As shown in the figure, a slider of mass $m$ is propelled upward at A along a fixed quarter section

Question

As shown in the figure, a slider of mass $m$ is propelled upward at A along a fixed quarter section of a ring of radius $R$ which lies in a vertical plane. The slider is observed to have a speed $u$ as it passes position B. The line OB makes an angle $	heta$ with the vertical, where O is the center of the ring section. Assume that friction is absent everywhere. Which of the following statements is/are correct?

Professor PC Thomas Classes · Answer

Let O be the origin $(0,0)$. The slider moves from A $(-R, 0)$ to the highest point $(0, R)$. The position B is at $(-R\sin	heta, R\cos	heta)$ as $	heta$ is the angle with the vertical. The tangential direction (direction of velocity) is $\hat{t} = \cos	heta\hat{i} + \sin	heta\hat{j}$. The force of gravity is $\vec{F}_g = -mg\hat{j}$. The tangential component of gravity is $\vec{F}_g \cdot \hat{t} = -mg\sin	heta$. Thus, the tangential acceleration is $-g\sin	heta$, and the rate of decrease of speed is $g\sin	heta$. The inward radial direction is $\hat{n} = \sin	heta\hat{i} - \cos	heta\hat{j}$. The inward component of gravity is $\vec{F}_g \cdot \hat{n} = mg\cos	heta$. By Newton's second law in the radial direction, $N_{inward} + mg\cos	heta = \frac{mu^2}{R} \implies N_{inward} = \frac{mu^2}{R} - mg\cos	heta$. The magnitude is $|N_{inward}|$. If $u > \sqrt{gR\cos	heta}$, then $N_{inward} > 0$, meaning the ring pushes towards O. To reach the top (height $R$ above A), $v_{top} \ge 0$. By energy conservation, $\frac{1}{2}mv_A^2 = mgR + \frac{1}{2}mv_{top}^2 \implies v_A \ge \sqrt{2gR}$.