Page 137 - Schaum's Outline of Theory and Problems of Applied Physics

122 ROTATIONAL MOTION [CHAP. 10

Table 10-1 Comparison of Linear and Angular Quantities

Linear Quantity Angular Quantity

1
1
Distance s = v 0 t + at 2 Angle θ = ω 0 t + αt 2
2 2

v 0 + v f ω 0 + ω f
s = t θ = t
2 2
Velocity v = v 0 + at Angular velocity ω = ω 0 + αt
2
2
2
2
v = v + 2as ω = ω + 2αθ
0
0
Acceleration a = v/ t Angular acceleration α = ω/ t
Mass m Moment of inertia I
Force F = ma Torque τ = Iα
Momentum p = mv Angular momentum L = Iω
Work W = Fs Work W = τθ
Power P = Fv Power P = τω
1
1
Kinetic energy KE = mv 2 Kinetic energy KE = Iω 2
2 2
SOLVED PROBLEM 10.20
Why do all helicopters have two propellers?
If a single propeller were used, the helicopter itself would have to rotate in the opposite direction in order to
conserve angular momentum.

SOLVED PROBLEM 10.21
2
A skater is spinning at 1 rev/s with her arms outstretched so that her moment of inertia is 2.4 kg·m . She
2
then pulls her arms to her sides, which reduces her moment of inertia to 1.2 kg·m . (a) What is her new
angular velocity? (b) How much work did she have to perform to pull her arms in?
(a) From conservation of angular momentum,

I 1 ω 1 = I 2 ω 2
2
I 1 2.4kg·m
ω 2 = ω 1 = (1 rev/s) = 2 rev/s
I 2 1.2kg·m 2
The skater spins twice as fast as before.
(b) The work done equals the difference between the initial and final kinetic energies of rotation. To find the KE
values, the rad/s must be used as the unit of angular velocity. We have

ω 1 = (1 rev/s)(2π rad/rev) = 2π rad/s
ω 2 = (2 rev/s)(2π rad/rev) = 4π rad/s

and so

1 2 1 2 2
I
KE 1 = 2 1 ω = (2.4kg·m )(2π rad/s) = 47 J
1
2
1 2 1 2 2
I
KE 2 = 2 2 ω = (1.2kg·m )(4π rad/s) = 95 J
2
2
The work done is therefore
W = KE 2 − KE 1 = 95 J − 47 J = 48 J