# Rotational inertia

1. Oct 30, 2005

Hi,
I have hard time understanding the rotational inertia .
A string is wrapped around a cylindar spool of radius 1cm. The axis of the spool is fixed. A length of string of .8 m is pulled off in 1.5 s at a constant tension of 20N.
What is the rotational inertia of the spool?

Can I have some help with this problem?
I don't know how to start. The only definition of rotational inertia I have is I=mR2!
Thank you
B.

2. Oct 30, 2005

### Fermat

The eqn of motion you have to use for rotational motion is,

T = Iα

You know the tension applied, so what is the torque ?

You now have to find α, the angular acceln.
Use the other info you have to work out α.

Then solve for I = T/α.

3. Oct 30, 2005

Thank you very much Fermat.

However, I have some problem to find the angular acceleration.
From the data from the problem, I found the circumference of the spool:
2*pi*R=0.0628 m
then, I find the number of revolutions for the .8 m string.
I find .8/.0628=12.73 rev.-- 13 revolutions

The angular velocity is w=12.73*(2*pi)/1.5s=53.32 rad/s
so the angular acceleration is: a=w/1.5=35.54rad/s^2

Am I right?

B

4. Oct 31, 2005

### Fermat

Sorry it took so long to respond to your post.

The angular velocity you got is actually the average value over that time period. You need to find the final angular velocity, at the end of the 1.5s time period, before you can use that last formula in your post.

You're almost there.

5. Nov 1, 2005

Hey sorry I was busy with 3 exams.

I tried to compute the angular velocity but sorry I don't get it. can you give me more hints please?

thanks

6. Nov 1, 2005

### lightgrav

The most straightforward approach to this is Work and Energy
... The Work done by the string becomes KE of rotation ...
KE = ½ I w^2 (w = omega_final)

7. Nov 2, 2005

### Fermat

You had the average angular velocity. Under constant acceleration, this is just half of the final velocity! (provided the initial velocity is zero)

$$v_{av} = 0.8m\ /\ 1.5s$$
$$v_{av} = 0.5333\ m/s$$
$$v_f = 2v_{av} = 1.0667\ m/s$$

$$\omega = v/r$$
$$\omega_f = v_f/r_{spool}$$
$$\omega_f = 1.0667/0.01$$
$$\omega_f = 106.667\ rad/sec$$

To get angular acceln,

$$\alpha = \omega_f/t$$
$$\alpha = 106.667\ /\ 1.5$$
$$\alpha = 71.111\ rad/s^2$$

Now you should get a pretty small value for the inertia of the spool.

Last edited: Nov 2, 2005