SATHEE: Rotation 3 Question 12

Rotation 3 Question 12

18. Two thin circular discs of mass $m$ and $4 m$ , having radii of $a$ and $2 a$ , respectively, are rigidly fixed by a massless, rigid rod of length $l = \sqrt{24} a$ through their centers. This assembly is laid on a firm and flat surface and set rolling without slipping on the surface so that the angular speed about the axis of the rod is $ω$ . The angular momentum of the entire assembly about the point "

(are) true?

(2016 Adv.)

(a) The magnitude of the $z$ -component of $L$ is $55 m a^{2} ω$

(b) The magnitude of angular momentum of centre of mass of the assembly about the point $O$ is $81 m a^{2} ω$

(d) The magnitude of angular momentum of the assembly about its centre of mass is $17 m a^{2} \frac{ω}{2}$

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Answer:

Correct Answer: 18. (c, d)

Solution:

(a) $L_{z} = L_{C M - O} \cos θ - L_{D - C M} \sin θ$

$\begin{aligned} = \frac{81 \sqrt{24}}{5} a^{2} m ω \times \frac{\sqrt{24}}{5} - \frac{17 m a^{2} ω}{2} \times \frac{1}{\sqrt{24}} \\ = \frac{81 \times 24 m a^{2} ω}{25} - \frac{17 m a^{2} ω}{2 \sqrt{24}} \end{aligned}$

(b) $L_{C M - O} = (5 m) \frac{9 l}{5} ω \frac{9 l}{5} = \frac{81 m l^{2} ω}{5}$

$\begin{aligned} = \frac{81 m l^{2}}{5} \times \frac{a ω}{l} \\ L_{C M - O} & = \frac{81 m l a ω}{5} = \frac{81 \sqrt{24} a^{2} m ω}{5} \end{aligned}$

point $O$ .

Angular velocity of CM w.r.t. $Z$ -axis $= ω_{0} \cos θ$

$ω_{C M - z} = \frac{a ω}{1} \frac{\sqrt{24}}{5} = \frac{a ω}{\sqrt{24} a} \frac{\sqrt{24}}{5}$

$\begin{aligned} ω_{C M - z} & = \frac{ω}{5} \\ (d) L_{D - C M} & = \frac{m a^{2}}{2} ω + \frac{4 m (2 a)^{2}}{2} ω = \frac{17 m a^{2} ω}{2} \end{aligned}$

Rotation 3 Question 12

Answer:

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Rotation 3 Question 12

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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