TS EAMCET 04 May 2019 Shift 2 Solved Paper

The total initial kinetic energy of the rolling ring is given by,
Ki‌‌=Krot+KLinear
‌‌=‌

1

2

Iω2+‌

1

2

mv2
‌‌=‌

1

2

mR2(‌

v

R

)+‌

1

2

mv2
‌‌=mv2
Substitute 10kg for m and 1.5m∕s for v in equation (I)
Ki‌‌=(10kg)(1.5m∕s)2
‌‌=22.5J
Now the change in kinetic energy is given by,
∆KE‌‌=Kf−Ki
‌‌=0−22.5J
‌‌=−22.5J