# Newton's Second Law for Rotation

1. May 1, 2008

### frig0018

1. The problem statement, all variables and given/known data
A pulley, with a rotational inertia of 1.0 x 10^-3 kg*m^2 about its axle and a radius of 10 cm, is acted on tangentially at its rim. the force magnitude varies in time as F=0.50t + 0.30t^2, with F in newtons and t in seconds. The pulley is initially at rest . At t=3.0 s what are its (a) angular acceleration and (b) its angular speed?

2. Relevant equations

(a) $$\alpha$$= (r*F(3)) / I = 420 rad/sec^2
(b) unsure, but tried using $$\omega$$= $$\omega$$$$_{}0$$ +$$\alpha$$t = 1260 rad/s which is wrong and several other variations of that equation.

3. The attempt at a solution

(see above)

2. May 1, 2008

### Nabeshin

The kinematic equation you used is incorrect because this is only for cases of constant alpha. In this case, since force is changing, torque is changing. And since torque is changing, alpha is changing. Think of the general (geometric?) relationship between alpha and omega.

3. May 2, 2008

### frig0018

so is the geometric relationship that when alpha increases, so does omega? do i need to use an integral?
dw= (integral)alpha dt? but how can i differentiate this if there are no variables?
i come up with dw=alpha*t + c but when i plug in t=3, i still get 1260...

4. May 2, 2008

### alphysicist

Hi frig0018,

alpha is not 420 rad/s^2 except at t=3. To integrate over t you need alpha as a function of (the variable) t.

5. May 2, 2008

### frig0018

they are just asking for alpha at t=3 s, which is 420 rad/s^2. I am fairly sure of this because it is the answer in the back of the book... However, i am struggling to figure out omega at t=3 s which should be 500 rad/s but i need the work to back it up and i'm not sure which equations to use.

6. May 2, 2008

### Nabeshin

Your method for obtaining the alpha at t=3 is correct, and your idea of integrating is also correct, it seems you're just stuck on finding the function? Well, look at the way in which you solved for alpha and you'll see you have a general formula for alpha. That should help :)

7. May 2, 2008

### alphysicist

Yes, your answer for part a looks correct, but my previous comment:

was referring to part b. In post #3 you tried to integrate (alpha dt), with the integral running from t=0 to t=3. But if you integrate from t=0 to t=3, you can't just plug in the value of alpha that is true only at t=3. You need alpha as a function of t.

They give you the force as a function of t: F= 0.50t + 0.30t^2. What is alpha as a function of t?