Question

Round disc of moment of inertia i2 about its axis perpendicular to its plane and passing through its centre is placed over another disc of moment of inertia i want rotating with an angular velocity omega about the same access the final angular velocity is a combination of discuss

Answer 1

The angular momentum is ω' = L1 ω/ I1 + I2

Explanation:

Angular momentum of a disc of moment of inertia I1 and rotating about its axis with angular velocity is

L1=I1ω

When a round disc of moment of inertia I2 is placed of first disc, then angular momentum of the combination is

L2=(I1+I2)ω'

Angular momentum remains conserved, in the absence of any external force.

L1 = L2

I1ω=(I1+I2)ω'

ω' = L1 ω/ I1 + I2

Thue the angular momentum is ω' = L1 ω/ I1 + I2

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Answer 2

U will easily get it

Explanation: