Question

A string is wrapped over the edge of a uniform
disc and the free end is fixed with the ceiling.
The disc moves down, unwinding the string. Find
the downward acceleration of the disc.​

Answer 1

hope this answer might help you

Answer 2

A string is wrapped over the edge of a uniform disc and the free end is fixed with the ceiling. The disc moves down, unwinding the string, The downward acceleration of the disc is $2 g / 3$.

Explanation:

• As shown in the image above, the disc is connected to the string with tension T and its acceleration is taken as ‘a’.

         $\text { Force } \rightarrow m g-T=m a$

• When the disc is rotating, there will be an angular acceleration, which can be taken as α.  
• Unwinding point on the disc is r α in the upper direction, and ‘a’ in downward direction.
• Rotation of the body and the string’s rotation happen at the same time, which means there is no acceleration, which means downward ‘a’ and upward r α gets cancelled.
• This is the reason why the unwinding point doesn’t slip.

        $\mathrm{So}, T \times R=1 \alpha$

       $T R=\left(\frac{1}{2}\right) m R^{2} \times \frac{a}{R}$

       $T=\left(\frac{1}{2}\right) m a$

      When this value is added to the equation of force, we get

      $M g-\left(\frac{1}{2}\right) m a=m a$

      $M g=\frac{3}{2} m a \rightarrow 2=\frac{2 g}{3}$.