JEE Main Physics Class 11 System of Particles and Rotational Motion Part 1 Questions

When body rolls dawn on inclined plane with velocity V0 at bottom then body has both rotational and translational kinetic energy.
Therefore, by law of conservation of energy,
P.E.=K.Etrans +K.Erotational
=

1

2

mV02+

1

2

Iω2
=

1

2

mV02+

1

2

mk2

V02

R02

......(i) [∵I=mk2,ω=

V

R0

]
When body is sliding down then body has only translatory motion.
∴P.E.=K.Etrans
=

1

2

m(

5

4

v0)2 ......(ii)
Dividing (i) by (ii) we get

P.E

P.E

=

1 2 mv02[1+ K2 R02 ]

1 2 × 25 16 ×mV02

=

25

16

=1+

K2

R02

⇒

K2

R02

=

9

16

or, K=

3

4

R0