Page 132 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 10] ROTATIONAL MOTION 117

Fig. 10-4. Four directions along which a force F can be applied to a wrench. In (a) the moment arm L is longest, hence the
torque τ = FL is a maximum. In (d) the line of action of F passes through the pivot O,so L = 0 and τ = 0.

The torque exerted by a force is also known as the moment of the force. A force whose line of action passes
through O produces no torque about O because its moment arm is zero.
Torque plays the same role in rotational motion that force plays in linear motion. A net force F acting on a
body of mass m causes it to undergo the linear acceleration a in accordance with Newton’s second law of motion
F = ma. Similarly a net torque τ acting on a body of moment of inertia I causes it to undergo the angular
2
acceleration α (in rad/s ) in accordance with the formula
τ = Iα
Torque = (moment of inertia)(angular acceleration)
In the SI system, the unit of torque is the newton-meter (N·m); in the British system, it is the pound-foot
(lb·ft).

SOLVED PROBLEM 10.9

The radius of gyration of a body about a particular axis is the distance from that axis to a point at which
the body’s entire mass may be considered to be concentrated. Thus the moment of inertia of a body of
2
mass M and radius of gyration k is I = Mk . (a) The radius of gyration of a hollow sphere of radius R and

2
mass M is k = R. What is its moment of inertia? (b) Find the radius of gyration of a solid sphere.
3
2 2
2
2
(a) I = Mk = M R = MR 2
3 3
2
2
(b) From Fig. 10-3 the moment of inertia of a solid sphere is I = MR . Hence its radius of gyration is
5

2
I 2 MR 2
k = = = R
M 5 M 5
SOLVED PROBLEM 10.10
The radius of gyration of a 200-lb flywheel is 1 ft. Find its moment of inertia.
2
The mass of the flywheel is M = w/g = 200 lb/(32 ft/s ) = 6.25 slugs, and so its moment of inertia is
2
2
I = Mk = (6.25 slugs)(1ft) = 6.25 slug·ft 2