Question

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Answer 1

Answer:

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Answer 2

BEFORE

Let the moment of inertia of the disc is

I = mR²

angular velocity of the disc= w

angular momentum (L) is written as

L=I ×w

L = mR² w

AFTER

Now, as 2 masses are attached to the ends of the

circular ring the moment of inertia becomes

I ' =mR² + MR² +MR²

I ' = mR² + 2 M R ²

I ' = (m+2M)R²

angular velocity = w/2

angular momentum

L ' = I ' × w /2

L ' = (m+2M)R² w/2

By conservation of angular momentum

L = L '

mR² w = (m+2M)R² w/2

m = m/2 +2M/2

m= m/2 + M

m/2m + M/ m = 1

1/2 + M/m = 1

M/m = 1 - 1/2

M /m = 2- 1 /2

M /m = 1/ 2

m/M = 2