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MHT-CET : Physics Entrance Exam

MHT - CET : Physics - Rotational motion Know More

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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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