(ii)  Angular Momentum of a Rigid Body Rotation about a Fixed Axis

Uniform Pure Rolling

 vp = vQ

 v - Rw = 0

 v = Rw

If vp > vQ or v > Rw, the motion is said to be forward slipping and if vp < vQ or v < Rw, the motion is said the backward slipping

Accelerated  Pure Rolling

v = Rw

at = Ra





Rolling on Rough Incline plane






Isolid  <  Ihollow  or  asolid  > ahollow

tsolid   <   thollow



A block of mass m and a cylinder of mass 2m are released on a rough inclined plane, inclined at an angle  with the horizontal. The coefficient of friction for all the contact surface is 0.5 find the accelerations of the block and the cylinder.

Assume pure rolling.

If the block and cylinder move independently on the incline, their acceleration are





 So, both the bodies will move in contact

with each other with common acceleration a which we have to calculate.

Free body diagram of block is as shown.






From Eqs. (i) and (ii), we get


The free body diagram of cylinder is as shown.





From Eqs. (iv) and (v), we get


Using this in Eq. (iii), we get

Case:- Ladder → mass = m, Length = L

For equilibrium

fN = 0

N1mg = 0

Torque about C. M. of Ladder


Substitute value of N1 and f



Case:- Topping and Sliding

For translational equilibrium



For rotational equilibrium




 block has a tendency to slide before topple  

 block has a tendency to topple before slide 

 block slides down 

 block topples