Question 762: Centre of Mass and Rotational Motion

Question

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 $	au$ about an axis normal to the plane of the paper passing through the point Q. Which of the following options is/are correct?

Accepted Answer

(a) If force is applied normal to surface at P, then line of action of force will pass from Q and thus $	au$ = 0.
(b) Wheel can climb.
(c) $	au$ = F(2Rcos$	heta$) $-$ mgRcos$	heta$
$ \Rightarrow $ $	au$ $\propto$ cos$	heta$
Hence, as $	heta$ increases, $	au$ decreases. So its correct.
(d)
$	au$ = Fr
$ \bot $
$-$ mgcos$	heta$ ; $	au$ increases with $	heta$.

Answer

If the force is applied normal to the circumference at point P, then $	au$ is zero

Answer

If the force is applied tangentially at point S, then $	au$ $
e$ 0 but the wheel never climbs the step

Answer

If the force is applied at point P tangentially, then $	au$ decreases continuously as the wheel climbs

Answer

If the force is applied normal to the circumference at point X, then $	au$ is constant

Centre of Mass and Rotational Motion

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