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: 11
        

a sphere of mass m is at rest on a rough horizontal surface. The coefficient of friction between the ground and sphere is u. then, max value of F so that the sphere will not slip is??

7 years ago

Answers : (1)

vikas askiitian expert
509 Points
							

let  a force F is applied at a distance r from the center ...

friction force will be in foreward direction

for translational motion

 F - f =ma ..........1

for rotational motion

Fr  - fR = I(alfa) ..........2                               (R is radius of sphere)

for no slipping a = (alfa)R & Isphere = 2mR2/5

Fr - fR = 2maR/5 .................3

from 1 & 3

f = F(5r - 2R)/3R ......4

for rolling ,

 f < or = umg

F(5r-2R)/3R = umg          

F= 3umgR/(5r-2R)

this value of force is maximum .......

7 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

Get Extra Rs. 551 off

COUPON CODE: SELF20


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. 594 off

COUPON CODE: SELF20

Ask Experts

Have any Question? Ask Experts

Post Question

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