# Physics: Torque on a Disk problem - Multiple forces

1. Nov 14, 2013

### Zeus5966

1. The problem statement, all variables and given/known data

A uniform disk with mass m = 9.27 kg and radius R = 1.42 m lies in the x-y plane and centered at the origin. Three forces act in the +y-direction on the disk: 1) a force 344 N at the edge of the disk on the +x-axis, 2) a force 344 N at the edge of the disk on the –y-axis, and 3) a force 344 N acts at the edge of the disk at an angle θ = 39° above the –x-axis.

Diagram: https://www.smartphysics.com/Content/smartPhysics/Media/Images/Mechanics/15/torqueondisk.png
1) What is the magnitude of the torque on the disk about the z axis due to F1?
2) What is the magnitude of the torque on the disk about the z axis due to F2? = 0, this was correct
3) What is the magnitude of the torque on the disk about the z axis due to F3? = 379.6, this was correct
4) What is the x-component of the net torque about the z axis on the disk?
5) What is the y-component of the net torque about the z axis on the disk?
6) What is the z-component of the net torque about the z axis on the disk?
7) What is the magnitude of the angular acceleration about the z axis of the disk?
8) If the disk starts from rest, what is the rotational energy of the disk after the forces have been applied for t = 1.8 s?

2. Relevant equations
torque = F * r * sin(theta)
torque = I * alpha

3. The attempt at a solution

1) torque = F * r * sin(theta)
so torque = (344N) * (1.42) * sin(90)
the answer I get is 488N*m but it says it's incorrect. I'm also not sure how to convert the sin(90) to the z axis instead of the x-y axis.

4) I can't actually find this until I get 1 correct, however, it would be the sum of all torque, so
Ʃtorque = torquey + torquex + torquez where torquez and torquey are to be transferred into the y direction, but multiplying by sin(90)?

5/6, same method as above

7) The equations I used are torque = I * alpha, torque = F * r, I = 1/2 * mdisk * radius2

rearranging, I * alpha = F * r so
alpha = F * r / I where
I = 1/2 * mdisk * radius2
to end up with

alpha = F*R/(1/2 * mdisk * radius2)
alpha = 344 * 1.42 / (1/2 * 9.27 * 1.422)
alpha = 52.267 rad/s2 <- seems too high

8) Krotational = (1/2) * I * ω2 where ω = alpha * t and I = 1/2 * mdisk * radius2

so K = (1/2) * (1/2 * mdisk * radius2) * (alpha * t)2
K = 551.278J

Thanks for the help

2. Nov 14, 2013

### Simon Bridge

Did you sketch out the situation?
I'm guessing the on-axis forces are pointing along the axis, and the one at the -x axis has components in the +y and -x direction.

It would be helpful if the question were clearer about the directions.

You appear to be applying the equation without understanding it.

The radial component $F_r$ points along the line joining the center or rotation to the point of application of the force.
The tangential component $F_{\!\perp}$ is perpendicular to that.
The magnitude of the torque is given by: $$\tau = rF_{\!\perp}$$ ... where $r$ is the distance from the center to the point of application.

You can also get the direction by $$\vec{\tau}=\vec{r}\times\vec{F}$$

3. Nov 14, 2013

### Zeus5966

I did sketch it out, however it still makes no sense due to the confusing directions. I'm not quite sure what you mean by Fperpendicular. How would I find that? And why would there be a difference? The force, to my understanding is already tangent to the disk at force 1 judging by the wording and the way my sketch and the picture that I linked look.

Last edited: Nov 14, 2013
4. Nov 14, 2013

### Simon Bridge

(my bf) Excuse - must have had my eyes closed.
Sorry I thought you were doing something else.

Now I get it - all the forces are in the +y direction ... the last one is not on an axis.

I'm guessing you are submitting this to a computer?
In that case it's probably just being fussy - try using more decimal places and try using a minus sign - though, usually, anti-clockwork torques (viewed down the axis) are positive.

To work out which direction the torque points - remember that the torque is a cross product.
Are all the forces in the x-y plane?
Use the right-hand-screw rule on the vectors.

5. Nov 14, 2013

### Zeus5966

Apparantly the computer can't tell the difference between 488.48 and 488.5. I was trying to figure out what I was doing wrong, just a fussy website. As it stands, after reading the question four times, the reason they say about the z-axis, to my understanding is that the disk is rotating around the z axis, so all of the forces act in the x-y plane. My problem is, is that I don't quite know how to find the sum of all torque on a rotating axis, when the two torque are against each other. F3 being clockwise, and F1 being counter-clockwise.

6. Nov 14, 2013

### Simon Bridge

You use the right-hand screw rule to determine if the torque is positive (pointing up the z axis) or negative (pointing down the z axis).

On that video - r is blue, F is green, so that $\vec \tau = \vec r \times \vec F$

Last edited by a moderator: Sep 25, 2014
7. Nov 14, 2013

### Zeus5966

I just realized I answered my own question in my question. Thank you for the help Simon

8. Nov 14, 2013

### Simon Bridge

Well done - that is often the case :)