Answer:
The rotational kinetic energy takes 0.430 seconds to become half its initial value.
Explanation:
By the Principle of Energy Conservation and the Work-Energy Theorem we know that flywheel slow down due to the action of non-conservative forces (i.e. friction), the energy losses are equal to the change in the rotational kinetic energy. That is:
(1)
Where:
- Energy losses, measured in joules.
,
- Initial and final rotational kinetic energies, measured in joules.
By definition of rotational kinetic energy, we expand the equation above:
(2)
Where:
- Moment of inertia of the flywheel, measured in kilograms per square meter.
,
- Initial and final angular speed, measured in radians per second.
If we know that
,
and
, then the initial angular speed is:
(3)




(4)




Under the assumption that flywheel is decelerating uniformly, we get that the time taken for the flywheel to slowdown is:
(5)
If we know that
,
and
, then the time needed is:


The rotational kinetic energy takes 0.430 seconds to become half its initial value.