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 a small element of the chain of length $ds = R d\theta$ at an angle $\theta$ from the vertical.

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

The tangential component of the gravitational force acting on this element is $dF_t = dm g \sin\theta = \lambda R g \sin\theta d\theta$.

The total tangential force acting on the chain is obtained by integrating $dF_t$ from $\theta = 0$ to $\theta = \dfrac{l}{R}$ (since the chain has length $l$ and one end is at the top of the sphere):

$F_t = \int_{0}^{l/R} \lambda R g \sin\theta d\theta$

$F_t = \lambda R g \left[-\cos\theta\right]_{0}^{l/R}<br><br>$F_t = \lambda R g \left[1 - \cos\left(\dfrac{l}{R}\right)\right]<br><br>Since the chain is inextensible, all its elements move with the same tangential accelerationa_t$. Using Newton's second law for the entire chain:<br><br>$F_t = m a_t$<br><br>Substituting$\lambda = $\dfrac{m}{l} into the force equation:<br><br>\dfrac{m}{l}$R g \left[1 - \cos\left(\dfrac{l}{R}\right)\right] =m a_t$<br><br>$a_t = \dfrac{Rg}{l}\left[1 - \cos \left(\dfrac{l}{R}\right)\right]