Torque and Forces of grindstone

[NinjaIQ]

Homework Statement

Young University Physics 12e: Mastering Physics Problem 10.53

A 70.0-kg grindstone is a solid disk 0.540 m in diameter. You press an ax down on the rim with a normal force of 150 N. The coefficient of kinetic friction between the blade and the stone is 0.60, and there is a constant friction torque of 6.50 Nm between the axle of the stone and its bearings.

Part A
How much force must be applied tangentially at the end of a crank handle 0.500 m long to bring the stone from rest to 120 rev/min in 9.00 s ?

Part B
After the grindstone attains an angular speed of 120 rev/min, what tangential force at the end of the handle is needed to maintain a constant angular speed of 120 rev/min?

Part C
How much time does it take the grindstone to come from 120 rev/min to rest if it is acted on by the axle friction alone?

Homework Equations

Torque = Radius(R) x Force(F) or Moment of Inertia(I) x Angular Acceleration(alpha)

The Attempt at a Solution

Not even sure how to do part A

Do I need to include the negative torque done by the ax along with the friction on the bearings in my calculation for part A?

 

[tiny-tim]

Part A
How much force must be applied tangentially at the end of a crank handle 0.500 m long to bring the stone from rest to 120 rev/min in 9.00 s ?

Do I need to include the negative torque done by the ax along with the friction on the bearings in my calculation for part A?

Hi NinjaIQ!

You do know what a crank handle is, don't you?

You push tangentially on one end of the handle, the other end is a pivot (so the handle moves in a circle), and so you produce a torque at the pivot. That torque is transferred directly to the grindstone.

In other words: torque produced by crank handle equals torque received by grindstone.

oh, and yes … the whole purpose is to sharpen the axe … so include axe and bearings!

 

[Moetasim]

Yes, you will need to include the negative torque from the axle and the friction on the bearings in your calculation for part A. This is because these forces will act in the opposite direction of the applied tangential force and will need to be overcome in order to accelerate the grindstone to 120 rev/min in 9.00 seconds.

To solve for the required tangential force, you can use the equation for torque:

Torque = Radius x Force

In this case, the radius is given as 0.500 m and the torque needed to overcome the friction and axle forces is 6.50 Nm. So, the equation becomes:

6.50 Nm = 0.500 m x Force

Solving for Force, we get:

Force = 6.50 Nm / 0.500 m = 13 N

Therefore, a tangential force of 13 N must be applied at the end of the crank handle to bring the grindstone from rest to 120 rev/min in 9.00 seconds.