IIT JEE , AIEEE Forum - Physics , Mechanics

apurviitjee2008

a hollow sphere is released from an inclined plane with a certain angle with respct to horizontal(assume any angle)

find the coefficient of friction reqd to prevent slipping

if the coefficient of friction is half the value found out, find the kinetic energy of the block as it slides a length l down the inclined plane

elessar_iitkgp

Draw the FBD
N = mgcos

-----------(1)
mgsin

- f = ma -------------(2)
fR = (2/3)mR²

----------------(3)
For no slipping
a =

R ------------(4)
From (3) and(4)
f = (2/3)ma ------(5)
Adding (5) & (2)
a = (3/5) gsin

Substitute in (5) to get
f = (2/5)gsin

Now,
f

N
(2/5)gsin

mgcos

(2/5)tan

_min = (2/5)tan

The Problem should read " Find the minimum coefficient of friction to prevent slipping"
If

is half this minimum value, then slipping occurs.Friction will act with its maximum value. The equations are

N = mgcos

-----------(6)
mgsin

-

N = ma -------------(7)

N = (2/3)mR²

----------------(8)
Now applying work energy theorem
₀

^L (mgsin

-

N)dx = (1/2)mv² - 0
(mgsin

-

mgcos

)L = (1/2)mv²
v =

2g(sin

-

cos

)L

apurviitjee2008

sir but the answer to the second part is 7/8 tan theta