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System of Particles and Rotational Motion

Moment of inertia

Consider a badminton racket with length scales as shown in the

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Consider a badminton racket with length scales as shown in the figure.

If the mass of the linear and circular portions of the badminton racket are same (M) and the mass of the threads are negligible, the moment of inertia of the racket about an axis perpendicular to the handle and in the plane of the ring at, $\frac{r}{2}$ distance from the end A of the handle will be $Mr^{2}$ .

Name: Video solution 1: Consider a badminton racket with length scales as shown in the figure.If the mass of the linear and circular portions of the badminton racket are same (M) and the mass of the threads are negligible, the moment of inertia of the racket about an axis perpendicular to the handle and in the plane of the ring at, 2r distance from the end A of the handle will be Mr2.
Uploaded: 2022-10-17T13:17:52.000Z
Duration: 6 min
Description: Video solution 1: Consider a badminton racket with length scales as shown in the figure.If the mass of the linear and circular portions of the badminton racket are same (M) and the mass of the threads are negligible, the moment of inertia of the racket about an axis perpendicular to the handle and in the plane of the ring at, 2r distance from the end A of the handle will be Mr2.

Views: 6,062 students

Updated on: Feb 25, 2023

Exam:

JEE Mains 2021 Aug 26 Shift 1

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Alexander Discussed

Consider a badminton racket with length scales as shown in the figure.

If the mass of the linear and circular portions of the badminton racket are same (M) and the mass of the threads are negligible, the moment of inertia of the racket about an axis perpendicular to the handle and in the plane of the ring at,

\frac{r}{2}

distance from the end A of the handle will be

Mr^{2}

6 mins ago

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6 mins ago

Text solutionVerified

I_{A} = [\frac{Mr ^{2}}{2} + M^{r} (7 r - \frac{r}{2})^{2}] + [\frac{M ( 6 r ) ^{2}}{12} + (M (3 r - \frac{r}{2})^{2})] = {\frac{Mr ^{2}}{2} + M \frac{( 169 ) r ^{2}}{4}} + {M r^{2} + \frac{M 25 r ^{2}}{4}} = \frac{171 Mr ^{2}}{4} + \frac{37 Mr ^{2}}{4} = \frac{208}{4} M^{2} = 52 Mr^{2}

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Question Text	Consider a badminton racket with length scales as shown in the figure. If the mass of the linear and circular portions of the badminton racket are same (M) and the mass of the threads are negligible, the moment of inertia of the racket about an axis perpendicular to the handle and in the plane of the ring at, $\frac{r}{2}$ distance from the end A of the handle will be $Mr^{2}$ .
Updated On	Feb 25, 2023
Topic	System of Particles and Rotational Motion
Subject	Physics
Class	Class 11
Answer Type	Text solution:1 Video solution: 2
Upvotes	317
Avg. Video Duration	10 min

Text solutionVerified

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