Question

A thin circular ring of mass m and radius R is rotating about its axis with a constant angular velocity ω . two objects each of mass M are attached gently to the opposite ends of diameter of the ring . the ring is Now rotates with an angular velocity ω' is =? [AIEEE 2006]​

Answer 1

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Answer 2

Answer :

Mass of ring = m

Radius of ring = R

Angular velocity of ring = ω

Mass of each object = M

We have to find final angular velocity of ring$.$

★ Since no external torque acts on the system, angular momentum of system is conserved.

• L = I ω

→ L denotes angular momentum

→ I denotes moment of inertia

→ ω denotes angular velocity

★ Moment of inertia of ring of mass m and radius R rotating about axis passing through its centre and perpendicular to the plane of ring is given by

• I = mR²

➙ L (initial) = L (final)

➙ I ω = I' ω'

where I' = I + I｡

• I｡denotes moment of inertia of two particles.
• I｡ = MR²+MR² = 2MR²

➙ mR²ω = (mR² + 2MR²) ω'

➙ mω = (m + 2M) ω'

➙ ω' = mω/(m + 2M)

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