Question

A wheel of moment of inertia 10kg-m2 is rotating at a speed of 10 rotations per minute. Find the work done in increasing its speed to five times its initial value.

Answer 1

As no external torque is acting on the wheel, assume the system to be isolated and then apply the law of conservation of energy to calculate the change in kinetic energy of the system before and after the work is done.

Complete step by step answer:
In the question there is given a wheel whose moment of inertia is 10kg-m2 and is rotating at an angular speed of 10 rotations per minute. We must find out the work done in increasing its speed to five times its initial value. But before that we must understand the angular mechanics.
Angular mechanics is like normal mechanics, only with everything being in rotation in the angular mechanics. For displacement there is angular displacement, for speed there is angular speed, for mass there is moment of inertia, for acceleration there is angular acceleration etc. and all of them are related to each other with some factor.
Similarly, the kinetic energy has an angular kinetic energy and is given as
K.E = \dfrac{1}{2}m{v^2} \\\ K.{E_{angular}} = \dfrac{1}{2}I{\omega ^2} \\\
m = mass, v = speed, I = angular momentum, ω = angular speed
Using this angular kinetic energy, we can find out the work done to change the speed of a rotating wheel. Since in the question there is no mention of any external torque on the system, we can assume it to be isolated and therefore we can use the law of conservation of energy which says that the total energy is conserved for an isolated system.
T.{E_{initial}} = K.{E_{initial}} = \dfrac{1}{2} \times 10 \times {10^2} \\\ T.{E_{final}} = K.{E_{final}} + W = \dfrac{1}{2} \times 10 \times {50^2} + W \\\
Therefore, the total energy initially for the initial speed will be the same as after the speed has been increased to five times by doing work on the system.
T.{E_{initial}} = T.{E_{final}} \\\ \dfrac{1}{2} \times 10 \times {10^2} = \dfrac{1}{2} \times 10 \times {50^2} + W \\\ W = \dfrac{1}{2} \times 10 \times \left( {{{10}^2} - {{50}^2}} \right) = \dfrac{1}{2} \times 10 \times \left( { - 2400} \right) = - 12000kg{m^2}/{\min ^2} \\\ W = - \dfrac{{12000}}{{60 \times 60}}kg{m^2}/{\sec ^2} = - 3.34J \\\
Work is negative because work is done on the system from an external factor. If work was done by the system, then it would be positive.