rotational motion part-b

8/18/2019 Rotational Motion Part-B

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PHYSICS 27LOCUS

The following table contains the expressions for moments of inertia of uniform bodies of various shapes:



PHYSICS 28LOCUS

Find the moment of inertia of a uniform ring of massm and radius r about an axis, AA' touching the ringtangentially and lying in the plane of the ring only,as shown in figure 7.33 (a).

A

Á

C r

r i n g

fig. 7.33(a)

Solution: The axis AA' is parallel to the axis passing through the centre of mass of the ring andlying in the plane of the ring only, as shown in figure7.33 (b). Using parallel axis theorem, we get,

2cm I I mh= +

A

Á

C r

r i n g

fig. 7.33(b)

I I cm

22

4mr

mr = +

254

mr =

A disc of radius R/2 has been removed from a disc of radius R andmass M , as shown in figure 7.34. Find the moment of inertia of theremaining body about an axis which is perpendicular to the planeof the body and touches it tangentially as shown in figure (axis is

perpendicular to the plane of the body).

fig. 7.34

axis

R R /2

Solution: We have,

remaining initial removed I I I = −

( )22initial removed3 3

22 2

m R m R = −

223 3

2 2 4 4 M R

MR= − ( )2removed 2 ( 2) 4 M

m R M R

π π

= × = !

24532

MR=

Find the moment of inertia of a thin rod AB of length l about an axis perpendicular to the length of the rod and passing through its ends

A. Mass per unit length for the rod at a distance r from the end A isgiven as

0( )r r λ λ α = +fig. 7.35(a)

A B

axis

l

where 0λ and α are positive constants.



PHYSICS 29LOCUS

Solution: Let us divide the rod into several elements so that each element can be considered as a point mass. Suchan element of mass dm and length dr is shown in figure 7.35 (b) at a distance r from the axis of rotation. The momentof inertia of the element

2dI dm r = ⋅2

( )dr r λ = ⋅ ⋅( ) 20( )r dr r λ α = + ⋅ ⋅

fig. 7.35(b)

A

axis

l

r

dm

dr

2 30r dr r dr λ α = ⋅ + ⋅

Therefore, the moment of inertia of the rod is

I dI =∫ 2 3

00 0

l l

r dr r dr λ α = ⋅ + ⋅∫ ∫ 3 4

0

3 4l l λ α = +

Find the moment of inertia of a uniform rod of mass m andlength l about an axis passing through one of the ends of the rod and making an angle θ with the length of the rod, asshown in figure 7.36(a).

fig. 7.36(a)

m, l

axis

θ

Solution: The moment of inertia of the element shown in figure 7.36 (b) is2( sin )dI dm r θ = ⋅

2( sin )m

dr r l

θ = sin is the distance of

the element from the axisr θ

!

22sinm r dr

l θ =

fig. 7.36(b)

m, l

axis

!

dm

dr

r

Therefore, the moment of inertia of the rod is

I dI =∫ 2

2

0

sin l mr dr

l θ = ⋅∫ [ ]is same for all elementsθ

2 3sin3

m l l

θ = ⋅

2 2sin3

ml θ =

ALTERNATE METHOD:

Consider the situation shown in figure 7.36 (c). The rod AB can be made from the rod A' B' by displacing its eachelement parallel to the axis of rotation and by different distances.



PHYSICS 30LOCUS

Therefore, the moment of inertia of the rod AB

the moment of inertia of the rod A B= ′ ′

2( sin )

3

m l θ

=

As the distribution of

mass in is uniform A B

′ ′ 2 2sin

3ml θ =

axis

!

B'

B

A'

Astretch

fig. 7.36(c)

m, l

NOTE: Whenever you use this approach of displacement of mass parallel to the axis must be

careful. Analyze the situation shown in figure 7.37.A uniform cone is compressed along the length parallel to its axis to get a disc. Here we cannotsay that the moment of inertia of the cone about its

axis is2

2 MR

because the disc we have got is not

a uniform one. If we have the moment of inertia of this nonuniform disc then of course that will be equalto the moment of inertia of the cone aboutthe axis of symmetry.

After Compression

Disc (nonuniform)

Cone

fig. 7.37: When we compress the cone along the length parallel to its axis, we get a disc which is not uniform.Density of the disc is highest at its centre and it decreaseswith radial distance from the centre.

CompressCompress

Find the moment of inertia of the body shown infigure about the axis passing through the point Oand perpendicular to the plane of the body. Theshown body has a mass M and it is one-fourth

part of a uniform disc of radius R.

R M

axis

Ofig. 7.38(a)

Solution: If we combine four such bodies to form the complete disc, then,moment of inertia of the complete disc is

0 4 I I = ...(i)where I is the moment of inertia of each of the four parts of the disc.Again, mass of the disc is 4M and radius is R, therefore,

2

0 (4 ) 2 R

I M = ⋅ ...(ii)Radius = R

M

M M

M O

axis

fig. 7.38(b)

From equations (i) and (ii), we have,2

4 42

R I M =

2

2 MR

I =

moment of inertia of the given body.=



PHYSICS 31LOCUS

ALTERNATE METHOD:The distribution of mass with respect to the given axis of rotation in the given body has the same fashion as that in a

uniform disc, hence, the moment of inertia of the given body is2

.2

MR

RADIUS OF GYRATION: The radius of gyration, k , of a body about an axis is the distancefrom the axis where the whole mass of the bodycan be assumed to be concentrated so that themoment of inertia about that axis remains the same.If k be the radius of gyration of the body shown infigure 7.39, then the entire mass can be assumedto be concentrated at a distance k from the axis.As the moment of inertia is same in both thesitutations, we have,

2 I Mk = fig. 7.39

M, I

axis axis

K M

⇒ =k M I ...(7.22)



PHYSICS 32LOCUS

1. (a) Given that ˆˆ ˆr ix jy kz = + +" and ˆˆ ˆ x y z F iF jF kF = + +"

find the torque r F τ = × "" " .

(b) Show that if r "

and F "

lie in a given plane then τ has no component in that plane.

2. Torque is defined as(a) a force tending to cause rotation(b) the product of the force and the perpendicular distance from the axis of rotation to the line of action of

the force(c) the product of the force and the angular displacement(d) the product of the force and the angular velocity(e) the rotational work done

3. In show the line of a action and the moment arm of each force about the origin O. Imagine these forces to beacting on a rigid body pivoted at O, all vectors shown being in the plane of the figure, and find the magnitude

and the direction of the resultant torque on the body.

F F r !

!

!

11 2

r !

2θ 1

θ 2O

4. Starting from Newton’s third law, prove that the resultant internal torque on a system of particles is zero.

5. Can a given rigid body have more than one moment of inertia?

6. Compute the torque about the origin for the force ˆ F mgj= −"

and ˆ ˆr xi yj= +" and show that this torque isindependent of the coordinate y.

7. A particle of mass m is projected with a speed u at an angle θ with the horizontal. Find the torque of theweight of the particle about the point of projection when the particle is at the highest point.

8. A force of magnitude F is applied horizontally in the nega-tive x direction to the rim of a disk of radius R, as shown infigure. Write F

"

and r "

in terms of the unit vectors ˆ ˆ,i jand k̂ and compute the torque produced by the force aboutthe origin at the center of the disk.

y

F

c R

x

9. Five particles, each of which has a mass of 0.24 kg, are fixed at positions that are equally spaced along ameter stick with one of these particles at each end. What is the moment of inertia about an axis that is perpendicular to the meter stick (which has negligible mass) and through the center of mass of this rigid body?(a) 0.30 kg • m² (b) 0.15 kg • m²(c) 0.25 kg • m² (d) 0.20 kg • m²(e) 0.35 kg • m².

10. Two particles 1 2( 0.20 kg, 0.30 kg)m m= = are positioned at the ends of a 2.0 m long rod of negligiblemass. What is the moment of inertia of this rigid body about an axis perpendicualr to the rod and through thecenter of mass?

(a) 0.48 kg • m² (b) 0.50 kg • m²(c) 1.2 kg • m² (d) 0.80 kg • m²(e) 0.70 kg • m².



PHYSICS 33LOCUS

11. If M = 0.50 kg, L = 1.2 m, and the mass of each connecting rod shown in negligible, what is the moment of inertia about an axis perpendicular to the paper through the center of mass? Treat the mass as particles.(a) 3.7 kg • m²(b) 2.8 kg • m²(c) 3.2 kg • m² M

L L M 3 M

(d) 2.3 kg • m²(e) 3.9 kg • m².

12. The rotational inertia of an object about an axis depends on the(a) angular velocity about the aixs (b) angular acceleration about the axis(c) mass distribution about the axis (d) torque about the axis(e) linear acceleration about the axis

13. To increase the moment of inertia of a body about an axis, you must(a) increase the angular acceleration (b) increase the angular velocity(c) decrease the angular velocity (d) make the body occupy less space(e) place part of the body farther from the axis

14. In the figure R1 = R2 and cm is the center of mass. The rotationalinertia about an axis through point P 1 is I 1, the rotational inertiaabout an axis through point P 2 is I 2, and the rotational inertia aboutan axis through the cm is I cm. The relationship among the momentsis

(a) 1 2 cm I I I = > (b) 1 2 cm I I I = <(c) 1 2 cm I I I > > (d) 1 2cm I I I < >(e) 1 2 cm I I I = =

15. A homogeneous solid cylinder of mass m, length L, and radius R rotates about an axis through point P which

is parallel to the cylinder axis. If the moment of inertia about the cylinder axis is21

2mR , the moment of

inertia about the axis through P is

(a) 20.4mR (b)21

2mR

(c)22

3mR (d) mR2

(e) 1.5 mR2

16. Four particles of mass m are connected by massless rodsto form a rectangle of sides 2 a and 2 b as shown. Theassembly rotates about an axis in the plane of the figurethrough the centre (Shown in figure). Find the moment of inertia of this system about this axis.

Axis of rotation

2b

2a



PHYSICS 34LOCUS

R17. Use the parallel - axis theorem and to find the moment of

inertia of a solid sphere about an axis tangent to the sphere.

18. Figure shows a pair of uniform spheres each of mass 500 g and radius 5 cm. They are mounted on a uniformrod which has length L = 30 cm and mass 60 g.

(a) Calculate the moment of inertia of this system about an axis perpendicular to the rod through the enter of the rod, using the approximation that the two spheres can be treated as point particles a distance20 cm from the axis of rotation and that the mass of the rod is negligible.

(b) Calculate the moment of inertia exactly and compare your result with your approximate value.

10cm10cm L = cm30

m = cm60

Axis of rotation

500g 500g

19. Find the moment of inertia of a rectangular plate of mass m and dimensions a×b, as shown in figure, aboutthe axis passing through one of its vertices and perpendicular to its plane.

b

a

axistop view

b

a

20. From the result obtained in the previous question, find the moment of inertia of the cuboid about an axis passing through one of its edges, as shown in figure.

mass = m

b

ah

axis

21. The moment of inertia of a uniform semicircular wire of mass M and radius r about a line perpendicular tothe plane of the wire through the centre is

(a) Mr ² (b) 12 Mr ²

(c)14

Mr ² (d)25

Mr ².



PHYSICS 35LOCUS

22. A body having its centre of mass at the origin has three of its particles at ( a , 0, 0). (0, a , 0), (0, 0, a ). themoments of inertia of the body about the X and Y axes are 0.20 kg-m² each. The moment of inertia aboutthe Z -axis.(a) is 0.20 kg-m² (b) is 0.40 kg-m²(c) is 0.20 2 kg-m² (d) cannot be deduced with this information.

23. A thin wire of length L and uniform linear mass density ρ is bent into a circular loop with center at O asshown. The moment of inertia of the loop about the axis XX ′ is :

(a)3

28 Lρπ (b)

3

216 Lρπ

X X '90°

O

(c)3

2

516

Lρπ (d)

3

2

38

Lρπ

24. The surface density (mass/area) of a circular disc of radius α depends on the distance from the centre as( ) p r A Br = + . Find its moment of inertia about the line perpendicular to the plane of the disc through its

centre.

25. Find the moment of inertia of a uniform square plate of mass m and edge a about one of its diagonals.

26. Find the radius of gyration of a circular ring of radius r about a line perpendicular to the plane of the ring and passing through on of its particles.

27. The radius of gyration of a uniform disc about a line perpendicular to the disc equals its radius. Find thedistance of the line from the centre.

28. From a circular disc of radius R and mass 9 M a small disc of radius R/3 isremoved from the disc. The moment of inertia of the remaining disc aboutan axis perpendicular to the plane of the disc and passing through O is :

(a) 24 MR (b)240

9 MR O R

R/3

2 /3 R

(c) 210 MR (d)237

9 MR

29. The moment of inertia of a thin square plate ABCD, of uniform thickness about anaxis passing through the centre O and perpendicular to the plane of the plate is :(a) 1 2 I I +(b) 3 4 I I + O

C D

A B1

2

3

4

(c) 1 3 I I +(d) 1 2 3 4 I I I I + + +where 1 2 3, , I I I and 4 I are respectively moments of inertia about axes 1, 2,3, and 4 which are in the planeof the plate.

30. Let I be the moment of inertia of a uniform square plate about an axis AB that passes through its centre andis parallel to two of its sides. CD is a line in the plane of the plate that passes through the centre of the plateand makes and angle θ with AB. The moment of inertia of the plate about the axis CD is then equal to :(a) I (b) 2sin I θ(c)

2

cos I θ (d)2

cos ( / 2) I θ31. A 0.5 kg particle falls under the influence of gravity. (a) It is at 10 y m= and 2 x m= at time 1t . What is the

torque about the origin exerted by gravity on the particle at this time? (b) At some later time the particle is at0 y = , 2 . x m= What is the torque about the origin at this time?



PHYSICS 36LOCUS

ANGULAR ACCELERATION PRODUCED BY A TORQUE:In terms of moment of inertia, equation (7.18) can be written as

net# = I # ...(7.19 a)

The vector form of the above equation is

net# =" "

I # ...(7.19 b)The angular acceleration produced is along the direction of applied net external torque. The magnitude of the

produced angular acceleration is directly proportional to the magnitude of the net external torque and inversely proportional to the moment of inertia of the body. The above relation looks the translational equation net . F ma=

" "

Here, you should not forget that I α Γ = is not an independent rule. It is derived from F = ma only. We can establishan analogue between translational and rotational variables. By doing so concept developed so far for translationalmotion would help to solve the problems involving rotational motion. The possible analogue is as follows

/ /

( ) distance traversed, angle turned,s

( ) average speed, average angular speed,t t

( ) intantaneous speed, instantaneous a

i S ii v

dsiii v

dt

θ θ

ω ∆ ∆= =∆ ∆=

TRANSLATIONAL ROTATIONAL

VARIABLE EXPRESSION VARIABLE EXPRESSION

ngular speed,

( ) average acceleration, average angular acceleration,

( ) instantaneous acceleration, instantaneous angular acceleration,

( ) mass, moment of inertia,

(

d dt

dviv a

dt t dv d

v adt dt

vi m I

v

θ ω

ω α

ω α

=

∆= = ∆= =

) force, torque,

( )

ii F

viii F ma I α

Γ

= Γ =( ) linear momentum, Angular momentum,

( )

( )

( ) conservation of linear momentum: conservation of angular momentum:

When 0, constant. When 0, constant.

translational ki( )

ix p l

x p mv l I

dp dl xi F

dt dt xii

F p l

xiii

ω = =

= Γ =

= = Γ = =

2 21 12 2

netic energy, rotational kinetic energy,

( ) work done, work done,

k mv k I

xiv d F ds d d

ω

ω ω θ

= =

= ⋅ = Γ ⋅Angular quantities involved in analogues (ix) to (xiv) would be discussed later in this topic.

Find the angular acceleration of the rod given in example 4 at the moment

(a) when it is released from rest in the horizontal position;

(b) when it makes an angle θ with the horizontal.



PHYSICS 37LOCUS

Solution: After the moment when the rod is released from the restin the horizontal position , it would rotate in the vertical plane abouta horizontal axis passing through the hinge and perpendicular to thelength of the rod. Initial angular velocity of the rod is zero but dueto nonzero torque of gravity it has some angular acceleration and

hence, it will acquire some angular speed as it rotates. Asdiscussed in example 4, the hinge force does not provide any

mg

Oα

Γ

r l ⊥ = /2

fig. 7.40(a)

torque about the axis under consideration and the weight of the rod tries to rotate it in the clockwise sense, i.e.,it provides a torque perpendicularly, inward to the plane of the paper. An approach using r F ×

"" to find the torque

would also give the same result. Hence, angular acceleration of the rod,

gravitynet

I I α

Γ Γ = =

22

3

mg l

ml ⋅=

32

g l

=

(b) When the rod makes an angle θ with the horizontal, its angular acceleration,

mgnet

I I α

Γ Γ = =

mg r I ⊥⋅=

α

mg

θO

r l ⊥ = 2 cos/ θ

fig. 7.40(b)

2

cos2

3

l mg

ml

θ ⋅ ⋅= [from figure 7.40 (b)]

3cos

2 g l

θ = ⋅

In the previous case, find the angular velocity of the rod when it has turned through an angle θ after the moment when

it was released from rest in the horizontal position. Also find the angular velocity when the rod becomes vertical.

Solution: From the result obtained in part (b) of the previous example, at some angle θ , the angular acceleration of the rod,

3cos

2 g l

α θ = ⋅

⇒3

cos2

d g dt l ω

θ =

⇒

3cos

2

d d g

d dt l

ω θ θ

θ ⋅ =

[Using chain rule.]

⇒0 0

3. cos

2 g

d d l

ω θ

ω ω θ θ = ⋅∫ ∫ [ ]at = 0, = 0θ ω !



PHYSICS 38LOCUS

⇒ [ ]2

00

3sin

2 2 g l

ω θ ω θ

=

⇒

2 3sin2 2

g l

ω θ =

⇒3 sin g

l θ

ω =

When the rod becomes vertical, 2,θ π = and hence, angular velocity,3 g l ω =

In the previous example, find the hinge force on the rod at θ = 0.

Solution: Just after the moment when the rod was released from the rest in the horizontal position, it is shown infigure 7.41(a). Let the vertical component of the force on the rod from the hinge be 1 R and the horizontal componentof the same be 2 R , as shown in figure. The subsequent motion of the centre of mass of the rod is a nonuniformcircular notice on the vertical circular path of radius l /2 with the centre at the hinge, as suggested in the figure.

Initially the rod is at rest and hence radial component of the acceleration of the centre of mass of the rod, ω ²r , is zero.Hence, applying net cm F ma= along the radial direction, we get

net, radial cm, radial F ma=

⇒ 2 0 R m= ×0=

Applying the same along the tangential direction, we have, a cm =l 2

α .

mg

R 2

R 1

Oα

ω = 032

α = g l

fig. 7.41(a)

net, tangential cm, tangential F ma=

⇒ 1 2l

mg R m α − =

⇒ 1 2l

R mg mα = −

34

mg mg = −

14

mg =

Net hinge force 2 21 21

.4

R R mg = + = 2[ 0] R =!



PHYSICS 39LOCUS

In the previous example, find the magnitude of the net hinge force on the rod when the rod has turned through anangle θ.

Solution: If ω be the angular velocity of the rod when it has turned through an angle θ , the centre of mass of the rod

has2

2l

ω and2l

α as radial and tangential components of its acceleration, respectively, as shown in figure 7.41(b).

Applying ext cm F Ma= on the rod along the radial direction, we have,2

2 sin 2l

R mg mθ ω − =

⇒ 23

sin sin2

R mg mg θ θ = +ω

α

θ

ω 2 l

2C

l 2

α .mg

R 2 R 13 sin

3 cos2

g l

g l

θω

α θ

=

=

fig. 7.41(b): Rod when it makes an angle with thehorizontal. , are perpendicular and radialcomponents, respectively, of the reaction force acting

θ R R1 2

on the rod from the hinge.

You can also assume reaction force as andacting at some angle with the rod.NOTE: R

5sin

2mg θ =

Applying the same along the tangential direction, we have,

1cos 2l

mg R mθ α − = ⋅

⇒ 1 cos 2l

R mg mθ α = −

3cos cos4mg mg θ θ = −

1cos

4mg θ =

Therefore, net force on the rod from the hinge can be obtained by solving

2 21 2 R R R= +

If the disc given in example 5 has mass M and it is free to rotate about its symmetrical axis passing through O, findits angular acceleration.

Solution: If α be the angular acceleration of the disc, then, using net , I α Γ = we have,

net

I α

Γ =

2

3

( 2)

FR

MR=

6 F MR

=

As the net torque is in clockwise sense, α has the same sense of rotation.



PHYSICS 40LOCUS

A uniform disc of radius 0.12 m and mass 5 kg is pivoted so that itrotates freely about its axis. A thin, massless and inextensible stringwrapped around the disc is pulled with a force of 20 N, as shown

in figure 7.42(a) .(a) What is the torque exerted on the disc about its axis?

20 N

5 kg

R

fig. 7.42(a)

(b) What is the angular acceleration of the disc?(c) If the disc starts from rest, what is the angular velocity after 3s?

Solution: It is obvious that the string force gives a torque to the disc in the clockwise direction. As the torque given by the force from the axle is zero. Net torque on the disc is,

net torque of the string forceΓ = F r

⊥

= ⋅ F R= ⋅ 20 N

5 kg

R

fig. 7.42(b)(20 N) (0.12 m)= ⋅2.4 N-m.=

As the net torque on the disc is in clockwise direction, the disc has angular acceleration in the same direction. If α be the magnitude of the angular acceleration,

net net2( 2) I MR

α Γ Γ = =

22

2.4 2 rad/s5 (0.12)

×= ×

266.66 rad/s$At t = 0 if the disc has zero angular velocity, then, at some time t , its angular velocity,

ω ω = in0

t α +

⇒ t ω α =∴ At t = 3 s,

200 rad/sω =

A uniform disc of radius R and mass M is mounted on an axissupported in fixed frictionless bearing. A light string is wrappedaround the rim of the disc and a body of mass m is supported bythe string, as shown in figure 7.43(a).

(a) find the angular acceleration of the disc;

(b) find the magnitude of the tangential acceleration of the point on the rim where the string separates

from the rim.(c) if the system is released from rest at t = 0, find the

speed of the block at some time t (>0). m

M, R, Disc

fig. 7.43(a)



PHYSICS 41LOCUS

Solution: Analyze the situation according to the information provided in the figure 7.43(b). You should also note thefollowing points:• Only tension force of the string, T , produces a torque on the disc about its centre O. Torque of the weight of

the disc and that of the reaction force from the bearing are zero about O.• If α be the angular acceleration of the disc (in the clockwise direction) then the point P on the disc has a

tangential acceleration α R in the vertically downward direction at the moment shown in figure. The stringunwinds at the same acceleration and the block has the same acceleration in the vertically downwarddirection. Therefore, if a be the acceleration of the block, then,

a Rα = ...(i) Now, applying net I α Γ = on the disc about its symmetrical axis, we have,

net I α Γ =

⇒

2

.2

MRT R α = ⋅

m

α

α R P

T

T

mg

afig. 7.43(b)

⇒2

M RT α =

⇒2

MaT = [Using (i) ...(ii)

Using net F ma= for the block in the vertical direction, we have,

net F ma=⇒ mg T ma− = ...(iii)

Adding (ii) and (iii), we get,

2 M

mg m a = +

⇒ 2m

a g Rm M = ⋅ +

If v be the speed of the block at some time t , then, we have,

v u at = + [ ] is constanta!at = [ ]0u =!

2m gt

m M = ⋅ +

Find the acceleration of 1m and 2m in an Atwood’s Machine,shown in figure 7.44(a), if there is friction present between thesurface of pulley and the thread does not slip over the surface of

the pulley. Moment of inertia of the pulley about its symmetricalaxis is I and its radius is R. The pulley can rotate freely about itssymmetrical axis. m 2

m 1

I, R

fig. 7.44(a)



PHYSICS 42LOCUS

Solution: Due to friction between the pulley and the thread tensions in the parts of the thread on the two sides of the pulley are different. Let that in the right part it is 1T and that in the left part is 2 ,T as shown in figure 7.44(b). Forcesacting on the two blocks and the pulley are also shown in figure 7.44(b). Force on the pulley from the support andits weight are not shown because they do not produce torque on the pulley about its symmetrical axis of rotation. If the block 1m comes down with an acceleration a then 2m would go up with the same acceleration because theyare connected by the same string, as shown in the same figure.7.44(b).

m 2 m 1a a

T 2 T 1

m g 2 m g 1

T 2 T 1

α

fig. 7.44(b)

T 2 T 1

α

fig. 7.44(c)

R R

α

R R

α R

α R

aa

fig. 7.44(d)

If we assume that the pulley gets an angular acceleration in the clockwise sense then the torque of 1T would be positive and that of 2T would be negative, as suggested in figure 7.44(c).Again, as any point on the rim of the pulley has a tangential acceleration α R, the block 1m comes down and the

block 2m goes up with the same acceleration, as shown in figure 7.44(d).Therefore, we can write,

a Rα = ...(i)

Using net F ma= for the two blocks, we have,

1 1 1m g T m a− = ...(ii) [for 1m ]

2 2 2T m g m a− = ...(iii) [for 2m ]

Using net I α Γ = for the pulley, we have,

1 2T R T R I α + ⋅ − ⋅ = ...(iv) torque of support forceand weight are zero !

Substituting from (i) in (iv), we get,

1 2 2

aT T I

R− = ...(v)

Adding (ii), (iii) and (v), we get,

1 2 1 2 2( ) I

m m g m m a R

− = + +

⇒1 2

21 2

m ma g

m m I R

−= + +



PHYSICS 43LOCUS

A thin uniform rod AB of mass m = 1.0 kg moves translationallywith acceleration a = 2.0 m/s² due to two antiparallel force 1 F and

2 F acting on it perpendicularly to its length, as shown in figure7.45. The distance between the points at which these forces areapplied is x = 20 cm. Besides, it is known that 2 5.0 N. F = Findthe length of the rod.

x

A

B

F 1

F 2a = 2.0 m/s

fig. 7.45

Solution: Before analyzing the details of the given situation, let us analyze the rotational effect of two antiparallelforces. Consider the situations shown in figure 7.45.

AΓ 2

Γ 1

C

F 2

F 1

Γ 1Γ 1

F 2

F 1

(a): and are producing torques about in opposite directions.

F F A

21 (b): and are producing torques about in opposite directions.

F F B

21 (c): and are producing torques about in the same direction

F F C

21

Γ 2 Γ 1

F 2

F 1

B

If we analyze the torques of the two forces about every point in their plane containing them, then, we arrive at the

conclusion that if the point lies between the lines of action of 1 F and 2 F then torques of the forces about that pointadd up together otherwise they are in opposite directions.If the magnitudes of the two forces are equal then such a pair is called as a couple . If the magnitude of each force is

F and the distance between their lines of application is d , then, the net torque about any point in their plane is F .d ,as shown in figure 7.47.

O Γ 1Γ 1

OΓ 2

Γ 1

Γ 2Γ 1

F =F 2

F = F 1

O

d

l

F = F 1

F =F 2

d

l d

d

F = F 1

F =F 2

net 1 2

( ) F l d Fl

F d

Γ = Γ − Γ

= + −

= ⋅

net 2 1

( ) F l d Fl

F d

Γ = Γ − Γ

= + −

= ⋅

net 1 2

1 1 2 2

1 2( )

F d F d

F d d

F d

Γ = Γ + Γ

= ⋅ + ⋅

= +

= ⋅fig. 7.47: Torque of a couple.

(a) (b) (c)

d 2

1



PHYSICS 44LOCUS

Now, let us discuss the given case. As the rod is in pure translationmotion, net torque on it about any point must be zero. Therefore,the centre of mass of rod can not lie between the lines of action of the forces because in that case torques produced by then aboutthe centre of mass do not cancel each other.

Let us assume that the centre of mass of the rod lies at a distance yaway from the line of action of 2 , F as shown in figure 4.48. As therod translates towards right, 2 F must have a greater magnitudethan 1. F

Using net , F ma= we have

x

F 1

F 2a= 2.0 m/s²

fig. 7.48

y

C

2 1 F F ma− =⇒ 1 2 F F ma= −

(5 1 2) N= − × = 3 N

Again, as the net torque on the rod about C must be zero,

1 2

magnitude of the torque magnitude of the torque produced by about produced by about F C F C =

the two torques haveopposite directions

!

⇒ 1 2( ) F x y F y+ =⇒ 2 1 1( ) F F y F x− =

⇒

1

2 1

F y x

F F =

−

320

5 3= − cm

= 30 cm∴ Length of the rod, 2( ) 1.0 ml x y= + =

A force ˆ ˆ F Ai Bj= +"

is applied to a point whose radius vector relative to the origin of coordinates O is equal toˆ ˆ

,r ai bj= +"

where a , b, A, B are constants, andˆ ˆ,i j are the unit vectors of the x and y axes. Find the torque Γ

"

and the arm length l of the force F "

relative to the point O.

Solution: Torque of F "

about O is

r F Γ = ×" ""

ˆ ˆ ˆ ˆ( ) ( )ai bj Ai Bj= + × +r

l= r sin θ

F

!

!

O

θ

fig. 7.49

ˆ( )aB bA k = −Arm length of F

" with respect to O is

sinl r θ = is the distance of the point of application offrom and is the angle between and .r F O r F θ

"

""

r F r

r F

×=

⋅

"""

"" sinr F r F θ × = ⋅ ⋅ " "" "

!



PHYSICS 45LOCUS

2 2

2 2 2

aB bAa b

a b A B 2−= +

+ ⋅ +

2

aB bA

A B2

−=+

A uniform cylinder of radius R is spun about its axis to the angular velocity 0ω and then placed into a corner, as shown in figure 6.50(a).The coefficient of kinetic friction between the corner walls and thecylinder is equal to k . How many turns will the cylinder accomplish

before it stops?

R

fig. 7.50(a)Solution: All forces acting on the cylinder are shown in figure6.50(b). As the cylinder rotates, its surface slips over the corner walls and hence frictional forces acting on it, 1 f and 2 , f are kineticin nature. Normal contact forces acting on the cylinder from thecorner walls, 1 N and 2 , N and the weight of the cylinder, mg ,

pass through the centre of the cylinder and hence, these forces produce no torque about the centre C . Only frictional forces produce torque about C and the torques produced by them are inopposite direction of the direction of the angular velocity of the

cylinder and hence, they retard the rotational motion of the cylinder fig. 7.50(b)

C

f 1

N 1

N 2α

ω

f 2 +ve

As the cylinder does not translate, net force on it in both vertical and horizontal directions must be zero. Therefore,

1 2 N f mg + =⇒ 1 2 N N mg µ + = ...(i) 2 2[ ] f N µ =!and 2 1 N f =⇒ 2 1 N N µ = ...(ii) [ ]1 1 f N µ =!

Substituting for 2 N in equation (i) from equation (ii), we have,

21 1 N N mg µ + =

⇒ 1 21

mg N

µ =

+ ...(iii)

Substituting for 1 N in equation (ii) from equation (iii), we have,

2 21

mg N

µ µ

=+ ...(iv)

If we define the anticlock wise sense of rotation as the +ve direction of rotation, then, the clock wise sense becomesthe –ve direction for the same. Hence, angular acceleration in the present case becomes negative for this choice of

refrence direction. The angular acceleration,1 2net torque due to and about

moment of inertia about the axis of rotation f f C

α =



PHYSICS 46LOCUS

1 22

2

f R f R

mR⋅ + ⋅= −

1 22( ) N N

mRµ µ +

= −

1 22

( ) N N mR

µ = − +

2

2 1

1mg

mRµ µ

µ += −+

22 1

1

g Rµ µ

µ += −+

If the cylinder had the angular velocity 0ω at t = 0and at some time t it has an angular velocity ω , and in this durationit has turned through an angle θ , then,

2 20 2ω ω αθ = +

20 2

4 1

1

g Rµ µ

ω θ µ

+= −+

If the cylinder stops having rotated through an angle 0 ,θ then at 0 ,θ θ = ω = 0. Therefore,

2 20 02

4 10

1

g Rµ µ

ω θ µ

+= −+

⇒

2 20

0(1 )

4 (1 ) R g

ω µ θ

µ µ += +

Therefore, the number of rotations accomplished by the cylinder, before it stops,

0

2n

θ π

=

2 20 (1 )

8 (1 ) R g

ω µ π µ µ

+= +



PHYSICS 47LOCUS

1. State Newton’s three laws of motion in words suitable for rotating bodies.

2. Can an object rotate if there is no torque acting?

3. If the angular velocity of a body is zero at some instant, does this mean that the resultant torque on the bodymust be zero?

4. A uniform rod of mass M = 1.2 kg and lengt h = 0.80 m is free to rotate about one end as shown. Themoment of inertia of the rod about an axis perpendicular to the rod an through the centre of mass is given

by 2 /12. ML If a force ( F = 5.0 N, θ = 40°) acts as shown, what is the resulting angular acceleration aboutthe pivot point?(a) 16 rad/s²(b) 12 rad/s²(c) 14 rad/s²

L θF

Pivot(d) 10 rad/s²(e) 33 rad/s².

5. Two small masses, mA = 4.0 × 10 –3 kg and mB = 2.0 × 10

–3 kg ,are connected by a 1.0 m rod of negligible mass. The angular acceleration about B produced by a force of 0.016 N appliedat A is is approximately(a) 4.0 rad/ s2

(b) 2.7 rad/ s2

(c) 11 rad/ s2

A B

F N = 0.016

1.0 m

(d) 12 rad/ s2

(e) 4.0 × 10 2 rad/ s2

6. A solid cylinder has a moment of inertia of 2 kg . m2. It is at rest at time zero when a net torque given by26 6t τ = + (SI units) is applied. After 2 s, the angular velocity of the cylinder will be

(a) 3.0 rad/s (b) 12 rad/s(c) 14 rad/s (d) 24 rad/s(e) 28 rad/s

7. In a laboratory experiment, various torques are applied toa rotor and the angular acceleration is measured. The resultsare plotted on the graph above. From the graph, the moment

of inertia of the rotor is(a) 20.010 kg m⋅(b) 20.011 kg m⋅(c) 20.0125 kg m⋅ (d) 20.0138 kg m⋅(e) 20.0225 kg m⋅

8. A thin, massless string is wrapped around a 0.25-m radius grindstonesupported by bearings that produce negligible frictional torque. A steadytension of 20 N in the string causes the grindstone to move from rest to a

speed of 60 rad/s in 12 s. The moment of inertia of the grindstone is(a) 21.0 kg m⋅ (b) 22.0 kg m⋅(c) 23.0 kg m⋅ (d) 24.0 kg m⋅(e) 25.0 kg m⋅



PHYSICS 48LOCUS

9. A wheel of radius R1 has an axle of radius 2 114

R R= . If a force F 1 isapplied tangent to the wheel, a force F 2, applied tangent to the axle that willkeep the wheel from turning, is equal to

(a)1

/ 4 F

(b) F 1(c) 4 F 1(d) 16 F 1(e) F 1/16

10. A wheel is rotating clockwise on a fixed axis perpendicular to the page. A torque that causes the wheel to slow downis best represented by the vector (a) 1

(b) 2(c) 3(d) 4(e) 5

11. A solid cylinder is spinning counterclockwise about alongitudinal axis when a net torque τ is applied, as shown.The cylinder

(a) speeds up

(b) slows down

(c) precesses about a vertical axis(d) precesses about a horizontal axis

(e) does none of these

12. Two masses M and m ( M > m) are hung over a disk2

disk 1

( ' )2

I M R=and are released so that they acclerate. If T 1 is the tension in the cord onthe left and T 2 is the tension in the cord on the right, then

(a) 1 2T T =(b) 2 1T T >(c) 2 1T T < (d) 2T Mg =(e) 2 /T Mg m=

13. The moment of inertia of the wheel in the figure is 0.50 kg.m2, and the bearing is frictionless. The acceleration of the15-kg mass is approximately

(a) 9.8 m/ s2

(b) 8.7 m/ s2

(c) 74 m/ s2

(d) 16 m/ s2



PHYSICS 49LOCUS

14. A mass 1( 5.0 kg)m = is connected by a light cord to a mass2( 4.0 kg)m = which slides on a smooth surface, as shown in the

figure. The pulley (radius = 0.20 m) rotates about a frictionlessaxle. The acceleration of 2m is 3.5 m/s². What is the moment of inertia of the pulley?(a) 0.29 kg • m²(b) 0.42 kg • m²

M

2 M M 2

M 1

(c) 0.20 kg • m²(d) 0.62 kg • m²(e) 0.60 kg • m².

15. A 1 Mg car is being unloaded by a winch, as shown in figure. At this moment , the winch gearbox shaft breaks, and the car falls from rest. The moment of inertia of the winch drum is 320 kg. m 2 and that of the pulley is 4 kg.m 2; the radius of the winch drum is 0.80 m and that of the pulley 0.30m. Find the speed of thecar as it hits the water.

Winchdrum

which

5.0 m

Pulley

16. A 2000 kg block is lifted by a steel cable which passes over a pulley to a motor-driven winch (see fig 9.22).The radius of the which drum is 30 cm, and the moment of inertia of the pulley is negligible.

(a) What force must be exerted by the cable to lift the block at a constant velocity of 8 cm/s?(b) What torque does the cable exert on the winch drum?(c) What is the angular velocity of the winch drum ?(d) What power must be developed by the motor to drive the winch drum?

r = cm30

T T

which

2000 kg

r

17. The system in figure is released from rest. The 30 kg body is 2 m above thefloor. The pulley is a uniform disk with a radius of 10 cm and mass 5kg.Find

(a) the speed of the 30 kg body just before it hits the floor and

the angular speed of the pulley at that time, 2m

30kg

20kg

r m m= kg r = cm

510

(b) the tensions in the strings and(c) the time it takes for the 30 kg body to reach the floor.



PHYSICS 50LOCUS

18. A triangular plate of uniform thickness and densityis made to rotate about an axis perpendicular tothe plane of the paper and (a) passing through A,(b) passing through B, by the application of thesame force, F , at C (mid-point of AB) as shown in

the figure. The angular acceleration in both the caseswill be the same. (True/false). C

F

B A

19. A uniform disk of mass M and radius R is pivoted so that it can rotate freely about an axis through its center and perpendicular to the plane of the disk. A small particle of mass m is attached to the rim of the disk at thetop directly above the pivot. The system is given a gentle start and the disk begins to rotate.

(a) What is the angular velocity of the disk when the particle is at its lowest point ?(b) At this point, what force must be exerted on the particle by the disk to keep it one the disk ?

20. A uniform boom 5.0 m long and having a total mass of 150 kg is connectedto the ground by a hinge at the bottom and is supported by a horizontal

cable, as shown in figure.(a) What is the tension in the cable ?(b) What is the angular acceleration of the boom the instant the cable is cut?

T

4.0cm

3.0 cm(c) It the cable is cut, what is the angular velocity of the boom whenit is horizontal?

(e) 0.53 m/ s2

21. two forces of magnitude of 50 N, as shown in the figure below, act on a cylinder of radius 4 m and mass6.25 kg. The cylinder sits on a frictionless surface. After 1 second, the velocity and angular velocity of thecylinder in m/s and rad/s are respectively:(a) v = 0 ; ω = 0(b) v = 0; ω = 4(c) v = 0; ω = 8 F

F

(d) v = 8; ω = 8(e) v = 16; ω = 9.

22. In the figure, the rotational inertia of the wheel and axleabout the center is 12.0 kg. m2, the constant force F is39.2 N , and the radius r is 0.800 m. The wheel starts fromrest. When the force has acted through 2.00 m, therotational velocity ω acquired by the wheel due to this forcewill be(a) 1.26 rad/s(b) 3.33 rad/s

2m

F

r

(c) 3.61 rad/s(d) 6.24 rad/s(e) 10.3 rad/s

23. A uniform disc of radius R is spinned to the angular velocity ω and then carefully placed on a horizontalsurface. How long will the disc be rotating on the surface if the friction coefficient is equal to µ? The

pressure exerted by the disc on the surface can be regarded as uniform.

24. A cubical block of mass M and edge a slides down a rough inclined plane of inclination θ with a uniformvelocity. The torque of the normal force on the block about its centre has a magnitude(a) zero (b) Mga

(c) Mga sin θ (d)12

Mga sin θ.


25/25

PHYSICS 51LOCUS

25. Three particles A, B and C , each of mass m, are connected to each other by three massless rigid rods to form a rigid, equilateral triangular body of side l . This body is placed on a horizontal frictionless table ( x- y plane) andis hinged to it at the point A so that it can move without friction about thevertical axis through A(see figure). The body is set into rotational motion on

the table about A with a constant angular velocity .ω (a) Find the magnitude of the horizontal force exerted by

the hinge on the body. B C

ω

y

A x

l F

(b) At time T , when the side BC is parallel to the x-axis, a force F is applied on B along BC (as shown).Obtain the x-component and the y-component of the force exerted by the hinge on the body, immediatelyafter time T .

26. One-fourth length of a uniform rod of mass m and length l is placedon a rough horizontal surface and it is held stationary in horizontal

position by means of a light thread as shown in the figure. Thethread is then burnt and the rod starts rotating about the edge. Findthe angle between the rod and the horizontal when it is about toslide on the edge. The coefficient of friction between the rod andthe surface is .µ

l /4

rotational motion part-b

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