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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