Question

A steady 4.0-lb force is applied normal to the handle of the hand-operated grinder. The gear inside the housing with its shaft and attached handle together weigh 4.26 lb and have a radius of gyration about their axis of 2.97 in. The grinding wheel with its attached shaft and pinion (inside housing) together weigh 1.07 lb and have a radius of gyration of 1.95 in. If the gear ratio between gear and pinion is 5:1, calculate the speed N (in revolutions/min) of the grinding wheel after 5 complete revolutions of the handle starting from rest.

Answer 1

the speed N of the grinding wheel after 5 complete revolutions of the handle starting from rest is 3122.62 rev/min

Step-by-step explanation:

From the given information ;

According to the principles of Conservation of energy;

`M \Theta =((1)/(2)I_(gh) \omega_1^2)_(gear \ housing) +( (1)/(2) I_(ph) \omega^2)_(pinion \ housing)`

where;

`M \Theta =` gain in potential energy as a result of restoring friction

`((1)/(2)I_(gh) \omega_1^2)_(gear \ housing)` = kinetic energy as a result of rotation of the gear housing

`((1)/(2) I_(ph) \omega^2)_(pinion \ housing)` = Kinetic energy as a result of rotation of pinion housing

However; the equation can be re-written as:

`F*I* \Theta =((1)/(2)*(mr^2)* \omega_1^2)_(gear \ housing) +( (1)/(2) *(mr^2)*\omega^2)_(pinion \ housing)`

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;

`4.0*(6.0)/(12)*5*2 \pi =[ ((1)/(2)*((4.26)/(32.2))*((2.97)/(12))^2*((\omega)/(5))^2+( (1)/(2))*((1.07)/(33.2)) *((1.95)/(12))^2* \omega ^2)]`

62.83 =
`1.6208*10^(-4) \omega^2 + 4.255*10^(-4) \omega^2`

62.83 =
`5.8758*10^(-4) \ \omega^2`

`\omega^2 = (62.83)/(5.8758*10^(-4) )`

`\omega^(2)= 106930.12 \\ \\ \omega = √(106930.12)`

`\omega = 327 \ rad/s`

The spinning of the wheel is
`\omega = (2 \pi N)/(60)`

`N = (\omega *60)/(2 \pi)`

`N = (327 *60)/(2 \pi)`

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