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
f – N = 0
N1 – mg = 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
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Kaysons Academy
Course : JEE
Start Date : 04.06.2023
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Kaysons Academy
Course : JEE
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End Date : 30.06.2023
Types of Batch : Live Online Class