Page 129 - Schaum's Outline of Theory and Problems of Applied Physics

114 ROTATIONAL MOTION [CHAP. 10

The first step is to find the equivalent in radians of 200 , which is
◦

2π rad
◦
θ = (200 ) = 3.49 rad
360 ◦
The radius of the record is 15 cm. Since θ = s/r, a point on its rim travels
s = rθ = (15 cm)(3.49 rad) = 52 cm

when it turns through 200 .
◦

SOLVED PROBLEM 10.5
A wheel 80 cm in diameter turns at 120 rev/min. (a) What is the angular velocity of the wheel in radians
per second? (b) What is the linear velocity of a point on the rim of the wheel in meters per second?

(a) The angular velocity of the wheel is
2π rad/rev

ω = (120 rev/min) = 12.6 rad/s
60 s/min
(b) Since the radius of the wheel is r = 40 cm = 0.4 m, a point on its rim has the linear velocity
v = ωr = (12.6 rad/s)(0.4m) = 5.04 m/s

SOLVED PROBLEM 10.6
A steel cylinder 60 mm in radius is to be machined in a lathe. If the desired linear velocity of the cylinder’s
surface is to be 0.7 m/s, at how many revolutions per minute should it rotate?
From the formula v = ωr we obtain, with r = 60 mm = 0.06 m,
v 0.7 m/s
ω = = = 11.7 rad/s
r 0.06 m
and so, since 1 rev/min = 0.105 rad/s,

60 s/min
ω = (11.7 rad/s) = 112 rev/min
2π rad/rev

ANGULAR ACCELERATION
A rotating body whose angular velocity changes from ω 0 to ω f in the time interval t has the angular acceleration α
(Greek letter alpha)of
ω f − ω 0
α =
t
angular velocity change
Angular acceleration =
time
A positive value of α means that the angular velocity is increasing; a negative value means that it is decreasing.
Only constant accelerations are considered here.
The formulas relating the angular displacement, velocity, and acceleration of a rotating body under constant
angular acceleration are analogous to the formulas relating linear displacement, velocity, and acceleration given
in Chapter 3. If a body has the initial angular velocity ω 0 , its angular velocity ω f after a time t during which its
angular acceleration is α will be

ω f = ω 0 + αt