A body of mass 1 kg is rotating in a vertical circle of radius 1 m. What will be the difference in its kinetic energy at the top and bottom of the circle? (take g=10 ms-2 )

Question

A body of mass 1 kg is rotating in a vertical circle of radius 1 m. What will be the difference in its kinetic energy at the top and bottom of the circle? (take g=10 ms-2 ) (A) 10 J (B) 20 J (C) 30 J (D) 50 J

Tardigrade Admin · Accepted Answer

Velocity at the top is √gr and that at the bottom is √5gr . Hence, required difference in kinetic energy =(1/2)m[5gr-gr]=2gmr =2× 10× 1× 1=20 J

Q. A body of mass $1 k g$ is rotating in a vertical circle of radius $1 m$ . What will be the difference in its kinetic energy at the top and bottom of the circle? (take $g = 10 m s^{- 2}$ )

10 J

20 J

30 J

50 J

Velocity at the top is $g r$ and that at the bottom is $5 g r$ .
Hence, required difference in kinetic energy
$= \frac{1}{2} m [5 g r - g r] = 2 g m r$
$= 2 \times 10 \times 1 \times 1 = 20 J$

Q. A body of mass 1kg is rotating in a vertical circle of radius 1m. What will be the difference in its kinetic energy at the top and bottom of the circle? (take g=10ms−2 )