The kinetic energy of a body rotating at 300 revolutions per minute is 62.8 J. Its angular momentum ( textin kg m2s-1) is approximately:

Question

The kinetic energy of a body rotating at 300 revolutions per minute is 62.8 J. Its angular momentum ( textin kg m2s-1) is approximately: (A) 1 (B) 2 (C) 4 (D) 8

Tardigrade Admin · Accepted Answer

Kinetic energy o rotating body K=(1/2)Iω 2=62.8 J Angular momentum, L=Iω =( (1/2)Iω 2 )× (2/ω ) =(62.8× 2/(2× π × 300/60)) =4 kg m2s-1

Q. The kinetic energy of a body rotating at 300 revolutions per minute is 62.8 J. Its angular momentum $(in kg m^{2} s^{- 1})$ is approximately:

1

2

4

8

Kinetic energy o rotating body $K = \frac{1}{2} I ω^{2} = 62.8 J$ Angular momentum, $L = I ω$ $= (\frac{1}{2} I ω^{2}) \times \frac{2}{ω}$ $= \frac{62.8 \times 2}{( 2 \times π \times 300/60 )}$ $= 4 k g m^{2} s^{- 1}$

Q. The kinetic energy of a body rotating at 300 revolutions per minute is 62.8 J. Its angular momentum (in kgm2s−1) is approximately:

1

2

4

8

Kinetic energy o rotating body K=21​Iω2=62.8J Angular momentum, L=Iω =(21​Iω2)×ω2​ =(2×π×300/60)62.8×2​ =4kgm2s−1

Q. The kinetic energy of a body rotating at 300 revolutions per minute is 62.8 J. Its angular momentum $(in kg m^{2} s^{- 1})$ is approximately:

Kinetic energy o rotating body $K = \frac{1}{2} I ω^{2} = 62.8 J$ Angular momentum, $L = I ω$ $= (\frac{1}{2} I ω^{2}) \times \frac{2}{ω}$ $= \frac{62.8 \times 2}{( 2 \times π \times 300/60 )}$ $= 4 k g m^{2} s^{- 1}$