Question

In the figure, the disc does not slip on the surface S. The pulley P has mass, and string does not slip on it. The string is wound around the disc and it is connected to a block B. Select the correct statement(s)

• A. The acceleration of the block B is double the acceleration of the centre of disc.
• B. The force of friction exerted by disc on S acts to the left
• C. The horizontal and the vertical sections of the string have the same tension
• D. The sum of the kinetic energies of disc and B is less than the loss in the potential energy of B as it moves down.

Answer 1

Answer: (A), (B), (D)

The correct options are
A The acceleration of the block B is double the acceleration of the centre of disc.
B The force of friction exerted by disc on S acts to the left
D The sum of the kinetic energies of disc and B is less than the loss in the potential energy of B as it moves down.

Let there be no friction.
$Ma=T_1\Rightarrow a=\frac{T}{M}$
Also ${\tau }_0=I\alpha$
$T_{1}R=\frac{1}{2}M{R}^2\alpha \Rightarrow \alpha R=\frac{2T_1}{M}$
Acceleration of point $A,a_A=\alpha R-a=\frac{T_1}{M}$ towards left
Thus point A tends to slide in left, hence point A will experience a friction force in right direction and in reaction apply equal friction force on S towards left.
Let COM moves with acceleration a and rotates with an angular acceleration $\alpha .$ As there is no slipping at A and D. i.e $a_A=0$ and $a_D=a^{\prime }$
$\Rightarrow a=R\alpha$
$a+R\alpha =a^{\prime }$
$a+a=a^{\prime }\Rightarrow a^{\prime }=2a$
Also the rotating pulley has some K.E
Work energy theorem : $mgh=K.E_D+K.E_B+K.E_{pulley}$
Hence, $K.E_D+K.E_B<$ Loss in $P.E_B$