# Moment of Inertia help pls

1. Dec 23, 2004

### Beretta

A pulley of moment of inertia 0.021kg-m^2 and radius 0.12m is acted upon by a force which varies in time as F= 0.23t + 0.12t^2 where F is in Newton and t is in second. suppose that the pulley is initially rotating at 0.18 r/s and the force acts tangentially to the pulley. Find the magnitude of the angular velocity of the pulley two seconds after the force began to act on the pulley.

I was thinking if I can use F(tan) r = I (alpha)

alpha = (0.12m(0.23t+0.12t^2))/0.21kgm^2 to get the accelration and then plug it in

w = wi + (alpha)t to get the velocity?

Is my reasoning correct?

What if I integrate the Force to get the work W, and then W = Kf - Ki?

Last edited: Dec 23, 2004
2. Dec 23, 2004

### Staff: Mentor

Good!
That equation assumes constant alpha, which is not the case here. You have to integrate to find the final omega.

3. Dec 23, 2004

### Beretta

You mean intergrate in the Force to get K = 1/2 Iw^2? I'm a bit lost! What do you mean?

Last edited: Dec 23, 2004
4. Dec 23, 2004

### Staff: Mentor

You have $\alpha (t)$. So realize that $d\omega/dt = \alpha (t)$, so:
$$\Delta \omega = \int_{0}^{2} \alpha(t) dt$$

5. Dec 23, 2004

### dextercioby

Nope,as Doc said,u need to integrate this relation

$$d\omega=\alpha dt$$

between corresponding limits.

Well,you figure it out.Is it possible without finding $\theta (t)$ and reversing it.??
$$\frac{I}{2}(\omega_{f}^{2}-\omega_{i}^{2})=\int_{\theta_{0}}^{\theta_{f}} F(t(\theta))r d\theta$$

Daniel.

6. Dec 23, 2004

### Beretta

Do I need to go through this?

Can't I just use $$\Delta \omega = \int_{0}^{2} \alpha(t) dt$$
But do I need to subtract the initial omega that is given from the final omega?

One more question please: when I did multiply the force equation by the radius, should I do the dot product or just multiply it commutatively. the reason I'm asking is the units when multiplying 0.12m by 0.23t + 0.12t^2 of course its m/t where t is in seconds. but after that I have to divide the answer by the moment of inertia which is kg m^2. Logicaly I know the answer should be in m/s^2 since its accelration and I also know that accelration could be written in N/kg, but here I have m(t)/kg m^2.

Last edited: Dec 23, 2004
7. Dec 23, 2004

### Staff: Mentor

$$\omega_f = \omega_i + \Delta \omega$$

Also realize that the units of "0.23t + 0.12t^2" are Newtons (when t is in seconds), and the units of torque are N-m. The units of $\omega$ will be radians/sec.

8. Dec 23, 2004

### Beretta

I keep thinking about rotation in terms of motion. Yes, radian/sec not m/s.
I really appreciate all your help guys. Thank you very much.

9. Dec 23, 2004

### dextercioby

Yes,it's the only way to get the answer.

U can.Doc gave you the integral formula,i gave u the differential formula.

Nope,correct integration will give the result.

The problem gives the force without units.It should be Newtons.But the constants are given numerically:they should be measured,first in Newtons per second,and the second in Newtons per second squared.

Daniel.