Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
        From  a uniform disc of  radius R a small disc of radius R/3 is cut . The mass of the remaining portion is M . find the moment of inrtia of such a disc about the axis passing through the geometrical centre of original disc and perpendicular to the the plane of the disc .
7 months ago

Kapil Khare
80 Points
							Area of the remaining disc = $\pi$[R2 – (R2/9)]                                        A = $\pi$(8R2/9)Mass of the remaing disc = MMass per unit area of the material = $\sigma$ = 9M/(8$\pi$R2)Mass of the part remaining = $\sigma$($\pi$R2/9) = M/8 So let us assume a M/8 and -M/8 plate on the part from where we removed the small disc.Now we have a disc of mass 9M/8 with radius R and another disc of mass -M/8 with radius R/3. So Moment of Inertia of larger disc about its axis =9MR2/16Now let us assume the distance between the centers of the two disc be d.Now, moment of inertia of smaller disc about its axis = -MR2/144Moment of inertia of smaller disc about center of original disc = (-MR2/144) + (-Md2/8) So, total moment of inertia about the centre of original disc = (9MR2/16) – (MR2/144) – (Md2/8)                             I = (5MR2/9) – (Md2/8)

7 months ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Mechanics

View all Questions »

### Course Features

• 101 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution

### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions