Rotational Motion Question 39

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 will now rotate with an angular velocity:

[AIPMT 1998]

Options:

A) ω(M2m)(M+2m)

B) ωM(M+2m)

C) ωM(M+m)

D) ω(M+2m)M

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Answer:

Correct Answer: B

Solution:

Angular momentum remains conserved in the universe.

According to conservation of angular momentum

L= constant

or lω=constant

l1ω1=l2ω2

Initial moment of inertia

l1=MR2

and angular velocity

ω1=ω

Hence, l1ω1=MR2ω

When two objects of mass m are attached to opposite ends of a diameter, the final readings are

l2=MR2+mR2+mR2

=(M+2m)R2

So, l2ω2=(M+2m)R2ω2 ….(iii)

From Eqs.(i), (ii) and (iii)

MR2ω=(M+2m)R2ω2

ω2=ωMM+2m



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