# Torque and Rotational Motion

1. Mar 13, 2006

### Swagger

A uniform 2.9-kg cylinder can rotate about an axis through its center at O. The forces applied are: F1 = 2.4 N, F2 = 3.9 N, F3 = 6.1 N, and F4 = 4.6 N. Also, R1 = 11.2 cm and R3 = 4.0 cm. Find the magnitude and direction (+: counterclockwise; -: clockwise) of the angular acceleration of the cylinder. (see attachment)

I'm having a hard time with this one. I know that:
Torque (net) = Inertia * Angular Acceleration
and

I've got:
T(F1)=0.2688 Nm
T(F2)=-0.4368 Nm
T(F3)=-0.0244 Nm
T(F4)=0 Nm

Tnet=-0.1924 Intertia=0.0181888
and the angular accerleration= -10.58 rad/s^2 (INCORRECT)

What am I doing wrong?

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2. Mar 13, 2006

### nrqed

I can't see the figure yet so I can't tell. But the torque is F times r times sin theta. That may (or may not, depending on what direction the forces are acting) explain a discrepancy.

Pat

3. Mar 13, 2006

### Staff: Mentor

The diagram seems to show a single radius. What are R1 and R2? What's the radius of the cylinder?

OK.
Don't forget the angle that the force is applied.

Torque = Force * Radius * sin(angle)

4. Mar 13, 2006

### Swagger

All angles a 90deg, that would just be multiplying by 1, right? Also, is my F3 torque correct? I did .004 * 6.1

5. Mar 13, 2006

### nrqed

Your T(F3) is off by a factor of 10. That will probably do it.

6. Mar 13, 2006

### Staff: Mentor

Perhaps you are defining things differently, but only F1 and F2 are applied at 90 degrees to the radial line through the point of application.

7. Mar 13, 2006

### Swagger

UGH! That was it...I got it. Thanks!

8. Jan 10, 2009

### consemiu21

O k Doc I am just having trouble get my (T1), T(2)..., my numbers are off, so what is the formula for that?

9. Jan 10, 2009

### Staff: Mentor

Formula for what? You just quoted the formula for finding the torque.

Show what you did.