Answers (1)

Moment of inertia for continuous body -

I= \int r^{2}dm

- wherein

r is perpedicular distance of a particle of mass dm of rigid body from axis of rotation

 

 taking a strip of radius x  and thickness dx 

MI of ring dI = dmx^2

I = \int dI= \int_{R_1}^{R_2}\left [ \frac{ M}{\pi (R_{2}^{2}-R_{1}^{2})} 2 \pi x\times dx \right ]x^2\\\\I = \int dI = \frac{ M}{\pi (R_{2}^{2}-R_{1}^{2})} 2 \pi\left [ \frac{x^4}{4} \right ]_{R_1}^{R_2}\Rightarrow I = \frac{M}{2}(R_{1}^{2}+R_{2}^{2})

Preparation Products

Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Test Series JEE Main Sept 2020

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 4999/- ₹ 1999/-
Buy Now
Knockout JEE Main Sept 2020

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 12999/- ₹ 6999/-
Buy Now
Rank Booster JEE Main 2020

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 9999/- ₹ 4999/-
Buy Now
Exams
Articles
Questions