For all round rigid bodies, define the zone transition line as follows:
Suppose a Force 'F' is applied at a height 'h' abv. the ground;
1)if h is less than H then friction acts opposite to the dirn. of application of force.
2)For h is greater than H, the friction acts in the same direction as the applied force.
For a rigid body with parameters M,K,R, specify the location of the Z.T.L.
H= (k^2 + R^2) /R
ReplyDeletesame here
ReplyDeletehere too
ReplyDeletemy ans depends on friction force(f)
ReplyDeleteit is the above ans multiplied with (1+f/F)
but if I put f=0, then its the above result
is akash's ans is correct, so it means that friction does not act on the body when force acts along the zone transition line
right ?
the property if the height h is that here the friction reverses direction...
ReplyDeleteso friction is actually 0 for a continuous variation.
also, i know the result as :
1+(K^2/R^2) times the radius.
the value k^2/R^2 is a property of the type of rigid body.
it is 2/5, 2/3, 1/2, 1 for a solid sphere, hollow sphere, cylinder and ring respectively.