JEE Advanced 2017 · Paper 2 · Physics Q14

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PhysicsMultiple correct+4 / -2Question 14 of 54

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?

A wheel of radius R resting on the floor against a step of height R; the step's top corner Q touches the wheel at the level of its centre. P is the opposite end of the horizontal diameter through Q, S is the topmost point and X is a point on the lower-left of the rim.

If the force is applied at point

P

tangentially then

\tau

decreases continuously as the wheel climbs

If the force is applied normal to the circumference at point

X

then

\tau

is constant

If the force is applied normal to the circumference at point

P

then

\tau

is zero

If the force is applied tangentially at point

S

then

\tau \ne 0

but the wheel never climbs the step

Answer and worked solution

Official answer: C

Take torques about the step corner $Q$ , which lies at the height of the wheel's centre.

(C) A force normal to the rim at $P$ acts along the diameter $PQ$ , so it passes through $Q$ . Then $\tau=0$ . True.
(D) A tangential force at the top point $S$ has a non-zero moment about $Q$ . The key also accepts (D)'s claim that the wheel cannot climb, given the constraint of a constant force applied there.
(A), (B) do not hold as stated for a force of fixed direction as the wheel rotates. False.

Answer: (C) (the key accepts C, or C and D)

Topics:Torque and rotational dynamics Equilibrium of rigid bodies

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JEE Advanced 2017 · Paper 2, Physics question 14