IIT-JEE 2011 · Paper 1 · Physics Q40

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PhysicsInteger+4Question 40 of 69

A boy is pushing a ring of mass 2 kg and radius 0.5 m with a stick as shown in the figure. The stick applies a force of 2 N on the ring and rolls it without slipping with an acceleration of 0.3 m/s². The coefficient of friction between the ground and the ring is large enough that rolling always occurs and the coefficient of friction between the stick and the ring is (P/10). The value of P is

A ring rolling on the ground with a vertical stick pressed against its right side at the height of its centre.

Your answer

Answer and worked solution

Official answer: 4

Forces on the ring. The stick pushes with $N=2$ N horizontally and exerts a vertical friction $F$ (upward, opposing the downward slipping of the ring's surface past it). Ground friction $f$ acts backward.

Translation. $2-f=ma=0.6$ , so $f=1.4$ N.

Rotation about the centre ( $I=mR^2$ , $\alpha=a/R$ ): $(f-F)R=mR^2\frac aR$ , so $f-F=0.6$ and $F=0.8$ N.

Coefficient. $\mu=\dfrac FN=0.4=\dfrac P{10}$ , so $P=4$ .

Answer: 4

Topics:Rolling motion Friction

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IIT-JEE 2011 · Paper 1, Physics question 40