MHT-CET : Physics Entrance Exam

### MHT - CET : Physics - Rotational motion Know More

Kinetic Energy of A Rolling Body: Consider a body that is rolling on a horizontal plane without slipping. Its motion is composed of two motions
a) Rotation about a horizontal axis through its centre of mass and
b) translation of the centre of mass. Let M, R and
n be the mass, radius and uniform velocity of the body. If T is the time period of rotation of the body, it describes an angle of 2 p radians in this time and its centre of mass is displaced by 2 p R in the same time.

Thus angular velocity w =

 2p T

and the linear velocity of translation of the centre of mass,

n =

 2pR T

=

 2p T

. R

n = w R
The total kinetic energy of the rolling body, Ek is given by
Ek = K. E. of rotation + K.E. of translation

=

 1 2

Iw 2 +

 1 2

Mn2

=

 1 2

MK2w2 +

 1 2

Mv2

where K is the radius of gyration about a horizontal axis through the centre of mass

\ Ek =

 1 2

M

 K2v2 R2

+ v2

=

 1 2

Mv2

 K2 R2

+ 1

...(1)

or \ Ek =

 1 2

M [K2w2 + R2w2 ] =

 1 2

mw 2[K2 + R2 ].

In the case of a cylinder or a disc,

K2 =

 R2 2

or

 K2 R2

=

 1 2

\ Ek =

 3 4

Mn2 =

 3 4

MR2w2

Rolling can be considered as a combination of pure rotation and pure translation.
For a body rolling down an inclined plane AB making an angle
q with the horizontal.

Total K.E., Ek =

 1 2

mn2

 K2 R2

+ 1

Hence, the body loses an amount of potential energy, mgh.
Neglecting the loss of energy due to friction and air resistance,

 1 2

mV2

 K2 R2

+ 1

= mgh

\ n2 =

2gh

 K2 R2

+ 1

From the equation of motion, n2 - u2 = 2as, n2 = 2as since u = 0

\
2as =

2gh

 K2 R2

+ 1

If AB = S, h = s sin q

\
2as =

2g s sinq

 K2 R2

+ 1

or
a =

2g sinq

 K2 R2

+ 1

For a ring,

 K2 R2

= 1

For a disc,

 K2 R2

=

 1 2

For a sphere,

 K2 R2

=

 2 3

\ a ring : a cyl : a sphere =

 1 2

:

 2 3

:

 5 7

= 21 : 28 : 30

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