Answer to Question #101763 in Classical Mechanics for Sridhar

Question

A wheel is rotating freely at an angular speed on a shaft.A second wheel with twice the moment of inertia of the first and initially at rest is suddenly coupled to the first shaft. If K is the original rotational kinetic energy and Delta K is the loss in rotational kinetic energy then (∆K)/(K) is
Ans 2/3

Answer 1

Let the moment of inertia of the first wheel be $I$ and angular velocity be $\omega$

then initial kinetic energy =$(1/2)I\omega^2$

moment of inertia of second wheel = $2I$

second wheel is coupled with first wheel which means that angular momentum will remain conserved and both wheel will rotate with same angular velocity let's say $\omega'$

conserving angular momentum

$I*\omega=I*\omega'+2I*\omega'$

$\omega'=\omega/3$

now calculating the final kinetic energy

$(1/2)*I*(\omega/3)^2+(1/2)*2I*(\omega/3)^2$

$(1/6)*I*\omega^2$

loss in kinetic energy=$(2/6)*I*\omega^2$

loss in kinetic energy/initial kinetic energy=2/3