SATHEE: Rotational Motion Question 50

Rotational Motion Question 50

Question: A thin circular ring of mass M and radius r is rotating about its axis with a constant angular velocity $ω$ . Four objects each of mass m, are kept gently to the opposite ends of two perpendicular diameters of the ring. The angular velocity of the ring will be:

[AIPMT 2003]

Options:

A) $\frac{(M + 4 m) ω}{M}$

B) $\frac{(M - 4 m) ω}{M + 4 m}$

C) $\frac{M ω}{4 m}$

D) $\frac{M ω}{M + 4 m}$

Show Answer

Answer:

Correct Answer: D

Solution:

Key Idea: If external torque acting on the system is zero, hence angular momentum remains conserved.

$τ_{e x t} = 0$

or $\frac{d L}{d t} = 0$

or $L =$ constant

or $I ω =$ constant

$∴$ $I_{1} ω_{1} = I_{2} ω_{2}$ (i)

Here, $I_{1} = M r^{2}, ω_{1} = ω, I_{2} = M r^{2} + 4 m r^{2}$

Hence, Eq.(i) can be written as $M r^{2} ω = (M r^{2} + 4 m r^{2}) ω_{2}$

$∴$ $ω_{2} = \frac{M ω}{M + 4 m}$

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Rotational Motion Question 50

Question: A thin circular ring of mass M and radius r is rotating about its axis with a constant angular velocity ω . Four objects each of mass m, are kept gently to the opposite ends of two perpendicular diameters of the ring. The angular velocity of the ring will be:

Options:

Answer:

Solution:

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Question: A thin circular ring of mass M and radius r is rotating about its axis with a constant angular velocity $ω$ . Four objects each of mass m, are kept gently to the opposite ends of two perpendicular diameters of the ring. The angular velocity of the ring will be: