Question

Two gear wheels of the same thickness (as shown), but one with twice the diameter of the other, are mounted on parallel light axles far enough so that the teeth of the wheels do not mesh. The larger wheel is spun with angular velocity $\Omega$ and the wheels are then moved together so that they mesh. For the meshed rotating gears, if $\frac{Kineticenergylost}{OriginalKineticenergy}=\frac{1}{3+X}$, find the value of $X$.

• A. 2
• B. 12
• C. 4
• D. 3

Answer 1

The correct option is A 2

Let MI of larger wheel be $I$
the MI of smaller wheel will be $I/4$
since radius of smaller wheel is half of larger wheel and MI is directly proportional to the radius
So, Initial KE will be
$K{E}_i=\frac{1}{2}I{\Omega }^2$
Now after both the wheels are meshed their angular momentum will be conserved and their final angular velocities will be same let it be $\omega$
Therefore,
$I\Omega =I\omega +I\omega /4$
this gives
$\omega =4\Omega /5$
Now final KE will be
$K{E}_f=\frac{1}{2}I{\omega }^2+\frac{1}{8}I{\omega }^2$
Putting the value of $\omega$ we get,
$K{E}_f=\frac{4}{10}I{\Omega }^2$
$K{E}_{lost}=K{E}_i-K{E}_f=\frac{1}{10}I{\Omega }^2=K{E}_i/5$
Therefore, X=2
option (A) is correct