SATHEE: Rotational Motion Question 36

Rotational Motion Question 36

Question: A ring of mass M and radius R is rotating about its axis with angular velocity $ω$ . Two identical bodies each of mass m are now gently attached at the two ends of a diameter of the ring. Because of this, the kinetic energy loss will be:

[JEE ONLINE 25-04-2013]

Options:

A) $\frac{m (M + 2 m)}{M} ω^{2} R^{2}$

B) $\frac{M m}{(M + m)} ω^{2} R^{2}$

C) $\frac{M m}{(M + 2 m)} ω^{2} R^{2}$

D) $\frac{(M + m) M}{(M + 2 m)} ω^{2} R^{2}$

Show Answer

Answer:

Correct Answer: C

Solution:

[c] Kinetic energy Kinetic energy Solving we get loss in K.E.

By using angular momentum conservation

$M R^{2} \times ω = (M R^{2} + M R^{2} + M R^{2}) ω'$

$ω' = \frac{M ω}{M + 2 m}$

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Rotational Motion Question 36

Question: A ring of mass M and radius R is rotating about its axis with angular velocity ω . Two identical bodies each of mass m are now gently attached at the two ends of a diameter of the ring. Because of this, the kinetic energy loss will be:

Options:

Answer:

Solution:

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Question: A ring of mass M and radius R is rotating about its axis with angular velocity $ω$ . Two identical bodies each of mass m are now gently attached at the two ends of a diameter of the ring. Because of this, the kinetic energy loss will be: