Basic Kinematics - Physics for Game Programmers - page 70

The angular acceleration will have units of rad/s 2 . The tangential velocity, v r , of any point
on the disk is equal to the angular velocity, w , multiplied by the distance, r , from the origin to
the point.

v r = wr (4.30)

The tangential velocity will have units of m/s and will be in a direction perpendicular to the
line drawn from the origin to the point.
A rotating object will have an axis of rotation , a line about which the object is spinning. In
Figure 4-7, the axis of rotation is the y-axis. The axis of rotation won't always conveniently line
up along one of the coordinate axes, but in general it can be defined by any vector in three-
dimensional space.

Torque
According to Newton's second law, a net external force on an object causes a translational
acceleration of the object. The rotational counterpart of force is torque , which causes a rota-
tional acceleration of an object. Torque is defined as a force applied to an object at a certain
distance from an axis of rotation.

t = Fr

(4.31)

Torque has units of N-m . The distance, r , from the center of rotation to the applied force is
called the moment arm of the torque. To get a feeling for torque, consider the situation shown
in Figure 4-8, where a wrench is applied to a nut. If a force of 10 N is applied at a distance of 0.12 m
from the center of the nut, a torque of 1.2 N-m will be applied to the center of the nut.

F = 10 N

τ = 1.2 N-m

r = 0.12 m

Figure 4-8. Force applied at a distance from an axis of rotation generates a torque.

When computing torque, it is the force component perpendicular to the moment arm that
is used in the calculation. In Figure 4-8 the force vector and moment arm are perpendicular to
each other, so the full force value of 10 N is used in the calculation. If the wrench were at an
angle of 30 degrees, as shown in Figure 4-9, and the force was still applied vertically, the force
vector and moment arm would no longer be perpendicular to each other. The perpendicular
force component, F p = F cos 30 , would be used in the torque calculation so the torque applied to
the nut would be equal to Fr cos 30 = 1.04 N-m .