7 CIRCULAR MOTION 7.6 The vertical pendulumv’
T
B
rr cos mg v (^) A
mg cos
mg
Figure 64: Motion in a vertical circle.
7.6 The vertical pendulum
Let us now examine an example of non-uniform circular motion. Suppose that
an object of mass m is attached to the end of a light rigid rod, or light string, of
length r. The other end of the rod, or string, is attached to a stationary pivot in
such a manner that the object is free to execute a vertical circle about this pivot.
Let θ measure the angular position of the object, measured with respect to the
downward vertical. Let v be the velocity of the object at θ = 0◦. How large do we
have to make v in order for the object to execute a complete vertical circle?
Consider Fig. 64. Suppose that the object moves from point A, where its
tangential velocity is v, to point B, where its tangential velocity is v J. Let us,
first of all, obtain the relationship between v and vJ. This is most easily achieved
by considering energy conservation. At point A, the object is situated a vertical
distance r below the pivot, whereas at point B the vertical distance below the
pivot has been reduced to r cos θ. Hence, in moving from A to B the object gains
potential energy m g r (1 − cos θ). This gain in potential energy must be offset by
a corresponding loss in kinetic energy. Thus,
1
m v^2 −
1
m vJ^2 = m g r (1 − cos θ), (7.46)
2 2