A chain having mass m and length l rests on the surface of a smooth...

Let the linear mass density of the chain be $\lambda = \dfrac{m}{l}$.

Consider an element of the chain of length $dl = R d\alpha$ at an angle $\alpha$ from the vertical.

The mass of this element is $dm = \lambda R d\alpha$.

Taking the center of the sphere as the reference level, the potential energy of this element is $dU = (dm)gh = (\lambda R d\alpha)g(R \cos\alpha) = \lambda R^2 g \cos\alpha d\alpha$.

Initially, the chain lies between $\alpha = 0 and$\alpha = \dfrac{l}{R}$.<br><br>The initial potential energy of the chain is:<br><br>$U_i = \int_0^{l/R} \lambda R^2 g \cos\alpha d\alpha = \lambda R^2 g [\sin\alpha]_0^{l/R} = \lambda R^2 g \sin \left(\dfrac{l}{R}\right)

When the chain has slid through an angle \theta, it lies between \alpha = \theta and \alpha = \theta + \dfrac{l}{R}.

The final potential energy of the chain is:

$U_f = \int_\theta^{\theta + l/R} \lambda$R^2 g \cos$\alpha$d$\alpha = \lambda$R^2 g [\sin$\alpha]_\theta^{\theta + l/R} = \lambda$R^2 g \left[\sin \left(\theta + \dfrac{l}{R}\right) - \sin\theta\right]

By the principle of conservation of mechanical energy, the gain in kinetic energy equals the loss in potential energy:

$K = U_i - U_f = \lambda R² g \sin \left(\dfrac{l}{R}\right) - \lambda R^2 g \left[\sin\left(\theta + \dfrac{l}{R}\right) - \sin\theta\right]$

$K = \lambda R^2 g \left[\sin \left(\dfrac{l}{R}\right) + \sin\theta - \sin\left(\theta + \dfrac{l}{R}\right)\right]<br><br>Substituting$\lambda = $\dfrac{m}{l}, we get:<br><br>$K = $\dfrac{mR^2 g}{l} \left[\sin \left(\dfrac{l}{R}\right) + \sin\theta - \sin\left(\theta + \dfrac{l}{R}\right)\right]$