# Insane rotation/torque problem!

1. Jun 24, 2004

### Brianjw

K this is the last problem for me in the Rotation/Torque section of my homework and its nuts. I've put it off after working on it for an hour and now I have to do it:

A 55.0-kg grindstone is a solid disk 0.520 m in diameter. You press an ax down on the rim with a normal force of 160 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 N*m between the axle of the stone and its bearings.

for a picture go here, trust me it helps alot:

http://members.cox.net/brianjw/wheel.JPG

It has 3 questions:

1)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?

For this one I know I need to get the sum of the torques. But the crank and the way the axe is being onto the stone is confusing me. Also, what formula do I use that has time in it? Can't seem to find any from this section.

The next 2 I can't really solve cause the first part is stumping me.

2)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?

3)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?

Thanks for any input, even where to start would be greatly appreciated. I don't expect someone to give me an answer, just a nudge in the right direction!

2. Jun 25, 2004

Force of friction, like usual, = Normal force*Coefficient of kinetic friction.

After that, use $\vec{\tau} = \vec{r} \times \vec{F} \Rightarrow |\tau| = rF$ and add them up.

The kinematics of rotating things have direct analogs to the kinematics of linear things. In linear, we have

$$a\Delta t = \Delta v$$

so that means in rotational we'll have

$$\alpha \Delta t = \Delta \omega$$

Last edited: Jun 25, 2004
3. Jun 25, 2004

### e(ho0n3

You are told that the axe is producing a normal force. Like cookiemonster already pointed out, you can use this normal force to find the force of friction and the torque caused by this friction. The force you apply to the crank is perpendicular to the crank. You are given the length of the crank so you can calculate the torque on it.

I think you can take it from there.

4. Jun 25, 2004

### Brianjw

I thought I understood but the numbers that I'm using aren't adding up:
$$Radius = .26m$$

$$U_{k of axe} = .60$$

$$Normal = 160N$$

$$Mass = 55.0Kg$$

$$I = .5*M*R^2$$

$$Constant Torque = 6.50 N*m$$

$$\sum\tau = .5m * F_{crank} - .26m*160N*.6 - 6.5N*m = I\alpha$$
I also use:
$$\alpha \Delta t = \Delta \omega$$

Using that I can get:

$$\alpha*9.0s = 2.0 Rev/s$$
which implies:
$$\alpha = \frac{2.0}{9.0} Rev/s^2$$
$$.5m * F_{crank} - .26m*160N*.6 - 6.5N*m = 1.859kg*m^2* \frac{2.0}{9.0}Rev/s^2$$
$$F_{crank} = 24.96N*m + 6.5N*m + 1.859kg*m^2* \frac{2.0}{9.0}Rev/s^2$$

I know I'm missing something because the units are the right way, but not sure wha tI'm missing...

Last edited: Jun 25, 2004
5. Jun 25, 2004

### e(ho0n3

You are computing the sum of the torques on the entire system (rim and crank) so the I on the RHS of the last equation should be the moment of inertia of the rim + crank. Since you don't know the mass of the crank, I suggest you reevaluate your methods.

6. Jun 25, 2004

### Brianjw

Hmm, didn't think about that. I didn't see another torque equation, that I thought would work, in my notes, but I did find an equation:

$$\tau = mr^2\alpha$$

Tried that and didn't work either. Hmm.

7. Jun 25, 2004

### Staff: Mentor

Express $\alpha$ in the proper units: radians/sec^2.