Question

A chain of mass m and length ($l_1 + l_2$) are placed on smooth hemisphere of radius R, as shown in figure. Find velocity of the chain after it completely slip on to the horizontal surface :

• A. $\sqrt{\frac{2gR^2}{l_1 + l_2}}[\sin{\frac{l_1}{R}} + \sin{\frac{l_2}{R}}]^{1/2}$
• B. $\sqrt{\frac{2gR^2}{l_1 + l_2}}$
• C. $\sqrt{\frac{gR^2}{l_1 + l_2}}\sin{\frac{l_1 + l_2}{R}}]^{1/2}$
• D. $\sqrt{\frac{gR^2}{l_1 + l_2}}[\cos{\frac{l_1}{R}} - \cos{\frac{l_1 + l_2}{R}}]^{1/2}$

Answer 1

Answer: (A)

$\sqrt{\frac{2gR^2}{l_1 + l_2}}[\sin{\frac{l_1}{R}} + \sin{\frac{l_2}{R}}]^{1/2}$

Answer 2

Concept: Conservation of Mechanical Energy. The smooth surface implies no energy loss due to friction.

1. Initial State: The chain is initially at rest, so its initial kinetic energy ($K_i$) is zero. $K_i = 0$

To find the initial potential energy ($U_i$), consider a small element of the chain of mass $dm$ at an angle $\theta$ from the vertical. The linear mass density of the chain is $\lambda = \frac{m}{l_1 + l_2}$. An element of length $dl = R d\theta$ has mass $dm = \lambda dl = \frac{m}{l_1 + l_2} R d\theta$. The height of this element above the horizontal surface (chosen as the reference for potential energy, $h=0$) is $h = R \cos\theta$. The potential energy of this element is $dU = dm \cdot g \cdot h = \left(\frac{m}{l_1 + l_2} R d\theta\right) g (R \cos\theta) = \frac{m g R^2}{l_1 + l_2} \cos\theta d\theta$.

The chain spans angular lengths corresponding to $l_1$ and $l_2$. Let $\theta_1 = l_1/R$ and $\theta_2 = l_2/R$. The chain extends from angle $-\theta_2$ to $\theta_1$ with respect to the vertical axis. Integrating $dU$ over the length of the chain: $U_i = \int_{-l_2/R}^{l_1/R} \frac{m g R^2}{l_1 + l_2} \cos\theta d\theta = \frac{m g R^2}{l_1 + l_2} [\sin\theta]_{-l_2/R}^{l_1/R}$ $U_i = \frac{m g R^2}{l_1 + l_2} (\sin(l_1/R) - \sin(-l_2/R))$ Since $\sin(-x) = -\sin(x)$, we get: $U_i = \frac{m g R^2}{l_1 + l_2} (\sin(l_1/R) + \sin(l_2/R))$

2. Final State: When the chain completely slips onto the horizontal surface, its entire length is at the reference height, so its final potential energy ($U_f$) is zero. $U_f = 0$ The chain moves with a uniform velocity $v$. Its final kinetic energy ($K_f$) is: $K_f = \frac{1}{2} m v^2$

3. Conservation of Mechanical Energy: According to the principle of conservation of mechanical energy: $K_i + U_i = K_f + U_f$ $0 + \frac{m g R^2}{l_1 + l_2} (\sin(l_1/R) + \sin(l_2/R)) = \frac{1}{2} m v^2 + 0$

Cancel $m$ from both sides: $\frac{g R^2}{l_1 + l_2} (\sin(l_1/R) + \sin(l_2/R)) = \frac{1}{2} v^2$ $v^2 = \frac{2 g R^2}{l_1 + l_2} (\sin(l_1/R) + \sin(l_2/R))$ $v = \sqrt{\frac{2 g R^2}{l_1 + l_2} (\sin(l_1/R) + \sin(l_2/R))}$