antilog 2.6818 1 0.8621 10

= 17antilog 2.6818 0.1379 10

= antilog 2.8197 1017

n = 6.604 10–2 1017 n = 6.604 1015 Hz Ans: The frequency of revolution of electron is

6.604 1015 Hz. Section 5: Motion of a Vehicle along a Curved

Unbanked Road Example 5.1 A vehicle driving at 54 km/hr can safely negotiate a curved road of 50 m radius. If the road is unbanked, find the co-efficient of friction between the road surface and the tyres. Solution: Given: r = 50 m, v = 54 km/h

= 54 5

18

= 15 m/s

To find: Coefficient of friction () Formula: v = rg


= 2(15)

50 9.8

= antilog [2 log (15) – log (50) – log (9.8)] = antilog [2 1.1761 – 1.6990 – 0.9912]

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= antilog [2.3522 – 2.6902]

= antilog 1.6620

= 0.459 Ans: The coefficient of friction between the road

surface and the tyres of the vehicle is 0.459. Example 5.2 A cyclist speeding at 18 km h1 on a level road takes a sharp circular turn of radius 3 m without reducing the speed and without bending towards the centre of the circular path. The coefficient of static friction between the tyres and the road is 0.1. Will the cyclist slip while taking the turn? Solution:

Given: v = 18 km/h = 18 5

18 = 5 m/s,

r = 3 m, = 0.1 To find: Whether the cyclist can take the turn or

not without slipping Formula: Maximum safe speed of cyclist

vmax = rg


vmax = 0.1 3 9.8 = 2.94

vmax = 1.715 m/s V > Vmax

Ans: As the actual speed is greater than the maximum safe speed, the cyclist will slip while taking the turn.

Example 5.3 A coin just remains on a disc rotating at 120 r.p.m. when kept at a distance of 1.5 cm from the axis of rotation. Find the coefficient of friction between the coin and the disc. Solution: Given: n = 120 r.p.m = 120 rev/60 s = 2 rev /s = 2 Hz r = 1.5 cm = 1.5 102 m g = 9.8 m/ s2 To find: Coefficient of frictions () Formula: mr2 = .m.g OR r2 = g Calculation: From formula,

= 2r

g

= 2(2 n) r

g

=

2 24 n r

g

….[ = 2n]

= 2 2 24 (3.14) (2) 1.5 10

9.8

= 2 216 (3.14) 1.5 10

9.8

= antilog log 16 2log 3.14

log 1.5 log 9.8 10–2

= antilog 1.2041 2 0.4969

0.1761 0.9912 10–2

= antilog 1.2041 0.998 0.1761

0.9912 10–2

= antilog 2.3740 0.9912 10–2

= antilog 1.3828 10–2

= 2.414 101 10–2 = 0.2414 Ans: The coeffcient of friction between the coin

and disc is 0.2414. Example 5.4 A coin kept at a distance of 5 cm from the centre of a turntable of radius 1.5 m just begins to slip when the turnable rotates at a speed of 90 r.p.m. Calculate the coefficient of static friction between the coin and the turntable. [g = 9.8 m/s2]. [Mar 16] Solution: Given: r = 5 cm = 0.05 m

n = 90 r.p.m. = 90

60 r.p.s.

g = 9.8 m/s2 To find: Coefficient of static friction (s)

Formula: s = 2r

g

Calculation:

Since, = 2n = 2 90

60

= 3 rad/s

s = 20.05 (3 )

9.8

= 20.45 (3.14)

9.8

= {antilog [(1.6532) + 2(0.4969) – (0.9912)]}

= {antilog(1.6558)} = 0.4527

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Ans: The coefficient of static friction between the coin and the turntable is 0.4527.

Example 5.5 A coin placed on a revolving disc at a distance of 4.8 cm from the axis of rotation does not slip. If the coefficient of friction between the coin and the disc is 0.4, then what is the angular velocity of the disc? Solution: Given : r = 4.8 cm = 0.048 m, = 0.4 To find: Angular velocity ()

Formula: v = rg = r


v = 0.4 0.048 9.8

v = antilog 1 1log 0.4 log 0.048

2 2

1log 9.8

2

= antilog 1 1

1.6021 2.68122 2

10.9912

2

= antilog 12 1.6021 1.3406 0.4956

2

= antilog 1.8011 1.8362

= antilog 1.6373

= 0.4338 0.43 m/s Also, from formula,

= v

r =

0.43

0.048

= antilog [log (0.43) – log (0.048)]

= antilog 1.6335 2.6812

= antilog [0.9523] = 8.96 rad/s Ans: The angular velocity of the disc is 8.96 rad/s. Example 5.6 The coefficient of friction of a road is when it is dry and the maximum speed of a car in a circular path along the road is 8 m/s. The coefficient of

friction becomes 2

when the road is wet. What

will be the maximum speed permitted in this situation?

Solution:

Given: dry = , wet = 2

vdry = 8 m/s To find: Maximum speed permitted (vdry)

Formula: v = gr

Calculation: From formula for dry road,

8 = gr ….(i)

From formula for wet road,

vwet = gr2

vwet = gr

2

….(ii)

Dividing eq. (ii) by eq. (i),

wetv

8 =

gr 1

2 gr

vwet = 8

2=

2

4 2

2

= 4 2

= 4 1.414 vwet = 5.656 m/s Ans: The maximum speed permitted when the road

is wet is 5.656 m/s. Section 6: Banking of Roads Example 6.1 A curve of radius 900 m is banked so that no friction is required at a speed of 30 ms1. Calculate the angle of banking. Solution: Given: v = 30 ms1, r = 900 m, g = 9.8 ms2 To find: Angle of banking ()

Formula: tan = 2v

rg


tan = 230

900 9.8

tan = 900

900 9.8 =

1

9.8 = 0.1021

….[using table of reciprocals] Angle of banking, = tan1(0.1021) = 550 6 Ans: The angle of banking of the curve is 6.

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Example 6.2 An aircraft executes a horizontal loop at a speed of 720 km h1 with its wing banked at 15. Calculate the radius of the loop. Solution: Given: v = 720 kmh1

= 720 1000

3600

ms1

= 200 ms1, = 15 To find: Radius of the loop (r)

Formula: tan = 2v

rgor r =

2v

g tan


r = 2(200)

9.8 tan15 =

2(200)

9.8 0.2679

r = antilog 2log (200) log(9.8) log(0.2679)

= antilog 2 2.3010 0.9912 1.4280

= antilog 4.6020 0.9912 0.5720

= antilog 3.6108 0.5720

= antilog 4.1828

= 1.524 104 m r = 15.24 km Ans: The radius of the loop is 15.24 km. Example 6.3 A motor cyclist is going along a banked circular path of length 1 km. He has to lean inwards making an angle of 21 49 with the vertical in order to keep his balance. Find his speed. Solution: Given: = 210 49, Distance(s) = Circumference

2r = 1 km = 1000 m or r = 1000

m2

= 159.23 m To find: Speed (v) Formula: v = rg tan


v = 0159.23 9.8 tan 21 49'

= 159.23 9.8 0.4003

v = antilog 1log(159.23) log(9.8) log(0.4003)

2

= antilog 12.2020 0.9912 1.6024

2

= antilog1

1.1010 0.4956 (2 1.6024)2

= antilog 1.5966 1.8012

= antilog 1.3978

v = 25 m/s Ans: The speed of the motor cyclist is 25 m/s. Example 6.4 Calculate the angle through which a cyclist bends from the vertical when he crosses a circular path

of circumference 34.3 m in 22 s.

[Take g = 9.8 ms2]. Solution: Given: Circumference = 34.3 m,

t = 22 s

To find: Bending angle of cyclist ()

Formula: tan =rg

v2

Calculation: r = 2

ncecircumfere

= 34.3

222

7

= 34.3 7

2 22

m

Velocity of cyclist,

v = Length of path

Time taken

= 22

3.34 ms1

From formula,

tan = 2

34.3 34.3 7g

2 2222

= 2

34.3

22

2 22

34.3 7

1

9.8

= 1

= tan–1 (1)

= 45 Ans: The angle through which the cyclist bends

from the vertical is 45.

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Example 6.5 Calculate the maximum speed with which a car can be safely driven along a curved road of radius 30 m and banked at 30 with the horizontal [g = 9.8 m/s2] [Mar 96] Solution: Given: r = 30 m, = 30, g = 9.8 m/s2 To find: Maximum speed (vmax)

Formula: vmax = rg tan


vmax = 30 9.8 tan(30 )

= 30 9.8 0.5774

vmax = antilog 1log (30) log(9.8) log(0.5774)

2

= antilog1

1.4771 0.9912 1.76152

= antilog 12.2298

2

= antilog[1.1149] vmax = 13.03 m/s Ans: The maximum speed with which the car can

drive safely along the curve of the road is 13.03 m/s.

Example 6.6 What is the greatest speed at which a car having a 1.5 m wide track can turn in a circular path of radius 24.5 m without overturning? Assume that the centre of gravity of the car is 0.49 m above the ground. Solution: Given: 2d = 1.5 m, d = 0.75 m, h = 0.49 m, r = 24.5 m To find: Greatest speed (Vmax)

Formula: vmax = h

grd


vmax = 49.0

75.05.248.9

vmax = antilog

1log(9.8) log(24.5) log(0.75) log(0.49)

2

= antilog1

0.9912 1.3892 1.8751 1.69022

= antilog1

2.2555 1.69022

= antilog 12.5653

2

= antilog 1.2827

vmax = 19.17 ms 1 Ans: The greatest speed at which the car can turn

without overturning is 19.17 ms 1. Example 6.7 Find the angle of banking of a railway track of radius of curvature 1600 m. The optimum velocity of the train is 20 m/s. Also find the elevation of the outer track above the inner track if the distance between the two tracks is 1.8 m. [g = 9.8 m/s2]. Solution: Given: r = 1600 m, v = 20 m/s, l = 1.8 m

To find: i. Angle of banking () ii. Elevation of outer track (h)

Formulae: i. tan = 2v

rg

ii. h = l sin Calculation: From formula (i),

= tan1 2v

rg

= tan1 400

1600 9.8

= tan1 1

4 9.8

= tan1 1

39.2

= tan1 [0.02551] ….[Using the table of reciprocals]

= 1 28 Let h be the elevation of the outer track above the inner track and l be the distance between them.

From formula (ii),

h = 1.8 sin(1 28) = 1.8 0.0256

h = 0.046 m Ans: i. The angle of banking of the railway

track is 128. ii. The elevation of the outer track above

the inner track is 0.046 m.

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Example 6.8 A car moves round a curved path of radius 50 m on a road banked through 14. If the coefficient of friction between the car tyres and the road surface is 0.3, what is the maximum permissible velocity of the car for safe travel ? Solution: Given: r = 50 m, = 14, = 0.3 To find: Maximum permissible velocity (vmax)

Formula: vmax = tan

rg1 tan


vmax = 0.3 tan14

50 9.81 0.3tan14

= 0.3 0.2493

50 4.91 0.3 0.2493

= 490 0.5493

0.9252

= 1anti log log(490) log(0.5493) log(0.9252)

2

= antilog1

2.6902 1.7398 1.96622

= antilog1

2.4310 1.96622

= antilog 12.4648

2

= antilog 1.2324

vmax = 17.08 m/s Ans: The maximum possible velocity of the car for

safe travel is 17.08 m/s. Example 6.9 A road is designed for vehicles to have a maximum velocity of 30 km/h. A turn on the road has a radius of 30 m. If the coefficient of friction between the vehicle tyres and the road surface is 0.2, what should be the angle of banking? Solution: Given: vmax = 30 km/hr

= 30 1000

60 60

m/s

= 25

3 m/s

r = 30 m, = 0.2 To find: Angle of banking ()

Formula: vmax =tan

rg1 tan


tan = 2max

2max

v rg

rg v

=

2

2

250.2 30 9.8

3

2530 9.8 0.2

3

=

62558.8

9125

2949

= 69.44 58.8

294 13.89

= 10.64

307.89

= antilog[log(10.64) log(307.89)] = antilog[1.0270 2.4882]

= antilog 2.5388

= 0.03457 = tan–1 (0.03457) = 159 Ans: The angle of banking of the road should be

159. Example 6.10 The maximum speed of a car to take a turn of radius 35 m on a banked road is 50 km/h. If the angle of banking is 21.2, what should be the coefficient of friction between the car tyres and the road surface so that the car can negotiate the turn? Solution: Given: r = 35 m, vmax = 50 km/hr

= 50 1000

60 60

m/s

= 125

9 m/s,

= 21.2 To find: coefficient of friction()

Formula: vmax =

1/2tan

rg1 tan

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= 2max

2max

v rg tan

rg v tan

….(i)

=

2

2

12535 9.8 tan 21.2

9

12535 9.8 .tan 21.2

9

=

2

2

12535 9.8 0.3879

9

12535 9.8 0.3879

9

Now,

2

125

9

= antilog 2 log(125) log(9)

= antilog 2 2.0969 0.9542

= antilog 2 1.1427

= antilog 2.2854

= 193 3.5 9.8 0.3879 = antilog[log (35) + log(9.8)

+ log(0.3879)]

= antilog 1.5441 0.9912 1.5887

= antilog[2.1230] = 132 Also,

2

125

9

0.3879 = 193 0.3879

= antilog[log(193) + log(0.3879)]

= antilog[2.2856 1.5887]

= antilog[1.8743] = 74.87 Substituting the values, we get,

= 193 132

343 74.87

….[from (i)]

= 61

417.87

= antilog[log(61) log(417.9)] = antilog[1.7853 2.6210]

= antilog 1.1643

= 0.146 Ans: The coefficient of friction between the car

tyres and the road surface should be 0.146.

Example 6.11 A motorcyclist rounds a curve of radius 25 m at the speed of 36 km/hr. The combined mass of motorcycle and motorcyclist is 150 kg. (g = 9.8 m/s2) a. What is centripetal force exerted on the

motorcyclist? b. What is upward force exerted on the

motorcyclist? c. What angle does the motorcycle make with

the vertical? [Feb 13 old course] Solution:

Given: r = 25 m, v = 36 5

18 = 10 m/s,

m = 150 kg To find: i. Centripetal force (F) ii. Upward force (N) iii. Angle with the vertical ()

Formulae: i. F = 2mv

r

ii. N cos = mg

iii. tan = 2v

rg


F = 2150 (10)

25

F = 600 N From formula (ii), N = 9.8 150 N = 1470 N From formula (iii),

= tan12v

rg

= tan1 210

25 9.8

= tan1 4

9.8

= tan1{antilog[log(4) log(9.8)]} = tan1 {antilog[0.6021 0.9912]}

= tan1 anti log 1.6109

= tan–1 [0.4083] = 22.2 = 2212 Ans: i. The centripetal force exerted on the

motocyclist is 600 N. ii. The upward force exerted on the

motocyclist is 1470 N. iii. The angle made by the motor cycle with

the vertical is 2212.

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Example 6.12 A vehicle enters a circular bend of radius 200 m at 72 km/h. The road surface at the bend is banked at 10. Is it safe? At what angle should the road surface be ideally banked for safe driving at this speed? If the road is 5 m wide, what would be the elevation of the outer edge of the road surface above the inner edge? Solution: Given: r = 200 m,

v = 72 km/h = 72 5

18 m/s

= 20 m/s. = 10, l = 5 m. To find: i. To decide if it is safe for the

vehicle to enter the circular bend. ii. The angle of banking for safe

driving = () iii. Elevation of the outer edge (h)

Formulae: i. v = rg tan

ii. h = l sin Calculation: From formula (i),

v = 1

2200 9.8 tan10

= 1

2100 19.6 0.1763

= antilog 1[log(100) log(19.6) log(0.1763)]

2

= antilog1

[2.0000 1.2923 1.2462]2

= antilog1

[2.5385]2

= antilog 1.2693

v 18.6 m/s As the given speed of the vehicle exceeds the

safety limit i.e. 18.6 m/s, it is unsafe for it to enter the circular bend.

Now, on squaring and rearranging formula (i) we get,

tan = 2v

rg

= 2(20)

200 9.8

= 1

4.9= 0.2041

….[Using table of reciprocals] = tan1 (0.2041) 11.53 11 + 60 0.53 1132

From formula (ii), h = 5 sin (1132) = 5 0.2 h = 1.0 m Ans: i. It is unsafe for the vehicle to enter the

circular bend. ii. For safe driving, the road surface should

be ideally banked at 1132. iii. The elevation of the outer edge of the

road surface above the inner edge would be 1.0 m.

Example 6.13 Find the angle of banking of the railway track of radius of curvature 1600 metre if the optimum velocity of the train is 20 m/s. Also find the elevation of the outer track above the inner track if the distance between the two tracks is 1.8 m.

[Mar 90] Solution:

Given: r = 1600 m, v = 20 m/s, = 1.8 m.

To find: i. Angle of banking () ii. Elevation of the outer track (h)

Formulae: i. tan = 2v

rgor = tan1

2v

rg

ii. sin = h

….[ is small, sin tan .]


= tan1 20 20

1600 9.8

= tan1 1

39.2

= tan1 (0.0255) ….[Using the table of reciprocals]

= 1 28 From formula (ii),

h

= 0.0255

h = 0.0255 1.8 h = 0.0459 m Ans: i. The angle of banking of the railway track

is 128. ii. The elevation of the outer track above the

inner track is 0.0459 m.

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Section 7: Conical Pendulum Example 7.1 A conical pendulum has a length of 100 cm and the angle made by the string with the vertical is 8. Find the period of circular motion of the bob. Solution: Given: l = 100 cm = 1m, = 8 To find: Period (T)

Formula: T = 2 cos

g

l


T = 2 3.14 1 cos8

9.8

= 6.28 0.9903

9.8

= antilog 1log(6.28) log(0.9903) log(9.8)

2

= antilog1

0.7980 [1.9957 0.9912]2

= antilog1

0.7980 [1.0045]2

= antilog1

0.7980 [2 1.0045]2

= antilog 0.7980 1.50225

= antilog 0.7980 1.5023

= antilog{0.3003} T = 1.996 sec. Ans: The period of circular motion of the bob is

1.996 s. Example 7.2 A stone of mass 2 kg is whirled in a horizontal circle attached at the end of 1.5 m long string. If the string makes an angle of 30o with vertical, compute its period. (g = 9.8 m/s2) [July 16] Solution: Given: m = 2 kg l = 1.5 m = 30 g = 9.8 m/s2 To find: Period (T)

Formula: T = 2cos

g

l


T = 2 3.14 1.5 cos30

9.8

= 6.28 1.5 0.8660

9.8

= antilog 1log 6.28 log1.5 log 0.8660 log 9.8

2

= 1antilog 0.7980 0.1761 1.9375 0.9912

2

= 1antilog 0.7980 0.1136 0.9912

2

= 1antilog 0.7980 0.8776

2

= antilog[0.3592] T = 2.2875 s Ans: Period of revolution is 2.287 s. Example 7.3 A conical pendulum has length 80 cm. Its bob of mass 200 g performs uniform circular motion in the horizontal plane so as to have radius of path 40 cm. Calculate i. the angle made by the string with the vertical. ii. the tension in the supporting thread. iii. the speed of bob. Solution: Given: m = 200g = 0.2 kg, l = 80 cm = 0.8 m, r = 40 cm = 0.4 m To find: i. Angle made by the string with the

vertical () ii. Tension in the thread (T) iii. Speed of the bob (V)

Formulae: i. tan =r

h

ii. T cos = mg

iii. v = rg tan

Calculation:

h = 2 2rl

= 0.64 0.16 = 0.48

= 0.4 1.732 0.693 m From formula (i),

tan = r

h=

0.4

0.693 = 0.5773

= 30.

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We have,

cos = h

l=

0.693

0.8 0.8663

From formula (ii), T 0.8663 = 0.2 9.8

T = 0.2 9.8

0.8663

= 2.262 N

From formula (iii),

v = 0.4 9.8 0.5773

= antilog1

[log(0.4) log(9.8) log(0.5773)]2

= antilog1

1.6021 0.9912 1.76142

= antilog 10.3547

2

= antilog{0.17735} = antilog{0.1774} v = 1.504 m/s Ans: i. The angle made by the string with the

vertical is 30. ii. The tension in the thread is 2.262 N. iii. The speed of revolving pendulum bob is 1.504 m/s.

Example 7.4 A bob of mass 100 g is suspended from a fixed support by a thread. The bob moves in a horizontal circle of radius 25 cm. The height of the fixed support from the centre of the circle is 0.55 m. What is the tension in the thread? Solution: Given : m = 100 g = 0.1 kg, r = 25 cm = 0.25 m, h = 0.55 m To find: Tension (T)

Formula: T = mg 2

r1

h


T = 0.1 9.8 2

0.251

0.55

= 0.98 2

51

11

= 0.98 121 25

121

= 0.98

11 146

= antilog1

[log(0.98) log(146) log(11)]2

= antilog1

1.9912 2.1644 1.04142

= antilog 1.9912 1.0822 1.0414

= antilog{1.0734 1.0414} = antilog{0.0320} T = 1.076 N. Ans: The tension in the thread is 1.076 N. Example 7.5 The string of a conical pendulum is of length 120 cm. If the radius of the circle in which the bob moves is 32 cm and the mass of the bob is 250 g, then what is the tension in the string? Solution: Given : l = 120 cm = 1.2 m, r = 32 cm = 0.32 m, m = 250 gm = 0.25 kg To find: Tension (T)

Formula: T = 1/22 2

mg

r

l

l


T = 1/22 2

0.25 9.8 1.2

1.2 0.32

T = 1

2

0.25 9.8 1.2

[1.3376]

= antiloglog(0.25) log(9.8) log(1.2)

1log(1.3376)

2

= antilog1

1.3979 0.9912 0.0792 0.12652

= antilog{0.4683 0.06325} = antilog(0.4683 0.0633} = antilog{0.4050} T = 2.541 N Ans: The tension in the string is 2.541 N. Example 7.6 A conical pendulum has a bob which moves in a circle of radius 15 cm. If the angle which the thread of the pendulum makes with the vertical is 6, what is the angular velocity of the bob? Solution: Given : r = 15 cm = 0.15 m, = 6 To find: Angular velocity ()

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Formula: = g tan

r


= 9.8 tan 6

0.15

=

9.8 0.1051

0.15

= antilog1

[log(9.8) log(0.1051) log(0.15)]2

= antilog1

[0.9912 1.0216 1.1761]2

= antilog 1

[0.0128 1.1761]2

= antilog1

0.83672

= antilog{0.41635} = antilog{0.4164} = 2.608 2.61 = 2.61 rad/s Ans: The angular velocity of the bob is 2.61 rad/s. Section 8: Vertical Circular Motion Example 8.1 A motor cyclist rides in a vertical hollow sphere of radius 7 m. Find the minimum angular speed required so that it does not lose contact with the sphere at the highest point. [g = 9.8 m/s2] Solution: Given: r = 7 m, g = 9.8 m/sec2 To find: Minimum angular speed ()

Formula: = g

r


= 9.8

7

= 1.4

= 1.18 rad/s Ans: The minimum angular speed required by the

motor cyclist is 1.18 rad/s. Example 8.2 A body weighing 0.4 kg is whirled in a vertical circle making 2 revolutions per second. If the radius of the circle is 1.2 m, find the tension in the string when the body is i. at the bottom of the circle ii. at the top of the circle.

Solution: Given: m = 0.4 kg, r = 1.2 m, n = 2 rps Angular speed, = 2n = 2 2 = 4 rad s1 To find: i. Tension in the string at the bottom

(Tbottom). ii. Tension in the string at the top

(Ttop). Formulae: i. Tbottom = m(42n2 r + g) ii. Ttop = m (42 n2 r g) Calculation: From formula (i), Tbottom = 0.4[4(3.14)2 (2)2 1.2 + 9.8] = 6.4 (3.14)2 1.2 + 3.92 = 75.71 + 3.92 Tbottom = 79.63 N From formula (ii), Ttop = 0.4[4(3.14)2 (2)2 1.2 9.8] = 75.71 – 3.92 Ttop = 71.79 N. Ans: i. The tension in the string at the bottom of

the circle is 79.63 N. ii. The tension in the string at the top of the

circle is 71.79 N. Example 8.3 A bucket containing water is tied to one end of a rope of length 2.5 m and rotated about the other end in a vertical circle in such a way that the water in it does not spill. What is the minimum velocity of the bucket at which this happens and how many rotations per minute is it making then? [g = 10 ms2] Solution: Given: r = 2.5 m To find: i. Minimum velocity of the bucket

(Vmim). ii. Number of revolutions per minute

(n).

Formulae: i. vmin = rg

ii. n = 2

Calculation: From Formula (i),

vmin at the top = rg = 2.5 10

vmin = 5 m/s Using,

= r

v =

5.2

5 = 2 rad s1

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From Formula (ii),

Number of rotations n = 2

= 2

2rps =

2

2 60 rpm =

60

r.p.m.

n 19 r.p.m. Ans: i. The minimum velocity of the bucket so

that the water in it does not spill is 5 m/s.

ii. The number of revolutions of the bucket per minute are 19.

Example 8.4 A flyover bridge is in the form of a circular arc of radius 30 m. Find the limiting speed with which a car can cross the bridge without losing contact with the road at the highest point. Assume the centre of gravity of the car to be 0.5 m above the road. Solution: Given: r = 30 m, h = 0.5 m, R = r + h = 30 + 0.5 = 30.5 m, g = 9.8 m/s2 To find: Limiting speed (V)

Formula: FC = 2mv

R= mg (Weight)

Calculation: From formula, v2 = Rg v = Rg

= 30.5 9.8

= 298.9 v = 17.29 m/s. Ans: The limiting speed of the car is 17.29 m/s. Example 8.5 A small body of mass 0.1 kg is tied at the end of a chord of length 1m and swings in a vertical circle with speed 2 m/s. When the chord makes an angle of 30 with the vertical, find the tension in the chord. Solution: Given: m = 0.1 kg, r = 1 m, v = 2 m/s, = 30 To find: Tension in the chord (T)

Formula: T = 2mv

r+ mg cos


T =20.1 (2)

1

+ 0.1 9.8 cos 30

= 0.1 4 + 0.98 3

2

= 0.4 + 0.98 0.8660 …. 3 1.732

= 0.4 + 0.849 T = 1.249 N Ans: The tension in the chord is 1.249 N. Example 8.6 A particle of mass 0.1 kg is tied at the end of a string and whirled in a vertical circle of radius 1 m at a constant speed of 5 m/s. Find the tension in the string at the highest point of its path. Solution: Given: m = 0.1 kg, r = 1 m, VH = 5 m/s To find: Tension at highest point (TH)

Formula: TH = 2Hmv

r mg


TH = m 2Hv

gr

= 0.1 2(5)

9.81

= 0.1 (25 9.8) = 0.1 (15.2) TH = 1.52 N Ans: The tension in the string at the highest point is

1.52 N. Example 8.7 A 4 kg ball tied at the end of a chord 1 m long, swings in a vertical circle. At what maximum speed can it swing if the chord can sustain a maximum tension of 163.6 N? Solution: Given: m = 4 kg, r = 1 m, TL = 163.6 N To find: Maximum speed (vL)

Formula: TL = 2Lmv

r + mg


2Lmv

r = TL mg

2L4 v

1

= 163.6 4 9.8

4 2Lv = 163.6 39.2

4 2Lv = 124.4

2Lv =

124.4

4

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2Lv = 31.1

vL = 31.1 vL = 5.58 m/s. Ans: The maximum speed of the ball is 5.58 m/s. Example 8.8 An aeroplane is flying in the sky with a speed of 360 km/hr in a vertical circle of radius 200 m. The weight of the pilot sitting in it is 75 kg. What is the force with which the pilot presses his seat when the aeroplane is i. at the highest position? ii. at the lowest position of the circle? Solution:

Given: v = 360 km/hr = 360 5

18m/s

= 100 m/s r = 200 m, m = 75 kg

To find: i. Force exerted by pilot at highest position (TH)

ii. Force exerted by pilot at lowest position (TL)

Formulae: i. TH = 2Hmv

r mg

ii. TL = 2Lmv

r + mg


TH = m 2Hv

gr

= 75 2100

9.8200

= 75 10000

9.8200

= 75 (50 9.8) = 75 40.2 TH = 3015 N From formula (ii),

TL = m 2Lv

gr

= 75 2100

9.8200

TL = 75 (50 + 9.8) = 75 59.8 TL = 4485 N Ans: i. The force exerted by the pilot on his seat

when the aeroplane is at the highest position is 3015 N.

ii. the force exerted by the pilot on his seat at the lowest position of the circle is 4485 N.

Example 8.9 A stone of mass 10 kg tied with a string of length 0.5 m is rotated in a vertical circle. Find the total energy of the stone at the highest position. Solution: Given: m = 10 kg, r = 0.5 m To find: Total energy (TE)

Formula: (T.E.)H = 5

2 mrg


(T.E.)H = 5

2 10 0.5 9.8

= 5 5 4.9 (T.E.)H = 122.5 J Ans: The total energy of the stone at the highest

position is 122.5 J. Example 8.10 A 500 g particle tied to one end of a string is whirled in a vertical circle of circumference 14 m. If the tension at the highest point of its path is 2 N, what is its speed? Solution: Given: m = 500 g = 0.5 kg, circumference = 14 m = 2r

r = 14

2m,

T = 2 N To find: Speed (v)

Formula: T = 2mv

r – mg


2mv

r = T + mg

v = 1/2

rT mg

m

= 1/2

14(2 0.5 9.8)

2 0.5

= [4.4586 6.9]1/2

v = 5.546 m/s Ans: The speed of the particle at the highest point is

5.546 m/s.

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Example 8.11 A stone of mass 100 g attached to a string of length 50 cm is whirled in a vertical circle by giving velocity at lowest point as 7 m/s. Find the velocity at the highest point. [Acceleration due to gravity = 98 m/s2] [Oct 15] Solution: Given: m = 100 g= 0.1 kg, r = 50 cm = 0.5m, g = 9.8 m/s2, vL = 7 m/s To find: Velocity at the highest point (vH)

Formula: vH = (H)2 T.E.

4grm

Calculation: Total energy at highest point,

T.E.(H)= K.E. at lowest point = 2L

1mv

2

T.E.(H) = 210.1 7

2

= 2.45 J From formula,

vH = 2 2.454 9.8 0.5

0.1

= 29.4

vH = 5.422 m/s Ans: The velocity at the highest point is 5.422 m/s. Example 8.12 A stone of mass 5 kg, tied to one end of a rope of length 0.8 m, is whirled in a vertical circle. Find the minimum velocity at the highest point and at the midway point. [g = 98 m/s2] [Oct 14] Solution: Given: m = 5 kg, r = 0.8 m, g = 9.8 m/s2 To find: i. Minimum velocity at the highest

point (vH) ii. Minimum velocity at midway

point (vM)

Formulae: i. vH = rg

ii. vM = 3rg


vH = 0.8 9.8 = 7.84 = 2.8 m/s From formula (ii),

vM = 3 0.8 9.8

= 3 2.8 = 4.85 m/s Ans: The minimum velocity at the highest point and

midway point is 2.8 m/s and 4.85 m/s respectively.

Example 8.13 A bridge over a railway track is in the form of a circular arc of radius 55 m. What is the limiting speed with which a car can cross the bridge so that no contact is lost if the centre of gravity of the car is 0.4 m above the road? Solution: Given : r = 55m, h = 0.4 m Total distance R = 55 + 0.4 = 55.4 m To find: Limiting speed (v)

Formula: F = 2mv

R = mg


v = Rg = 55.4 9.8

v = 23.3 m/s Ans: The limiting speed of the car is 23.3 m/s. Section 9: Kinematical Equations Example 9.1 On the application of a constant torque, a wheel is turned from rest through 400 radian in 10 s. Calculate its angular acceleration. Solution: Given: = 400 rad, 1 = 0, t = 10 s To find: Angular acceleration ()

Formula: = 1t + 2

1 t2


400 = 0 + 2

1 (10)2

= 2 400

100

= 8 rad s2. Ans: The angular acceleration of the wheel is

8 rad/s2. Example 9.2 The initial angular speed of a wheel is 4 rad/s. If its angular displacement is 200 rad, find its angular acceleration after 10 s. Solution: Given: 1 = 4 rad/s, = 200 rad, t = 10 s To find: Angular acceleration ()

Formula: = 1t + 1

2 t2


200 = 4 10 + 1

2 10 10

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200 = 40 + 50

= 160

50

= 3.2 rad/s2 Ans: The angular acceleration of the wheel is

3.2 rad/s2. Example 9.3 A particle moves along a circular path of length 15 cm with a constant angular acceleration of 4 rad/s2. If the initial angular speed of the particle is 5 rad/s, find the angular displacement of the particle in 5 sec. Solution: Given: Circumference = 15 cm = 0.15 m, = 4 rad/s2, 1 = 5 rad/s, t = 5s To find: Angular displacement ()

Formula: = 1t + 1

2 t2


= 5 5 + 1

2 4 (5)2

= 25 + 50 = 75 rad Ans: The angular displacement of the particle is

75 rad. Example 9.4 A gramophone turntable rotating at an angular velocity of 3 rad/s stops after one revolution. Find the angular retardation. Solution: Given: 1 = 3 rad/sec, = 1 rev. = 2 rad, 2 = 0 To find: Angular retardation () Formula: 2

2 = 21 + 2

Calculation: From formula, 0 = 32 + 2 2 0 = 9 + 4 4 = 9

= 9

4

= 9

4 3.142

= 9

12.568

= 0.716 rad/s2 Ans: The angular retardation of the turn table is

0.716 rad/s2.

Example 9.5 The spin dryer of a washing machine rotating at 15 r.p.s. slows down to 5 r.p.s. after making 50 revolutions. Find its angular acceleration.

[Mar 15] Solution: Given: n0 = 15 r.p.s., n = 5 r.p.s., No. of revolutions = 50 To find: Angular acceleration () Formulae: i. = 2n

ii. 2 = 20 2

Calculation: Using formula (i), = 2 5 = 10 0 = 2 5 = 30 In 1 revolution, angular displacement = 2 in 50 revolutions, angular

displacement = = 100 Using formula (ii), (10)2 = (30 )2 + 2(100 )

= 2 2900 100

200

= 4 rad/s2

= 12.56 rad/s2 Ans: Angular acceleration of spin dryer is

12.56 rad/s2. Example 9.6 A fly wheel gains a speed of 240 rpm in 3 s. Calculate the change in its angular speed in three seconds. Solution:

Given: n1 = 0 rpm = 0

60rps = 0 rps,

n2 = 240 rpm = 240

60= 4 rps,

t = 3s To find: Change in angular speed () Formula: w = 2 – 1= 2 (n2 – n1) Calculation: By using formula, = 2 (4) 2 (0)

= 8 rad/s = 8 3.14 = 25.12 rad/s Ans: The change in the angular speed of the

flywheel is 25.12 rad/s.

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Example 9.7 A wheel rotating at 500 r.p.m. slows down to 400 r.p.m. at a constant rate of 5 rad/s2. What is the angular displacement of the wheel? Solution:

Given: n1 = 500 r.p.m = 500

60 r.p.s.,

n2 = 400 r.p.m = 400

60r.p.s.,

= 5 rad/s2 To find: Angular displacement ()

Formula: 22 – 2

1 = 2

Calculation: 1 = 2n1

= 2 500

60 rad/s

1 = 52.33 rad/s 2 = 2n2

= 2 400

60 rad/s

2 = 41.87 rad/s From formula,

= 2 22 1

2

= 2 2(52.33) (41.87)

2 5

….(1)

Now, (52.33)2 (41.87)2 = antilog [2 log(52.33)] antilog[2 log(41.87)] = antilog[2 1.7187] antilog [2 1.6219] = antilog[3.4374] antilog[3.2438] = 2738 – 1753 = 985 Substituting the value in eq (1), we get

= 985

10

= 98.5 rad Ans: The angular displacement of the wheel is

98.5 rad.

Problems for Practice Section 1: Angular Displacement, Relation


1. Calculate the angular speed and linear speed

of the tip of a second hand of a clock, if the second hand is 4 cm long.

2. A flywheel turns at 600 rpm. Compute the

angular speed at any point on the wheel and the tangential speed 0.5 m from the centre.

3. A particle moves along a circular path of

radius 20 cm making 240 revolutions per minute. Find the angular and linear velocities of the particle.

4. What is the angular displacement of the

minute hand of a clock in 20 minutes? 5. Find the angular displacement of the tip of

second hand of clock whose length is 5 cm and sweeps an arc of length 7.25 cm.

6. If a body moves on a circular path of radius

10 m with a linear velocity of 2 m/s, find its angular displacement in 15 s.

7. Calculate the angular velocity of earth due to

it’s spin motion. 8. An aircraft takes a turn along a circular path of

radius 1500 m. If the linear speed of the aircraft is 300 m/s, find its angular speed and time taken by it to complete (1/5)th of the circular path.

9. A body is fixed to one end of a rope and the

other end of the rope is fixed to a peg on the ground. The body rotates with a uniform angular velocity of 5 rad/s around the peg. If the radius of the circle in which the body rotates is 50 cm, what is its linear velocity?

Section 2: Angular Acceleration 10. Determine the angular acceleration of a

rotating body which slows down from 500 r.p.m. to rest in 10 second.

11. The speed of a wheel increases from 600 rpm

to 1200 rpm in 20 s. What is its angular acceleration and how many revolutions does it make during this time?

12. The motor of an engine is rotating about its

axis with an angular velocity of 100 rpm. It comes to rest in 15 s after being switched-off. Assuming constant angular retardation, calculate the number of revolutions made by it before coming to rest.

13. A car is moving at a speed of 72 kmh1. The

diameter of its wheels is 0.50 m. If the wheels are stopped in 20 rotations by applying brakes, calculate the angular retardation produced by the brakes.

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Section 3: Centripetal and Tangential

Acceleration 14. If a particle has a radial acceleration of

123.62 m/s2 and a tangential acceleration of 91.41 m/s2, then what is its linear acceleration?

15. A particle moves in a circle of radius 5 cm and

has its velocity of rotation increased by 100 rotations in 5 seconds. Calculate its angular acceleration and tangential acceleration.

16. A car is moving along a circular road at a

speed of 20 m/s. The radius of the circular road is 10 m. If the speed is increased at the rate of 30 m/s2, what is the resultant acceleration?

17. A 0.5 kg mass is rotated in a horizontal circle

of radius 20 cm. Calculate the centripetal acceleration acting on it, if its angular speed of revolution is 0.8 rad/s.

18. To simulate acceleration of large rockets,

astronauts are spun at the end of a long rotating beam of length 9.8 m. What angular velocity is required to generate a centripetal acceleration 8 times the acceleration due to gravity?

(g = 9.8 m/s2) 19. A motor car is travelling at 20 m/s on a

circular curve of radius 100 m. It is increasing its speed at the rate of 5 m/s2. What is its acceleration?

Section 4: Centripetal and Centrifugal Forces 20. A body of mass 10 kg moves in a circle of

radius 1 m with constant angular speed of 2 rad/sec. Find the period of revolution and the centripetal force.

21. The breaking tension of a string of length 2

metre is 24 kg wt. A body of mass 2 kg is attached to its one end and whirled in a horizontal circle with the end fixed. Find the maximum frequency of revolution possible. Also find the velocity of the body when the string breaks.

22. A 1 kg mass tied at the end of a string 0.5 m

long is whirled in a horizontal circle, the other end of the string being fixed. The breaking tension in the string is 50 N. Find the greatest speed that can be given to the mass.

23. Find the centripetal force required to revolve a body of mass 0.2 kg along a circular path of radius 2 metre at a uniform rate of 300 revolutions per minute in the horizontal plane.

24. A coin kept on a horizontal rotating disc has

its centre at a distance of 0.25 m from the axis of rotation of the disc. If = 0.2, find the angular velocity of the disc at which the coin is about to slip-off. [g = 9.8 m/s2.]

25. A body of mass 1 kg is tied to a string and

revolved in a horizontal circle of radius 1 m. Calculate the maximum number of revolutions per minute so that the string does not break. Breaking tension of the string is 9.86 N.

26. A 0.5 kg mass tied to the end of a string is

whirled in a horizontal circle of radius 1.5 m with a angular speed of 4 rad/s. The maximum tension that the string can withstand is 250 N. What is the maximum speed with which the stone can be whirled? What is the tension in the string?

27. A coin placed on a revolving disc at a distance

of 25 cm from the axis does not slip-off when the disc revolves at 80 r.p.m. The coin is then placed 40 cm from the axis. How fast can the disc revolve without the coin slipping-off?

Section 5: Motion of a Vehicle along a Curved

Unbanked Road 28. With what maximum speed can a car be safely

driven along a curve of radius 40 m on a horizontal road if the coefficient of friction between the car tyres and road surface is 0.3? [g = 9.8 m/s2]

29. A car travelling at 18 km/hr just rounds a

curve without skidding. If the road is plane and the coefficient of friction between the road surface and the tyres is 0.25, find the radius of the curve.

30. A car moves on a level turn having a radius of

62 m. What is the maximum speed the car can take without skidding if the coefficient of static friction between the tyre and the road is 2.2?

31. What should be the coefficient of friction

between the tyres and the road, when a car travelling at 60 km h1 makes a level turn of radius 40 m?

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Section 6: Banking of Roads 32. A cyclist speeding at 6 ms1 in a circle of 18 m

radius makes an angle with the vertical. Calculate . Also determine the minimum possible value of the coefficient of friction between the tyres and the ground.

33. A bicycle and rider together have a mass of 90 kg. Find the angle which the rider must make with the horizontal in travelling around a curve of 36 m radius at 44/3 m/s.

34. A curved road of radius 90 m is to be banked so that a vehicle may move along the curved road with a uniform speed of 75.6 km/hr without any tendency to slip-off. What must be the angle of banking? [g = 9.8 m/s2]

35. The circumference of a track is 1.256 km. Find the angle of banking of the track if the maximum speed at which a car can be driven safely along it is 25 m/s.

36. A motorcyclist goes round a circular race course of diameter 320 m at 144 km h1. How far from the vertical must he lean inwards to keep his balance? [Take g = 10 ms2]

37. A train has to negotiate a curve of radius 400 m. By how much should the outer rail be raised with respect to the inner rail for a speed of 48 kmh1? The distance between the rails is 1 m.

38. The C.G. of a taxi is 1.5 m above the round and the distance between its wheels is 2 m. What is the maximum speed with which it can go round an unbanked curve of radius 100 m without being turned upside down? What minimum value of coefficient of friction would be required at this speed?

39. A turn on a road has a radius of 55 m and is banked at an angle of 15 with the horizontal. If the coefficient of friction between a car tyre and the road surface is 0.28, what is the maximum speed of the car on the turn?

40. An aircraft with its wings banked at 22 performs a horizontal loop at a speed of 250 m/s. What is the radius of the loop?

41. A railway track has a radius of curvature of 1.8 km. If the optimum velocity of a train on the track is 25 m/s, what is the angle of banking? If the elevation of the outer track above the inner track is 0.05 m, what is the distance between the two tracks?

42. What is the maximum speed at which a car can turn on a road which is banked at an angle of 10 if the radius of the turn is 25 m and the coefficient of friction between the tyres and the road surface is 0.4?

43. The maximum speed of a bus on turn of radius

50 m is 60 km/hr. If the coefficient of friction between the tyres and the road is 0.5, what is the angle of banking?

Section 7: Conical Pendulum 44. A conical pendulum has a bob of mass 200 g

and a length of 50 cm. If the radius of the circle traced by the bob is 25 cm, find the velocity of the bob and the period of pendulum. [g = 9.8 m/s2]

45. A conical pendulum has a length of 1.5 m and

a bob of mass 50 g. The bob completes 20 revolutions in 45 s. Find the radius of the circular path traced by the bob and the tension in the thread. [g = 9.8 m/s2]

46. The bob of a conical pendulum has mass 50 g

and it moves in a horizontal circle whose radius is 24 cm. If the length of the string is 75 cm, what is the tension in the string?

47. A conical pendulum has a bob of mass 300 g

and string of length 115 cm. If the angle made by the string with the vertical is 12, what is the period of circular motion of the bob? What is the tension in the string?

Section 8: Vertical Circular Motion 48. A 1.2 kg body attached to a string is whirled

in a vertical circle of radius 3 m. What minimum speed, vm must it have at the top of the circle so as not to depart from the circular path? Find its velocity at lowest point and at a midway position?

49. A vehicle weighing 4000 kg is going over a

convex bridge, the radius of curvature of which is 30 m. The height of the centre of gravity of the vehicle from the ground is 1.2 m. If the velocity of the vehicle is 50.4 km/h, calculate the thrust of the vehicle on the road at the highest point. Also find the greatest speed at which the vehicle can cross the bridge without losing contact with the road at the highest point. [g = 9.8 m/s2]

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50. A roadway bridge over a canal is in the form of an arc of radius 15m. What is the maximum speed with which a car can cross the bridge without leaving the ground at the highest point? [g = 9.8 m/s2]

51. The vertical section of a road over a bridge in

the direction of its length is in the form of an arc of a circle of radius 19.5 m. Find the limiting velocity at which a car can cross the bridge without losing contact with the road at the highest point, if the centre of gravity of the car is 0.5 m from the ground.

52. A motorcyclist rides in a vertical circle in a

hollow sphere of radius 5m. Find the minimum angular speed required so that he does not lose contact with the sphere at the highest point.

[g = 9.8 m/s2] 53. A stone of mass 0.3 kg is tied to one end of a

string 0.8 m long and rotated in a vertical circle. At what speed of the ball will the tension in the string be zero at the highest point of the circle? What would be the tension at the lowest point in this case?

[Given g = 9.8 ms2] 54. A bucket containing water is tied to one end of

a rope 8 m long and rotated about the other end in vertical circle. Find the minimum number of rotations per minute in order that water in the bucket may not spill.

[g = 9.8 m/s2] 55. What is the minimum speed required at the

bottom to perform a vertical loop, if the radius of the death-well in a circus is 25 m?

56. A particle of 0.1 kg is whirled at the end of

string in a vertical circle of radius 1m at constant speed of 7 m/s. Find the tension in the string at the highest position.

57. A stone weighing 0.5 kg tied to a rope of

length 0.5 m revolves along a circular path in a vertical circle. What maximum speed does it possess at the bottom where tension is 45 N?

58. A body weighing 0.4 kg tied to a string is

projected with a velocity of 15 ms1. The body starts whirling in a vertical circle. If the radius of the circle is 1.2 m, find tension in the string when the body is

i. at the top of the circle and ii. at the bottom of the circle.

59. A stone of mass 8 kg tied with a string of length 1m is rotated in vertical circle. Find the total energy of stone at the lowest point.

60. A body of mass 4 kg is rotating in a vertical circle at the end of a string of length 0.6 m. Calulate the difference in K.E. at the top and bottom of the circle.

61. A 0.6 kg bob attached to a string is whirled in a vertical circle at a speed of 15 rad/s. If the radius of the circle is 1.8 m., what is the tension in the string when the bob is at the bottom and at the top of the circle?

Section 9: Kinematical Equations 62. The spin drier of a washing machine revolving

at 15 r.p.s slows down to 5 r.p.s. while making 50 revolutions. Find the angular acceleration and time taken to complete the revolutions.

63. A particle moves along a circular path of radius 10 cm with a constant angular acceleration of 2 rad/s2. If the initial angular speed of the particle is 5 rad/s,

find the i. angular speed of the particle after 10 s. ii. angular displacement of the particle in

10 s. iii. tangential acceleration of the particle. 64. A particle moves in a circular path with an

angular velocity of 9 rad/s. The particle then accelerates at a constant rate of 3 rad/s2. In what time will the particle be displaced by an angle of 60 rad ?

65. A curved road having diameter 0.04 km is banked at an angle . If a car can travel at a maximum speed of 50 km/hr on the curved road and if the coefficient of friction between the tyres and the road is 0.26, what is the angle of banking?

Multiple Choice Questions

Section 1: Angular Displacement, Relation Between Linear Velocity and Angular Velocity

1. A body rotating in a horizontal circle has a

linear velocity of 25 m/s and its angular velocity is 0.5 rad/s. What is the radius of the circle in which it rotates?

(A) 0.05 m (B) 0.5 m (C) 0.25 m (D) 2.5 m

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2. The angular velocity of a point on the rim of a wheel is 8 rad/s. How many revolution per minute (r.p.m) does the wheel make ?

(A) 250 (B) 240 (C) 100 (D) 4 3. The angular velocity of the minute hand of a

clock is

(A) 60

rad/s (B)

3600

rad/s

(C) 2

60

rad/s (D)

2

3600

rad/s

4. The angular displacement of the second’s

hand of a clock in 30 s is (A) 6.28 rad (B) 3.14 rad (C) 1.07 rad (D) 60 rad 5. If the length of the minute hand of a clock is

15 cm, then its linear velocity is (A) 1.2 10–4 m/s (B) 2.62 10–4 m/s (C) 3.2 10–4 m/s (D) 3.6 10–2 m/s Section 2: Angular Acceleration 6. If the angular speed of a particle changes from

2.5 rad/s to 5 rad/s in 20 s, then its angular acceleration is

(A) 0.125 rad/s2 (B) 0.15 rad/s2 (C) 1.5 rad/s2 (D) 12.5 rad/s2 7. A body moves from rest through 100 rad in

15 s. Its angular acceleration is (A) 1.15 rad/s2 (B) 0.89 rad/s2 (C) 0.75 rad/s2 (D) 0.42 rad/s2 8. A particle moving with an angular speed of

4.2 rad/s accelerates to move with an angular speed of 6.4 rad/s. If its angular displacement is 50 rad, then its angular acceleration is

(A) 1.8 rad/s2 (B) 1.2 rad/s2 (C) 0.32 rad/s2 (D) 0.23 rad/s2 9. A body performing UCM changes from

50 r.p.m. to 100 r.p.m. in 20 s. Its angular acceleration is

(A) 0.52 rad/s2 (B) 0.26 rad/s2 (C) 0.12 rad/s2 (D) 0.02 rad/s2 Section 3: Centripetal and Tangential Acceleration 10. What is the centripetal acceleration of a

0.25 kg body which rotates in a horizontal circle of radius 12 cm with an angular velocity of 1.1 rad/s?

(A) 1.2 m/s2 (B) 0.8 m/s2 (C) 5.14 m/s2 (D) 0.15 m/s2

11. A particle moves in a horizontal circle of radius 2 cm with a constant angular speed of 14 r.p.s. What is its centripetal acceleration?

(A) 165.23 m/s2 (B) 154.59 m/s2 (C) 138.26 m/s2 (D) 101.19 m/s2 12. If the radial acceleration of a body is 29 m/s2

and its tangential acceleration is 12 m/s2, what is its linear acceleration?

(A) 31.38 m/s2

(B) 29.62 m/s2 (C) 24.19 m/s2

(D) 19.26 m/s2 13. A body revolves in a circle of radius 14 cm

with an angular velocity of 30 rad/s. If its tangential acceleration is 0.24 m/s2, then what is the ratio of its centripetal acceleration to its tangential acceleration?

(A) 600 : 1 (B) 550 : 1 (C) 525 : 1 (D) 1 : 400 14. A particle rotates in a horizontal circle with an

angular velocity of 20 rad/s. If its tangential acceleration is 0.2 m/s2 and the ratio of its centripetal acceleration to its tangential acceleration is 600 : 1, what is the radius of the circle in which the particle rotates?

(A) 20 cm (B) 30 cm (C) 15 cm (D) 10 cm Section 4 : Centripetal and Centrifugal Forces 15. A coin just slips when placed at a distance of

2 cm from the axis on a rotating turntable. At what distance from the axis will it just slip when the angular velocity of the turntable is halved?

(A) 1 cm (B) 2 cm (C) 4 cm (D) 8 cm 16. A 250 g mass tied to the end of a string is

whirled in a circle of radius 115 cm with a speed of 3 r.p.s. What is the tension in the string?

(A) 150.6 N (B) 112.50 N (C) 102.05 N (D) 89.65 N 17. A stone of mass 0.2 kg is tied to the end of a

string and whirled in a horizontal circle. The maximum tension in the string is 300 N. If the maximum velocity is 4.2 m/s, what is the radius of the circle in which it is whirled?

(A) 1.18 cm (B) 1.82 cm (C) 2.11 cm (D) 4.36 cm

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18. A 0.2 kg bob is tied to the end of a string and whirled in a horizontal circle of radius 60 cm. The string breaks when the tension is 200 N. What is the time period corresponding to the maximum velocity with which it can be whirled?

(A) 0.189 s (B) 0.154 s (C) 0.116 s (D) 0.009 s 19. A coin placed on a rotating turntable at a

distance of 12 cm from the axis does not slip when the turntable rotates steadily at a speed of r.p.m. If the coefficient of friction between the coin and the turntable is 0.4, what is the value of ?

(A) 54.6 r.p.m. (B) 60.3 r.p.m. (C) 72.9 r.p.m. (D) 90.1 r.p.m. 20. A 0.3 kg stone is tied to the end of a string and

whirled in a horizontal circle of radius 0.6 m. The string just breaks when the velocity of rotation is 10 m/s. What is the maximum tension which the string can withstand?

(A) 25 N (B) 40 N (C) 50 N (D) 100 N 21. A 0.5 kg mass is tied to the end of a string and

whirled in a circle of radius 75 cm. If the tension in the string is 15 kg wt, what is the linear velocity of rotation?

(A) 16.23 m/s (B) 14.85 m/s (C) 12.91 m/s (D) 8.23 m/s Section 5: Motion of a Vehicle along a Curved

Unbanked Road 22. A vehicle takes a turn on an unbanked road,

the radius of the turn being 42 m. If the coefficient of friction between the tyres and the road surface is 0.6, then the car travelling at which of the following velocities will slip while taking the turn?

(A) 8 m/s (B) 12 m/s (C) 15 m/s (D) 18 m/s 23. A car travels at a speed of 60 km/h on a level

road on which there is an unbanked turn of radius 50 m. For which of the following values of coefficient of friction between the tyres and the road surface will the car not skid while turning?

(A) 0.1 (B) 0.2 (C) 0.4 (D) 0.6

24. A road is being designed for vehicles to travel at 80 km/h. The coefficient of friction between the road surface and the tyres is 0.2. What should be the radius of an unbanked level turn on the road for vehicles to move without skidding?

(A) 251.90 m (B) 500 m (C) 128.2 m (D) 202.95 m Section 6: Banking of Road 25. The turn on a banked road has a radius of

25 m. If the coefficient of friction between a car tyre and the road surface is 0.32 and if the angle of banking is 17, what is the maximum velocity at which a car can turn?

(A) 12.61 m/s (B) 11.47 m/s (C) 13.04 m/s (D) 10.93 m/s 26. The centre of gravity of a car is 0.62 m above

the ground. It can turn along a track which is 1.24 m wide and has radius r. If the greatest speed at which the car can take the turn is 22.02 m/s, what is the value of r?

(A) 62.32 m (B) 52.01 m (C) 49.48 m (D) 38.62 m 27. A motorcyclist executes a horizontal loop at a

speed of 65 km/h while himself making an angle of 12 with the horizontal. What is the radius of the loop?

(A) 156.6 m (B) 161.2 m (C) 173.4 m (D) 180.9 m 28. A car executes a turn of radius 22 m on a

banked road while travelling at a speed of 45 km/h. If the height of the outer edge above the inner edge of the road is 1.1 m, what is the breadth of the road?

(A) 2.104 m (B) 1.875 m (C) 1.626 m (D) 1.213 m 29. The angle of banking of a turn of radius 75 m

on a road is 30. What is the speed at which a car can turn along this curve?

(A) 20.6 m/s (B) 22.3 m/s (C) 24.6 m/s (D) 28.3 m/s 30. If the angle of banking of a road is 32 and if a

turn has radius 72 m, what is the maximum speed at which a car can turn, given the coefficient of friction between the car tyres and the road is 0.34?

(A) 32.3 m/s (B) 29.4 m/s (C) 28.6 m/s (D) 25.1 m/s

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Section 7: Conical Pendulum 31. The mass of the bob of a conical pendulum is

100 g and the length of the string is 150 cm. If the radius of the circle in which the bob rotates is 22 cm and if the thread makes an angle of 15 with the vertical, calculate the velocity of the bob.

(A) 0.76 m/s (B) 0.84 m/s (C) 1.2 m/s (D) 2.4 m/s 32. The length of the string of a conical pendulum

is 90 cm and its bob moves in a circular path of radius 25 cm. What is the tension in the string if the bob has mass 150 g?

(A) 2.81 N (B) 1.53 N (C) 1.25 N (D) 0.92 N 33. A conical pendulum of length 120 cm moves

making an angle of 16 with the vertical. What is the period of circular motion of the bob?

(A) 2.155 s (B) 2.523 s (C) 3 s (D) 4.009 s 34. A conical pendulum has a bob of mass 200 g

and it moves in horizontal circle making an angle of 8 with the vertical. What is the tension in the string?

(A) 2.113 N (B) 1.979 N (C) 1.504 N (D) 1.216 N Section 8: Vertical Circular Motion 35. A gymnast hangs from one end of a rope and

executes vertical circular motion. The gymnast has a mass of 40 kg and the radius of the circle is 2.5 m. If the gymnast whirls himself at a constant speed of 6 m/s, what is the tension in the rope at the lowest point?

(A) 1200 N (B) 968 N (C) 782 N (D) 500 N 36. A 100 g mass attached to the end of a string is

rotated in a vertical circle of radius 40 cm. What is the total energy of the mass at the highest point?

(A) 1.5 J (B) 1.1 J (C) 0.98 J (D) 0.76 J 37. A body of negligible mass tied to the end of a

string is rotated in a vertical circle of radius 80 cm. At a horizontal point on the circle, what is the speed of the body?

(A) 4.85 m/s (B) 5.25 m/s (C) 6.58 m/s (D) 8.01 m/s

38. A fighter aircraft flying at 400 km/h executes a vertical circular loop of radius 100 m. The pilot has a weight of 60 kg. What is the force with which the pilot presses his seat when the aircraft is at the highest point?

(A) 8500.9 N (B) 7200.34 N (C) 6900.2 N (D) 6819.40 N 39. A stone of mass m tied to a string of length

60 cm is whirled in a vertical circle. If the total energy of the stone at the highest position is 250 J, what is the value of m?

(A) 20 kg (B) 18 kg (C) 17 kg (D) 15 kg 40. A bucket containing water is tied to one end of

a rope of length 1.8 m. It is rotated about the other end so that water does not spill out. What is the minimum velocity of the bucket at which this can happen?

(A) 4.2 m/s (B) 5 m/s (C) 6.2 m/s (D) 7.4 m/s 41. A motorcyclist riding in a vertical sphere does

not lose contact with the sphere at the highest point. If the minimum angular velocity at which the motorcyclist achieves this is 1.5 rad/s, what is the radius of the sphere?

(A) 3.21 m (B) 4.36 m (C) 4.63 m (D) 5.56 m 42. A 200 g mass is whirled in a vertical circle

making 60 revolutions per minute. What is the tension in the string at the top of the circle if the radius of the circle is 0.8 m?

(A) 4.35 N (B) 4.51 N (C) 5.4 N (D) 6.22 N 43. A road bridge is in the form of a circular arc of

radius 18 m. What is the limiting speed with which a car can traverse the bridge without losing contact at the highest point if the centre of gravity of the car is 0.4 m above the ground?

(A) 13.428 m/s (B) 14.314 m/s (C) 15.206 m/s (D) 16.009 m/s Section 9: Kinematical Equations 44. A disc starts from rest and then accelerates at

a constant rate of 12 rad/s2. If the angular displacement of the disc is 30 rad, then what is the final velocity with which the disc rotates?

(A) 32.71 rad/s (B) 26.83 rad/s (C) 24.01 rad/s (D) 18.23 rad/s

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45. A flywheel executing 600 r.p.m. stops after one complete rotation. What is its angular retardation?

(A) – 3.14 rad/s2 (B) – 31.4 rad/s2 (C) – 314 rad/s2 (D) – 3140 rad/s2 46. A wheel rotating at 10 rad/s accelerates to

12 rad/s. If its angular displacement is 40 rad, what is its angular acceleration?

(A) 1.5 rad/s2 (B) 0.55 rad/s2 (C) 0.44 rad/s2 (D) 0.15 rad/s2 47. An engine required 4 s to go from a speed of

600 rpm to 1200 rpm with a constant acceleration. The number of revolutions made by it in this time is

(A) 6.0 (B) 600 (C) 60 (D) 6000

Answers to Problems for Practice 1. 1.047 101 rad/s, 4.188 103 m/s 2. 62.84 rad/s, 31.42 m/s 3. 25.12 rad/s, 5.024 m/s 4. 2.093 rad 5. 1.45 rad 6. 3.0 rad 7. 7.27 105 rad/s 8. 0.2 rad/s, 6.28 s 9. 2.5 m/s 10. 5.23 rad/s2 11. 3.14 rad s2, 300 12. 12.5 13. 25.5 rad s2 14. 153.74 m/s2 15. 40 rad/s2 ; m/s2 16. 50 m/s2

17. 0.128 m/s2 18. 2.828 rad/s 19. 6.403 m/s2 20. 3.14 s, 40 N 21. 1.22 rev/s; 15.33 m/s 22. 5 m/s 23. 394.38 N 24. 2.8 rad/s 25. 30 26. vmax = 27.39 m/s, T = 12 N 27. 63.25 r.p.m. 28. 10.84 m/s 29. 10.2 m 30. vmax = 36.56 m/s 31. 0.71 32. 1132, 0.2041 33. 5838

34. 26 34 35. 17 41 36. 45 37. 0.0454 m 38. 25.56 ms1, 0.67 39. vmax = 17.12 m/s 40. r = 15,785 m 41. = 22, l = 1.412 m 42. vmax = 12.325 m/s 43. = 2 58 44. 119 cm/s; 1.32 s 45. 0.78 m; 0.57 N 46. T = 0.517 N 47. Period = 2.13 s Tension = 3 N 48. i. 5.422 m/s ii 12.124 m/s iii. 9.391 m/s 49. 14072 N; 17.49 m/s 50. 12.124 m/s 51. 14 m/s. 52. 1.4 rad/s 53. 2.8 ms1, 17.64 N 54. 10.57 rev/min 55. 35 ms1

56. 3.92 N 57. 6.332 m/s. 58. i. 71.08 N ii. 78.92 N 59. 196 J 60. 47.04 J 61. 237.12 N, 248.88 N 62. 12.56 rad/s2, 5 sec 63. i. 2 = 25 rad/s ii. = 150 rad iii. at = 0.2 m/s2 64. t = 4 s 65. = 29 53

Answers to Multiple Choice Questions

1. (B) 2. (B) 3. (D) 4. (B) 5. (B) 6. (A) 7. (B) 8. (D) 9. (B) 10. (D) 11. (B) 12. (A) 13. (C) 14. (B) 15. (D) 16. (C) 17. (A) 18. (B) 19. (A) 20. (C) 21. (B) 22. (D) 23. (D) 24. (A) 25. (C) 26. (C) 27. (A) 28. (B) 29. (A) 30. (B) 31. (A) 32. (B) 33. (A) 34. (B) 35. (B) 36. (C) 37. (A) 38. (D) 39. (C) 40. (A) 41. (B) 42. (A) 43. (A) 44. (B) 45. (C) 46. (B) 47. (C)

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