Rotation 3 Question 11

17. 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 tabout an

axis normal to the plane of the paper passing through the point $Q$. Which of the following options is/are correct? (2017 Adv.)

(a) If the force is applied normal to the circumference at point $P$, then $\tau$ is zero

(b) If the force is applied tangentially at point $S$, then $\tau \neq 0$ but the wheel never climbs the step

(c) If the force is applied at point $P$ tangentially, then $\tau$ decreases continuously as the wheel climbs

(d) If the force is applied normal to the circumference at point $X$, then $\tau$ is constant

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Answer:

Correct Answer: 17. (a, c)

Solution:

(a) If force is applied normal to surface at $P$, then line of action of force will pass from $Q$ and thus, $\tau=0$.

(b) Wheel can climb.

(c) $\tau=F(2 R \cos \theta)-m g R \cos \theta, \tau \propto \cos \theta$

Hence, as $\theta$ increases

$\tau$ decreases. So its correct.

(d)

$\tau=F r _{\perp}-m g \cos \theta ; \tau$ increases with $\theta$



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