TS EAMCET 18-July-2022 Shift 2 Solved Paper

When the cylinder reaches the ground, then its gravitational potential energy will get converted to translational as well as rotational kinetic energy.
∴‌‌U=KT+KR where, KT= Translational kinetic energy KR= Rotational kinetic energy
∴‌‌mgh=‌

1

2

mv2+‌

1

2

Iω2
mgh=‌

1

2

mv2+‌

1

2

×‌

mR2

2

×ω2‌‌[∵ICM=‌

mR2

2

]
⇒‌‌gh=‌

v2

2

+‌

R2

4

‌

v2

R2

‌‌(∵In pure Rolling v=Rω)
⇒10×30=‌

3v2

4

⇒v2=400
⇒‌‌v=20m∕ s