JEE Advanced · 2017 · 21 May 2017 · Paper 2 · Rotational Mechanics

A wheel of radius $R$ and mass $M$ is placed at the bottom of a fixed step of height $R$ as shown in the figure. A constant force is continuously applied on the surface of the wheel so that it just climbs the step without slipping. Consider the torque $\tau$ about an axis normal to the plane of the paper passing through the point $Q$. Which of the following options is/are correct?

1. (A)If the force is applied at point $P$ tangentially then $\tau$ decreases continuously as the wheel climbs
2. (B)If the force is applied normal to the circumference at point $X$ then $\tau$ is constant
3. (C)If the force is applied normal to the circumference at point $P$ then $\tau$ is zero✓ Correct
4. (D)If the force is applied tangentially at point $S$ then $\tau \ne 0$ but the wheel never climbs the step

Answer: (C)

Solution

When force F is applied tangentially at P and the wheel climbs, the net torque about Q is $\tau =F(2R\cos \theta )-mgR\cos \theta =R\cos \theta (2F-mg)$. As $\theta$ increases during climbing, $\cos \theta$ decreases, so $\tau$ decreases. For option (C), if force is applied normal at P, friction at the contact point produces non-zero torque about Q. Options (B) and (D) are also incorrect based on detailed torque analysis.

JEE Advanced · 2017 · 21 May 2017 · Paper 2 · One or more correct options