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
Menu
Grade: 12th pass
        A force F=-yi+xj acts on the rim of a rigid disc which can freely rotate about z axis along anticlockwise direction in x y plane .Center of disc is at origin .If the work done by the force is 32π in one complete revolution , find the radius of the disc?
7 months ago

Answers : (2)

Aman
29 Points
							
As we know that the work done by the torque is Torque.angle rotated.
So tor que about centre by applied force is. r*F. That is. Let r be radius (r j cap) *(force). Now doing vector product.  r j^  * (-yi^+xj^).  After solving we get. -ry k^ and work done in complete revolution is. (-ry k^) • 2pi -k^=32pi. The we get final answer r = 16/y.
Please report if any correction
7 months ago
Utkarsh shukla
12 Points
							Thanks dude I got my ans 4 let r=xi+yj and then use x^2+y^2 =r^2.        Torq=k(x^2+y^2)=kr^2Work=torq×2π=32πr=4
						
7 months ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

Get Extra Rs. 276 off

COUPON CODE: SELF10


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


Get Extra Rs. 297 off

COUPON CODE: SELF10

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details