A thin walled hollow spherical shell of mass m and radius r starts from rest and moves along atrack as shown in figure assume that shell successfully negotiates the loop. Statement-1: Normal reaction at A is greater when friction is absent on the track as compared to the case when sphere executes pure rolling. Statement-2 : Due to frictional torque shell have larger translational velocity at A in case of pure rolling.

Engineering

Physics

Conservation of Energy

Rolling

Circular Motion in Vertical Plane

Question

A thin walled hollow spherical shell of mass m and radius r starts from rest and moves along atrack as shown in figure assume that shell successfully negotiates the loop.

Statement-1: Normal reaction at A is greater when friction is absent on the track as compared to the case when sphere executes pure rolling.
Statement-2 : Due to frictional torque shell have larger translational velocity at A in case of pure rolling.

Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.

Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.

Statement-1 is false, statement-2 is true.

Statement-1 is true, statement-2 is false.

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We analyze a hollow spherical shell (moment of inertia $I = \frac{2}{3} m r^{2}$ ) negotiating a loop of radius R. At point A (top of the loop), the centripetal force is provided by weight and normal reaction: $N_{A} = \frac{m v^{2}}{R} - m g$ .

In pure rolling, some energy is used in rotation, resulting in a lower translational velocity $v$ at A compared to a frictionless case (where it only slides). Since $N_{A}$ is proportional to $v^{2}$ , the normal reaction is greater when friction is absent. Statement-2 correctly explains this, as the frictional torque in pure rolling causes the lower velocity.

Final Answer: Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.