Question 6.13: Motion in a vertical circle A fixed hollow sphere has centre...
Motion in a vertical circle
A fixed hollow sphere has centre O and a smooth inner surface of radius b. A particle P, which is inside the sphere, is projected horizontally with speed u from the lowest interior point (see Figure 6.5). Show that, in the subsequent motion,
v^{2}=u^{2}-2 g b(1-\cos \theta),
provided that P remains in contact with the sphere.
While P remains in contact with the sphere, the motion is as shown in Figure 6.5.
The forces acting on P are uniform gravity mg and the constraint force N, which is the normal reaction of the smooth sphere. Since N is always perpendicular to v (the circumferential velocity of P), it follows that N does no work. Hence energy conservation applies in the form
\frac{1}{2} m v^{2}-m g b \cos \theta=E,
where m is the mass of P, and the zero level of the potential energy is the horizontal plane through O. Since v = u when θ = 0, it follows that E=\frac{1}{2} m u^{2}-m g b and the energy conservation equation becomes
v^{2}=u^{2}-2 g b(1-\cos \theta) (6.17)
as required. This gives the value of v as a function of θ while P remains in contact with the sphere.