Question

A boy is standing on a horizontal massless rotating wheel with his hands stretched kinetic energy of boy iz 120j if he now folded his hands then m.I. Is redueced to 80% then a new k.E. Of boy

Answer 1

Answer:

New Kinetic Energy will be 150 J

Explanation:

If the moment of Inertia is I and the angular velocity ω then the kinetic energy

$K=\frac{1}{2}I\omega^2$

If the moment of inertia is reduced to 80%, new moment of Inertia

$I'=I\times \frac{80}{100}$

$I'=0.8I$

The angular momentum of the boy will be conserved

Therefore,

$I\omega=I'\omega'$

$\implies I\omega=0.8I\omega'$

$\implies \omega'=\frac{\omega}{0.8}$

Thus, the new Kinetic Energy

$K'=\frac{1}{2}I'\omega'^2$

$\implies K'=\frac{1}{2}\times0.8I\times(\frac{\omega}{0.8})^2$

$\implies K'=\frac{1}{0.8}\times \frac{1}{2}I\omega^2$

$\implies K'=\frac{5}{4}\times K$

$\implies K'=\frac{5}{4}\times 120$

$\implies K'=150$ Joule

Hope this helps.

Answer 2

Answer:

correct ANSWER is increase by 30J i.e.150J

Explanation:

This is totally correct