Answer:
the speed N of the grinding wheel after 5 complete revolutions of the handle starting from rest is 3122.62 rev/min
Explanation:
From the given information ;
According to the principles of Conservation of energy;
[tex]M \Theta =(\dfrac{1}{2}I_{gh} \omega_1^2)_{gear \ housing} +( \dfrac{1}{2} I_{ph} \omega^2)_{pinion \ housing}[/tex]
where;
[tex]M \Theta =[/tex] gain in potential energy as a result of restoring friction
[tex](\dfrac{1}{2}I_{gh} \omega_1^2)_{gear \ housing}[/tex] = kinetic energy as a result of rotation of the gear housing
[tex](\dfrac{1}{2} I_{ph} \omega^2)_{pinion \ housing}[/tex] = Kinetic energy as a result of rotation of pinion housing
However; the equation can be re-written as:
[tex]F*I* \Theta =(\dfrac{1}{2}*(mr^2)* \omega_1^2)_{gear \ housing} +( \dfrac{1}{2} *(mr^2)*\omega^2)_{pinion \ housing}[/tex]
where;
F = restoring Force
I = mass moment of the inertia
r = radius of gyration
Let assume the mass moment of inertia is 6.0 In around the the handle of the hand-operated grinder, since it is not given and a diagram is not attached ;
NOW;
[tex]4.0*\dfrac{6.0}{12}*5*2 \pi =[ (\dfrac{1}{2}*(\dfrac{4.26}{32.2})*(\dfrac{2.97}{12})^2*(\dfrac{\omega}{5})^2+( \dfrac{1}{2})*(\dfrac{1.07}{33.2}) *(\dfrac{1.95}{12})^2* \omega ^2)][/tex]
62.83 = [tex]1.6208*10^{-4} \omega^2 + 4.255*10^{-4} \omega^2[/tex]
62.83 = [tex]5.8758*10^{-4} \ \omega^2[/tex]
[tex]\omega^2 = \dfrac{62.83}{5.8758*10^{-4} }[/tex]
[tex]\omega^{2}= 106930.12 \\ \\ \omega = \sqrt{106930.12}[/tex]
[tex]\omega = 327 \ rad/s[/tex]
The spinning of the wheel is [tex]\omega = \dfrac{2 \pi N}{60}[/tex]
[tex]N = \dfrac{\omega *60}{2 \pi}[/tex]
[tex]N = \dfrac{327 *60}{2 \pi}[/tex]
N = 3122.62 rev/min
Thus; the speed N of the grinding wheel after 5 complete revolutions of the handle starting from rest is 3122.62 rev/min
