SATHEE: System of Particles and Rotational Motion

System of Particles and Rotational Motion - Result Question 42

45. 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

[2003]

(a) $\frac{(M - 4 m) ω}{M + 4 m}$

(b) $\frac{M ω}{4 m}$

(c) $\frac{M ω}{M + 4 m}$

(d) $\frac{(M + 4 m) ω}{M}$

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Answer:

Correct Answer: 45. (c)

Solution:

(c) Applying conservation law of angular momentum, $I_{1} ω_{1} = I_{2} ω_{2}$

$I_{2} = (M r^{2}) + 4 (m) (r^{2}) = (M + 4 m) r^{2}$

$\begin{aligned} . (Taking ω_{1} = ω and ω_{2} = ω_{1}) \\ \Rightarrow & M r^{2} ω = (M + 4 m) r^{2} ω_{1} \\ \Rightarrow & ω_{1} = \frac{M ω}{M + 4 m} \end{aligned}$

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System of Particles and Rotational Motion - Result Question 42

45. 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

Answer:

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45. 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