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Question: Increasing the speed of the flywheel from \(60\) to \(360\) revolution per minute it requires \(448\...

Increasing the speed of the flywheel from 6060 to 360360 revolution per minute it requires 448  J448\,\,J of energy. What is the moment of inertia of the wheel?
(A) 1.4  Kg m21.4\,\,Kg\,{m^2}
(B) 0.65  Kg m20.65\,\,Kg\,{m^2}
(C) 0.07  Kg m20.07\,\,Kg\,{m^2}
(D) 0.7  Kg m20.7\,\,Kg\,{m^2}

Explanation

Solution

Hint
To find the moment of inertia of the flywheel we can use the formula that is derived from a certain theorem that is the work done energy theorem which deals with total objects force and their kinetic energy of the object.
The for formula for the moment of inertia of the flywheel is;
⇒W=12I(ωf2−ωi2)\Rightarrow W = \dfrac{1}{2}I\left( {\omega _f^2 - \omega _i^2} \right)
Where, WW denotes the work done to rotate the flywheel, II denotes the moment of inertia of the flywheel, ωi{\omega _i} denotes the initial angular velocity of the flywheel, ωf{\omega _f} denotes the final angular velocity of the flywheel.

Complete step by step answer
The data given in the problem are;
The initial velocity of the flywheel is, ωi=60  rpm{\omega _i} = 60\,\,rpm.
The final velocity of the flywheel is, ωf=360  rpm{\omega _f} = 360\,\,rpm.
Kinetic energy of the flywheel is, K.E.=448  JK.E. = 448\,\,J.
The initial angular velocity:
⇒ωi=60×2π60  rad s−1\Rightarrow {\omega _i} = 60 \times \dfrac{{2\pi }}{{60}}\,\,rad\,{s^{ - 1}}
⇒ωi=2π  rad s−1\Rightarrow {\omega _i} = 2\pi \,\,rad\,{s^{ - 1}}
The final angular velocity:
⇒ωf=360×2π60  rad s−1\Rightarrow {\omega _f} = 360 \times \dfrac{{2\pi }}{{60}}\,\,rad\,{s^{ - 1}}
⇒ωf=12π  rad s−1\Rightarrow {\omega _f} = 12\pi \,\,rad\,{s^{ - 1}}
Form finding the moment of inertia of the flywheel apply the work done energy theorem, where;
⇒W=K.E.\Rightarrow W = K.E.
⇒K.E.=12I(ωi2−ωi2)\Rightarrow K.E. = \dfrac{1}{2}I\left( {\omega _i^2 - \omega _i^2} \right)
Substitute the value of the initial angular velocity, the final angular velocity and kinetic energy in the above formula.
⇒448=12I((12π)2−(2π)2)\Rightarrow 448 = \dfrac{1}{2}I\left( {{{\left( {12\pi } \right)}^2} - {{\left( {2\pi } \right)}^2}} \right)
⇒448×2=I(144π2−4π2)\Rightarrow 448 \times 2 = I\left( {144{\pi ^2} - 4{\pi ^2}} \right)
⇒896=I(π2(144−4))\Rightarrow 896 = I\left( {{\pi ^2}\left( {144 - 4} \right)} \right)
⇒896140×(227)2=I\Rightarrow \dfrac{{896}}{{140 \times {{\left( {\dfrac{{22}}{7}} \right)}^2}}} = I
⇒I=0.07  Kg m2\Rightarrow I = 0.07\,\,Kg\,{m^2}

Therefore, the moment of inertia of the flywheel is found as I=0.07  Kg m2I = 0.07\,\,Kg\,{m^2}.
Hence the option (C) I=0.07  Kg m2I = 0.07\,\,Kg\,{m^2} is the correct answer.

Note
The work energy theorem denotes that the total work done by the forces that is applied on an object is fairly the same as the change in kinetic energy of the body on which the force is applied on it.