AP EAMCET Engineering 23 Sep 2020 Shift 1 Paper

For a body rolling without slipping.
vCM=Rω .....(i)
Translational kinetic energy, KET=‌

1

2

mvCM2
Rotational kinetic energy, KER=‌

1

2

Iω2
=‌

1

2

(mK2)a2
(∵I=mK2, where k is radius of gyration.)
=‌

1

2

mk2(‌

vCM

R

)2 [from Eq. (i)]
=‌

1

2

mvCM2(‌

k2

R2

)
∴ Total kinetic energy, KE=KET+KER
=‌

1

2

mvCM2+‌

1

2

mvCM3‌

k2

R2

KE=‌

1

2

mvCM2(1+‌

k2

R2

)
Fraction of total energy associated with its rotation, i.e.
‌

KER

KE

=‌

‌ 1 2 mvCM2( k2 R2 )

1 2 mvCM2(1+‌ k2 R2 )

=‌

k2

k2+R2