Question

In the figure, the disc D does not slip on the surface S. The pulley P has mass, and the string does not slip on it. The string is wound around the disc. Then, which of the following statement(s) is/are true ?

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

Let there is no friction.

$Ma=T_1⟹a=\frac{T}{M}$

Also ${\tau }_O=I\alpha$

$T_{1}R=\frac{1}{2}M{R}^2\alpha ⟹\alpha R=\frac{2T_1}{M}$

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 $a_D=a^{\prime }$

$⟹a=R\alpha$

$a+R\alpha =a^{\prime }$

$a+a=a^{\prime }⟹a^{\prime }=2a$

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 $P.E_B$