Question

light string is wound around a uniform cylinder of mass m and radius r and mass m is supported from the free end of the string after the mass m is released its acceleration will be​

Answer 1

Explanation:

a=Rα      

Also, τ=Iα  and  τ=TR

Using the relations,  ⟹T=R2Ia

For hollow cylinder,  I=mR2⟹T=ma

Now for block,   ma=mg−T

ma=mg−ma

⟹a=2g

Answer 2

Given :

▪ Mass of cylinder = m

▪ Radius of cylinder = r

▪ Mass of block = m

To Find :

▪ Acceleration of the block of mass m.

Diagram :

✏ Please, see the attachment for better understanding.

Thinking Process :

→ Torque($\tau$) acting on a body and angular acceleration produced in it are related as

$\bigstar\:\underline{\boxed{\bf{\pink{\tau=I\alpha}}}}$

Where, I = moment of inertia of the cylinder about the axis through the centre.

→ Consider a solid cylinder, around which a rope is wounded as shown in the attachment.

→ Torque acting on the cylinder due to the force F is

$\bigstar\:\underline{\boxed{\bf{\green{\tau=rF}}}}$

Calculation :

$\dashrightarrow\sf\:rF=I\alpha\\ \\ \dashrightarrow\sf\:r(mg)={\huge(}\dfrac{mr^2}{2}{\huge)}{\huge{(}}\dfrac{a}{r}{\huge{)}}\\ \\ \dashrightarrow\sf\:2mg=ma\\ \\ \dashrightarrow\underline{\boxed{\bf{\purple{a=2g}}}}\:\red{\bigstar}$