Rolling and rotationn are considered toughest topics of mechanics or entire physics during exam preparation.What makes them tough is the the number of concepts involved.

We will discuss one by one some imp. points in these topics.

1. Simple rotation.

When a body moves in such a way that its distance from a certain point is fixed.It is said to be rotating.

eg. rotating disc of mp3 player.

the terms defined are

angular velocity  , angular acceleration ..... u must be well aware of them

the equations governing them are

  (final) =   (initial) +  t   ------- 1

   F= final     I= initial  ------ 2

    ------- 3

note how the similarity arises between one dimension motion equation and the above equations....

Also note

        and  hence also  

2. Rolling ....

When a body moves such that every part of that body can be thought of as moving linearly as well as rotating....then that body is said to be rolling.

Taking case of pure rolling.

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Now for pure rolling " The lowest point of contact is at INSTANTANEOUS REST RELATIVE TO THE SURFACE ON WHICH DISC IS MOVING"

So the forward velocity is v and backward velocity is    ... SO FOR IT TO BE AT REST.

v =   ---- 4         Remember this eqn . very well.....We will use it later.

3. KINETIC Energy involved in rolling.

Their are 2 types of kinetic energy attached with rolling.They are translational and rotational.

Supposer a rinr(R , m) rolls on ground with velocity V and angular velocity  and we need to find its Total KE ?

Assuming u ppl know abt Moment of Inertia of ring = MR²

So Trans. K.E = .5 M V²

Rotat. KE =       I = moment of inertia. (M R ²)

Now  since pure rolling     

so ROT. KE =     .5 * M R^2 * ( V²  / R ²)

  => Rot. KE  = .5 MV²

and hence TOTAL KE =  MV²     ( Rotational and trans. )

4. ring rolling on a plane.

question :  A ring ( M ,R) is rotating with angular velocity  and kept on ground swiftly.It starts moving and after some time starts rolling.We have to find final velocity and angular velocity of ring.

sol :

now look at diagram below.

Initially ( at t =0)  it was rotating purely  and finally the diagram is as shown

Now take a point on ground and conserve total momentum of ring about it

So we get      Iw + MvR = constant.

since initial velocity is zero    and I = MR² for a ring

 we get  ....   

Since its rolling  == > v= Rw   

so we get                OR       and so

5.  Another problem

1. What is the linear velocity at the end of a ramp for a sphere which rolls without slipping down the ramp? The angle of inclination of the ramp is theta and the sphere starts from rest at a height h above the end of the ramp.
   
   
   
   
   
   
   
   
   
   
   
   

   Solution: We can start by drawing the free-body diagram as shown in the figure above. Since there is rolling without slipping, static friction is acting and we therefore will not need to know what F_N is since the static friction can be any magnitude between 0 and mu_sF_N. The forces acting in the x direction yield:
   
   
   
   
   
   
   
    

   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   

   Mgsin(theta) - fs

   = MaCM

   
   
   
   
   
   
   
   
   
   
   
   

   The frictional force causes a torque about an axis going through the geometric center of the sphere.
   
   
   
   
   
   
   
    

   
   
   
   
   
   
   
   -f_sR = I*alpha
   
   
   
   

   where the minus sign results from the fact that the torque tends to make the sphere want to rotate in the clockwise direction and the convention in the text is that torques that make clockwise rotations are negative. If rolling without slipping is occuring, then alpha = -a_CM/R (again the minus sign results since positive a_CM corresponds to clockwise rotation). Since I for a sphere is 2/5 MR², we have
   
   
   
   
   
   
   
    

   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   

   -fsR	= -(2/5)MR2aCM/R ==>
   fs	= (2/5)MaCM

   
   
   
   
   
   
   
   
   
   
   
   

   Plugging this back into our force equation, we get
   
   
   
   
   
   
   
    

   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   

   Mgsin(theta) - fs	= MaCM
   Mgsin(theta) - (2/5)MaCM	= MaCM ==>
   aCM	= (5/7)g sin(theta)

   
   
   
   
   
   
   
   
   
   
   
    
   
    

Now we have a paradox? If in rolling motion the force of friction does no work, why do we have an acceleration here which is less than g sin(theta)? This will lead to a translational velocity that is less than the sqrt[2gh] that we expect if only gravity is doing work on the sphere. The answer, of course, is that some of the work done by gravity leads to increasing the translational velocity and some goes to increasing the rotational velocity. Hence, we can conjecture that the total kinetic energy of an object must be given by the sum of its rotational kinetic energy, ½ I*omega² and its translational kinetic energy ½ mv². To see that this works out for the case at hand, let's use the acceleration we derived to work out what the translational and rotational velocities are at the end of the ramp. Note that d is the total distance traveled along the ramp in the following equations.







 

For the last term, phi is the total angular displacement of the sphere as it rolls down the ramp. Since the translation of the center-of-mass is the same as the arc-length traveled for rolling without slipping, phi = d/R and alpha = a_CM/R, so omega² = 2a_CMd/R². This is precisely v²/R² as we expect for rolling without slipping. Furthermore, if we note that the height, h, is equal to d*sin(theta), then the total kinetic energy can be expressed as







 

So the total work done by gravity, Mgh, is just equal to the change in the total kinetic energy, provided we express the total kinetic energy as K_trans. + K_rot..

This was just a small effort to put forward my views.I hope goiit ppl were benefited from it.Your comments would be encouraging .....

I had already posted similar article bfore .....posting it again....will b posting similar article  on thermodynamics (physics) soon ....