KVPY2016PhysicsRotational Motion
Consider two masses with m ₁> m ₂ connected by a light inextensible string that passes over a pulley of radius R and moment of inertia I about its axis of rotation. The string does not slip on the pulley and the pulley turns without friction. The two masses are released from rest separated by a vertical distance 2 ~h . When the two masses pass each other, the speed of the masses is proportional to
Options
- Am₁-m₂ m₁+m₂+ 1 R²
- B(m₁+m₂ ) (m₁-m₂ ) m₁+m₂+ 1 R²
- Cm ₁+ m ₂+ 1 R ² ~m ₁- m ₂
- D1 R ² ~m ₁+ m ₂
Correct answer
C. m ₁+ m ₂+ 1 R ² ~m ₁- m ₂
Step-by-step solution
The total mechanical energy of system = conserved Hence array l KE _ i + PE _ i = KE _ f + PE _ f 0- m ₂ ~g 2 ~h = 1 2 ~m ₂ v ²+ 1 2 ~m ₁ v ²+ 1 2 I ²- m ₁ gh - m ₂ gh array Also = v R array l ( m ₁- m ₂ ) gh = 1 2 ( ~m ₁+ m ₂ ) v ²+ 1 2 I ( v R )² ( ~m ₁- m ₂ ) gh = v ² 2 [ ~m ₁+ m ₂+ I R ² ] v m ₁- m ₂ ~m ₁+ m ₂+ I R ² array