Question

A drum of radius $R$ and mass $M$ rolls down without slippping along an inclined plane of inclination $\theta$. Then identify the correct statement

• A. Frictional force acting on drum will dissipate energy as heat.
• B. Frictional force acting on drum will oppose the rotational motion of the drum.
• C. Frictional force acting on drum will oppose the translational as well as the rotational motion of drum.
• D. Gravitational Potential energy of drum is $100\%$ transformed into translational and rotational $K.E$ of drum.

Answer 1

The correct option is D Gravitational Potential energy of drum is $100\%$ transformed into translational and rotational $K.E$ of drum.


$\Rightarrow$ The work done by frictional force will be $\text{zero}$, $W_{f_s=0}$ since the point of contact $A$ is having zero displacement along the inclined plane (due to zero velocity).
So there is no dissipation of energy, hence total mechanical energy remains conserved.
$\therefore \text{Loss in P.E}=\text{Gain in}{K.E}_{\text{Trans}}+\text{Gain in}{K.E}_{\text{Rot}}$

$\Rightarrow$Since $\text{COM}$ of drum is moving down the incline, with velocity $v$. Hence static friction $f_s$ will be directed upwards along the inclined plane, such that it opposes relative motion between point of contact of drum $\&$ inclined plane.


Simultaneously, due to torque of $f_s$ in anticlockwise direction about $COM$ of drum, it will increase the angular velocity $\omega$.

$\Rightarrow$ Due to reduction of translational velocity $v$ and increase in $\omega$, the pure rolling condition
$v=R\omega$ will be achieved.
$\therefore$Only option $\left(d\right)$ is correct.