# Ok then since I am wrong then formula for torque is wrong

1. Apr 1, 2012

### sweetpete28

Force F is applied to the rim of a uniform disk (M=3.13kg, R=.193m). The disk is mounted on a fixed frictionless axis through its center, and the force is applied at an angle β=45.8° to the radius. The disk starts at rest, and reaches frequency f = 61.1 revolutions per second after rotating through an angle θ = 609 radians.

What is the magnitude of force F?

Torque = FRsinβ

Torque = I∂

∂ = ωf - ωi / t

ωf = 2∏f = 383.9026 rad/s

ωf = θ/t; 609 rad / 383.9026 rad/s = 1.586, so t = 1.586 s

∂ = 383.9026 / 1.586 = 242.005 rad/s^2

Torque = (1/2)(MR^2)(∂)
Torque = (1/2)(3.13)(.193)^2(242.005) = 14.1076

14.1076 = F(.193)(sin 45.8)
F = 102 N

But 102 N is wrong so I guess Torque really does not = FRsinβ

Can anyone help here....cause I have no clue why or how this could be wrong...

2. Apr 1, 2012

### Staff: Mentor

Good. That's the final angular velocity.
Careful. Do not assume that the angular velocity is constant. Redo this. (What's the average angular velocity?)

3. Apr 1, 2012

### tms

So $\omega = 2\pi$?

4. Apr 1, 2012

### Staff: Mentor

I think he meant ωf = 2∏f.

5. Apr 1, 2012

### sweetpete28

Thanks Doc Al!!

You're right!

ωavg = 0 + 383.9026 / 2 = 191.9513 rad/s so t= 609 rad / 191.9513 rad/s = 3.17267 s and ∂ = 383.9026 / 3.17267 = 121.002 rad/s^2 so....

Torque = (1/2)(3.13)(.193)^2(121.002) = 7.0538

and F = 7.0538 / (.193)(sin 45.8) = 51.0N......right????

6. Apr 1, 2012

Looks good!