Question

20. A boy is standing on a horizontal massless
rotating wheel with his hands stretched. Kinetic
energy of boy is 120 J. If he now folded his hands
then M.I is reduced to 80%. Then a new K.E. of
boy,
(1) Increases by 30 J
(2) Decreases by 30 J
(3) Increases by 20 J
(4) Remains constant​

Answer 1

$\huge\underline\mathfrak\red{Solution}$

rotating wheel with his hands stretched. Kinetic energy of boy is 120 J. If he now folded his hands ... is reduced to 80%. Then a new K.E. of boy, (1) Increases by

Answer 2

Answer:

Option 1) increased by 30 J.

Explanation:

Fro the kinetic energy we have that the kinetic energy is directly proportional to the moment of inertia which is K.E=1/2Iω^2. Since the moment of inertia is reduced to 80% so, the new moment of inertia will be (80/100)I=0.8I. So, the angular momentum which is conserved will be ω1 = ω/0.8.

Hence, the kinetic energy will be (1/2)*0.8I(ω1)^2 which will be (1/2)0.8I(ω/0.8)^2 So, the kinetic energy is (1/2)I*ω^2/0.8. Since, 1/2Iω^2 is the K.E. So, the new kinetic energy will be (1/0.8)*K.E or 1/0.8*120 which on solving we will get 150 J. So, the new kinetic energy is increased by 30 J.