Question 7.13: A propulsive force P of constant magnitude moves a slider S ...
A propulsive force P of constant magnitude moves a slider S of mass m in a smooth circular track in the vertical plane, as shown in Fig. 7.12. The slider starts from rest at the horizontal position A . Determine the speed of S as a function of θ. What is its angular speed after n complete turns?
The total force that acts on S in the Fig. 7.12 consists of the workless normal reaction force N exerted by the smooth tube, the conservative gravitational force F_C=mg, and the nonconservative propulsive force F_N=P=P_t which always is tangent to the path of S. The change in the potential energy of S is \Delta{V}=mg\sin{\theta}, the datum being at A, and the change in the kinetic energy from the initial rest position at A is \Delta{K}=\frac{1}{2}mv^{2}. Therefore, with \Delta {\mathscr{E}}=\Delta{K} + \Delta{V} and F_N . dx = Pds, the general energy principle (7.80) yields
\Delta{\mathscr{E}}=\mathscr{W}_N (7.80)
\frac{1}{2} mv^{2} + mgR \sin{\theta}=P\int_{0}^{R \theta} {ds=PR\theta}, (7.81a)
and hence the speed of S as a function of θ is given by
v(\theta)=\sqrt{\frac{2R}{m}(P \theta – mg \sin{\theta})}. (7.81b)
The angular speed of S is determined by v=R\omega. Thus, after n complete revolutions, θ = 2πn and (7.8I b) provides the angular speed
\omega (n)=\sqrt{\frac{4n \pi P}{mR}}. (7.8Ic)