SATHEE: Rotational Motion Question 60

Rotational Motion Question 60

Question: The moment of inertia of a uniform circular disc of radius R and mass M about an axis passing from the edge of the disc and normal to the disc is:

[AIPMT (S) 2005]

Options:

A) $\frac{1}{2} M R^{2}$

B) $M R^{2}$

C) $\frac{7}{2} M R^{2}$

D) $\frac{3}{2} M R^{2}$

Show Answer

Answer:

Correct Answer: D

Solution:

Key Idea: We should use parallel axis theorem.

Moment of inertia of disc passing through its centre of gravity and perpendicular to plane is

$I_{A B} = \frac{1}{2} M R^{2}$

Using theorem of parallel axes, we have,

$I_{C D} = I_{A B} + M R^{2}$

$= \frac{1}{2} M R^{2} + M R^{2}$

$= \frac{3}{2} M R^{2}$

Note: The role of moment of inertia in the study of rotational motion is analogous to that or mass in study of linear motion.

Play Store

App Store

Ask SATHEE

Welcome to SATHEE !
Select from 'Menu' to explore our services, or ask SATHEE to get started. Let's embark on this journey of growth together! 🌐📚🚀🎓

I'm relatively new and can sometimes make mistakes.
If you notice any error, such as an incorrect solution, please use the thumbs down icon to aid my learning.
To begin your journey now, click on "I understand".

Rotational Motion Question 60

Question: The moment of inertia of a uniform circular disc of radius R and mass M about an axis passing from the edge of the disc and normal to the disc is:

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu