ANGULAR MOMENTUM

(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

Case

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.

…..(i)

….(ii)

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

The free body diagram of cylinder is as shown.

And

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

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

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