Physics - Rotation Question with Solution | TestHub

PhysicsRotationTorque / angular accelerationMedium2 minQB

PhysicsMediummultiple choice

A rod bent at right angle along its centre line, is placed on a rough horizontal fixed cylinder of radius R as shown in figure. Mass of rod is 2 m and rod is in equilibrium. Assume that friction force on rod at A and B are equal in magnitude.

Options:(select one or more)

Answer:

A, C

Solution:

$f_{A} = f_{B} N_{A} + f_{B} = 2 m g N_{B} = f_{A}$

Torque equation

$f_{A} R + f_{B} R = m g R \Rightarrow f_{A} + f_{B} = m g \Rightarrow f_{A} = f_{B} = \frac{m g}{2}$

$\Rightarrow N_{B} = \frac{m g}{2} N_{A} + \frac{m g}{2} = 2 m g$

$N_{A} = \frac{3 m g}{2}$

Stream:JEESubject:PhysicsTopic:RotationSubtopic:Torque / angular acceleration

⏱ 2mℹ️ Source: QB

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