Question

In the figure shown, a thin light inextensible string is wrapped aroung a uniform disc. One end of the string is fixed to the ceiling of the lift and the other end is fixed to the circumference of the disc. Acceleration ${}^{\prime }{a}^{\prime }$ of the lift is such that the centre of the disc does not move with respect to the ground. Find the value of $\frac{a}{g}$. (Assume the string does not slip on the pulley)

• A. $2$
• B. $3$
• C. $4$
• D. $6$

Answer 1

The correct option is A $2$


According to condition, $a_{\text{C.O.M}}$ is zero in ground frame.
$\therefore T-mg=m{a}_{\text{C.O.M}}=0$
$\Rightarrow T=mg$
Lift is moving with acceleration $a{\text{m/s}}^2$ in upward direction.
$\therefore$ Due to string constraint, acceleration of point $P$ is also $a{\text{m/s}}^2$
We know,
Acceleration of point $P$
$a_P=a_{\text{C.O.M}}+\alpha \times R$
$a=0+\alpha R$
$\Rightarrow \alpha =\frac{a}{R}$
Torque about point $\text{O}$
$\tau =I\alpha$
$T\times R=\frac{m{R}^2}{2}\times \left(\frac{a}{R}\right)$
$mg\times R=\frac{m{R}^2}{2}\times \left(\frac{a}{R}\right)$
$\Rightarrow \frac{a}{g}=2$