Rotational Mechanics

First we have to draw FBDs.

Figure Applying newton's second law on cylinder

So $(T+F)=ma$ ..........................(1)

Let the angular acceleration of pulley be $\alpha$ in clockwise direction. Let clockwise direction be positive

So $(T-F)r=I\alpha$

Here $I=\frac { m{ r }^{ 2 } }{ 2 }$

So $\alpha=\frac{2(T-F)}{mr}$ ........(2)

Now we will get other equation be equating accelerations at the point of contact of cylinder and string.

Let the point 1 be on the string and point 2 be on the pulley. As the string is not slipping so acceleration of point 1 and 2 are equal.

Also acceleration of point 1 is equal to the acceleration of object.

Let $a_{1}$ be the acceleration of the object.

So $a_{1}=\frac{F}{m}$ .........................(3)

NOTE- The acceleration of the object will be towards left.

Now point 2 has two accelerations

1) Due to translation which is equal to $a$ towards right

2) Due to rotation which is equal to $r\alpha$ towards left.

Let us assume right side to be positive and left side to be negative.

Then $-a_{1}=a-r\alpha$

So $\frac{F}{m}=\frac{2(T-F)}{mr}-\frac{T+F}{m}$

So $F=\frac{T}{4}$ ..............(4)

Now from eq(3)

$a_{1} = -\frac{F}{m}$

From eq(4)

$a_{1} = -\frac{T}{4m}$