Page 134 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 10] ROTATIONAL MOTION 119

The rate at which work is being done when a torque τ acts on a body that rotates at the constant angular velocity
ω rad/s is

P = τω

Power = (torque)(angular velocity)

SOLVED PROBLEM 10.14
(a) Find the moment of inertia of a 1-kg phonograph turntable that is 34 cm in diameter. (b) What is its
kinetic energy when it rotates at 45 rev/min?
(a) The radius of the turntable is R = 17 cm = 0.17 m. Assuming that the turntable is a solid disk, we find for its
moment of inertia
1
2
1
2
I = MR = ( )(1kg)(0.17 m) = 0.0145 kg·m 2
2 2
(b) The angular velocity of the turntable is

2π rad/rev
ω = (45 rev/min) = 4.71 rad/s
60 s/min
and so its kinetic energy is
2
1
2
2
1
KE = Iω = ( )(0.0145 kg·m )(4.73 rad/s) = 0.162 J
2 2
2
2
(Note that 1 kg·m /s = 1N·m = 1 J.)
SOLVED PROBLEM 10.15
2
A flywheel of moment of inertia 10 slug·ft is rotating at 90 rad/s. (a) What constant torque is required
to slow it down to 40 rad/s in 20 s? (b) What is the angular displacement of the flywheel while it is being
slowed down? (c) How much kinetic energy is lost by the flywheel?
(40 − 90) rad/s
ω f − ω 0 2
(a) α = = =−2.5 rad/s
t 20 s
The minus sign means that ω is decreasing. The required torque is
2
2
τ = Iα = (10 slug·ft )(−2.5 rad/s ) =−25 lb·ft
2
2
1
2
(b) Either θ = ω 0 t + αt or ω = ω + 2αθ can be used to find θ. From the first formula,
2 f 0
1
1
2
2
2
θ = ω 0 t + αt = (90 rad/s)(20 s) − (2.5 rad/s )(20 s) = 1300 rad
2 2
(c) The KE lost by the flywheel appears as work done against the applied torque. Hence
4
KE = W = τθ =−(25 lb·ft)(1300 rad) =−3.25 × 10 ft·lb
As a check, we can find KE directly:
2
1

KE = KE f − KE 0 = I ω − ω 2
2 f 0
1
2
2
2
2
= (10 slug·ft )[(40 rad/s) − (90 rad/s) ] =−3.25 × 10 ft·lb
2
SOLVED PROBLEM 10.16
A solid cylinder rolls from rest down an inclined plane 1.2 m high without slipping. What is its linear
velocity at the foot of the plane?