Question

A chain of length l and mass m lies on the surface of a smooth sphere of radius R (R > l) with one end tied to the top of the sphere. The tangential acceleration of the chain when the chain starts sliding down is.

• A. $\frac{2Rg}{l}[2-cos\frac{l}{R}]$
• B. $\frac{Rg}{l}[2-cos\frac{l}{R}]$
• C. $\frac{Rg}{l}[1-cos\frac{l}{R}]$
• D. $\frac{Rg}{2l}[1-cos\frac{l}{R}]$

Answer 1

The correct option is C $\frac{Rg}{l}[1-cos\frac{l}{R}]$



Consider a small element of mass 'dm' as an angle $\theta$.

Torque acting on this small element due to force of gravity about centre O is given by $d\tau =\left(dm\right)fsin\theta \times R$

Nrt tique, $\tau ={\int }_0^{\frac{l}{R}}d\tau$

$dm=\frac{m}{l}Rd\theta$

$\tau ={\int }_0^{\frac{l}{R}}\frac{m}{l}Rd\theta Rgsin\theta =\frac{m}{l}{R}^{2}g{\int }_0^{\frac{l}{R}}sinthetad\theta$

$=-\frac{m}{l}{R}^{2}g[cos\frac{l}{R}-1]$

$\tau =I\alpha$ .........(i)

$I=m{R}^2$ .........(ii)

$\alpha =\frac{a_t}{R}$ ...........(iii)

(I is the moment of inertia of entire chain about O, $\alpha$ is angular acceleration and $a_t$ is tangential acceleration.)

From (i), (ii), and (iii)

$-\frac{m}{l}{R}^{2}g[cos\frac{l}{R}-1]=m{R}^2\times \frac{a_t}{R}$

$\therefore$ $a_1=\frac{Rg}{l}[1-cos\frac{l}{R}]$