Question 7.10: A particle of mass m is released from rest at A and slides d...
A particle of mass m is released from rest at A and slides down a smooth circular track of radius R shown in Fig. 7.9. At the lowest point B, the particle is projected horizontally and continues its motion until it strikes the ground at C. Determine the horizontal component of the particle’s velocity at C.
The free body diagram in Fig. 7.9 on the path from B to C shows that no horizontal forces act on the particle. Therefore, the linear momentum in the fixed horizontal direction i is conserved: p . i = m \dot{x}=\gamma, a constant. Consequently, on the entire path BC, specifically at C, the horizontal component of the particle’s velocity v . i = \dot{x} is a constant whose value is determined by its speed at point B. This value may be found by application of the work–energy principle.
The free body diagram of the particle on the circular path AB is shown in Fig. 7.9. The normal force N is workless on AB, while gravity does work \mathscr{W}_g=mgR in reaching B. Hence, the total work done by the forces acting on m is \mathscr{W}=mgR. The increase in the kinetic energy as the particle slides from rest at A to the end state at B is \Delta{K}=\frac{1}{2}mv_{B}^{2}. The work–energy principle \mathscr{W}=\Delta{K} determines the speed at B, and hence the horizontal component of the particle’s velocity at C is given by
\dot{x}=v_B=\sqrt{2gR}.