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16–2.

The angular acceleration of the disk is defined by a = 3t2 + 12 rad>s, where t is in seconds. If the disk is originally rotating at v0 = 12 rad>s, determine the magnitude of the velocity and the n and t components of acceleration of point A on the disk when t = 2 s.

SolutionAngular Motion. The angular velocity of the disk can be determined by integrating dv = a dt with the initial condition v = 12 rad>s at t = 0.

Lv

12 rad>sdv = L

2s

0(3t2 + 12)dt

v - 12 = (t3 + 12t) 2 2s

0

v = 44.0 rad>s

Motion of Point A. The magnitude of the velocity is

vA = vrA = 44.0(0.5) = 22.0 m>s Ans.

At t = 2 s, a = 3(22) + 12 = 24 rad>s2. Thus, the tangential and normal components of the acceleration are

(aA)t = arA = 24(0.5) = 12.0 m>s2 Ans.

(aA)n = v2rA = (44.02)(0.5) = 968 m>s2 Ans.

0.4 m

0.5 m

B

A

v0 � 12 rad/s

Ans:vA = 22.0 m>s(aA)t = 12.0 m>s2

(aA)n = 968 m>s2

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16–3.

The disk is originally rotating at v0 = 12 rad>s. If  it is subjected to a constant angular acceleration of a = 20 rad>s2, determine the magnitudes of the velocity and the n and t components of acceleration of point A at the instant t = 2 s.

SolutionAngular Motion. The angular velocity of the disk can be determined using

v = v0 + act; v = 12 + 20(2) = 52 rad>s

Motion of Point A. The magnitude of the velocity is

vA = vrA = 52(0.5) = 26.0 m>s Ans.

The tangential and normal component of acceleration are

(aA)t = ar = 20(0.5) = 10.0 m>s2 Ans.

(aA)n = v2r = (522)(0.5) = 1352 m>s2 Ans.

0.4 m

0.5 m

B

A

v0 � 12 rad/s

Ans:vA = 26.0 m>s(aA)t = 10.0 m>s2

(aA)n = 1352 m>s2

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16–5.

SOLUTION

Angular Displacement: At .

Ans.

Angular Velocity: Applying Eq. 16–1. we have

Ans.

Angular Acceleration: Applying Eq. 16–2. we have

Ans.a =dv

dt= 8 rad s2

v =du

dt= 20 + 8t 2

t= 90 s= 740 rad>s

u = 20(90) + 4 A902 B = (34200 rad) * ¢ 1 rev2p rad

≤ = 5443 rev

t = 90 s

The disk is driven by a motor such that the angular positionof the disk is defined by where t is inseconds. Determine the number of revolutions, the angularvelocity, and angular acceleration of the disk when t = 90 s.

u = 120t + 4t22 rad,

0.5 ft

θ

Ans:u = 5443 revv = 740 rad>sa = 8 rad>s2

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16–7.

D A

B

C

F

If gear A rotates with a constant angular acceleration ofstarting from rest, determine the time

required for gear D to attain an angular velocity of 600 rpm.Also, find the number of revolutions of gear D to attain thisangular velocity. Gears A, B, C, and D have radii of 15 mm,50 mm, 25 mm, and 75 mm, respectively.

aA = 90 rad>s2,

SOLUTIONGear B is in mesh with gear A. Thus,

Since gears C and B share the same shaft, . Also, gear D is inmesh with gear C. Thus,

The final angular velocity of gear D is

. Applying the constant acceleration equation,

Ans.

and

Ans. = 34.9 rev

uD = (219.32 rad)a1 rev

2p radb

(20p)2 = 02 + 2(9)(uD - 0)

vD

2 = (vD)0

2 + 2aD [uD - (uD)0]

t = 6.98 s

20p = 0 + 9t

vD = (vD)0 + aD t

20p rad>s

vD = a600 rev

minb a

2p rad1 rev

b a1 min60 s

b =

aD = arCrDbaC = a

2575b(27) = 9 rad>s2

aD rD = aC rC

aC = aB = 27 rad>s2

aB = arArBbaA = a

1550b(90) = 27 rad>s2

aB rB = aA rA

Ans:t = 6.98 suD = 34.9 rev

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16–9.

At the instant vA = 5 rad>s, pulley A is given an angular acceleration a = (0.8u) rad>s2, where u is in radians. Determine the magnitude of acceleration of point B on pulley C when A rotates 3 revolutions. Pulley C has an inner hub which is fixed to its outer one and turns with it.

SolutionAngular Motion. The angular velocity of pulley A can be determined by integrating v dv = a du with the initial condition vA = 5 rad>s at uA = 0.

LvA

5 rad>sv dv = L

uA

00.8udu

v2

2`5 rad>s

vA

= (0.4u2) `0

uA

vA2

2-

52

2= 0.4uA

2

vA = e20.8uA2 + 25 f rad>s

At uA = 3(2p) = 6p rad,

vA = 20.8(6p)2 + 25 = 17.585 rad>s

aA = 0.8(6p) = 4.8p rad>s2

Since pulleys A and C are connected by a non-slip belt,

vCrC = vArA; vC(40) = 17.585(50)

vC = 21.982 rad>s

aCrC = aArA; aC(40) = (4.8p)(50)

aC = 6p rad>s2

Motion of Point B. The tangential and normal components of acceleration of point B can be determined from

(aB)t = aCrB = 6p(0.06) = 1.1310 m>s2

(aB)n = vC2 rB = (21.9822)(0.06) = 28.9917 m>s2

Thus, the magnitude of aB is

aB = 2(aB)t2 + (aB)n

2 = 21.13102 + 28.99172

= 29.01 m>s2 = 29.0 m>s2 Ans.

50 mm

40 mm

60 mm

B

A

C

vA aA

Ans:aB = 29.0 m>s2

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16–17.

A motor gives gear A an angular acceleration of aA = (2 + 0.006 u2) rad>s2, where u is in radians. If this gear is initially turning at vA = 15 rad>s, determine the angular velocity of gear B after A undergoes an angular displacement of 10 rev.

SolutionAngular Motion. The angular velocity of the gear A can be determined by integrating v dv = a du with initial condition vA = 15 rad>s at uA = 0.

LvA

15 rad>sv dv = L

uA

0(2 + 0.006 u2)du

v2

2`vA

15 rad>s= (2u + 0.002 u3) `

uA

0

v2

A

2-

152

2= 2uA + 0.002 u3

A

vA = 20.004 u3A + 4 u + 225 rad>s

At uA = 10(2p) = 20p rad,

vA = 20.004(20p)3 + 4(20p) + 225

= 38.3214 rad>s

Since gear B is meshed with gear A,

vBrB = vArA ; vB(175) = 38.3214(100)

vB = 21.8979 rad>s

= 21.9 rad>s d Ans.

B175 mm

100 mm

A

aAvA

aB

Ans:vB = 21.9 rad>s d

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16–18.

A motor gives gear A an angular acceleration of aA = (2t3) rad>s2, where t is in seconds. If this gear is initially turning at vA = 15 rad>s, determine the angular velocity of gear B when t = 3 s.

SolutionAngular Motion. The angular velocity of gear A can be determined by integrating dv = a dt with initial condition vA = 15 rad>s at t = 0 s.

LvA

15 rad>sdv = L

t

02t3 dt

vA - 15 =12

t4 `t

0

vA = e 12

t4 + 15 f rad>s

At t = 3 s,

vA =12

(34) + 15 = 55.5 rad>s

Since gear B meshed with gear A,

vBrB = vArA ; vB(175) = 55.5(100)

vB = 31.7143 rad>s

= 31.7 rad>s d Ans.

B175 mm

100 mm

A

aAvA

aB

Ans:vB = 31.7 rad>s d