# Homework Help: Angular Speed of a disk

1. Nov 18, 2014

### ljucf

1. The problem statement, all variables and given/known data

A disk of radius 0.41 m and moment of inertia 2.8 kg·m2 is mounted on a nearly frictionless axle. A string is wrapped tightly around the disk, and you pull on the string with a constant force of 52 N.

What is the magnitude of the torque?
torque = 21.32 N·m

After a short time the disk has reached an angular speed of 4 radians/s, rotating clockwise. What is the angular speed 0.63 seconds later?

2. Relevant equations

torque = RFT

Rotational angular momentum = (MR2/2)ω

3. The attempt at a solution

The new angular momentum is the old angular momentum plus the angular impulse, torque times time interval.

(2.8/2)4 + (21.32)(.63) = 19.03 radians/s

However, this answers is wrong, and I can't figure out what I am doing wrong.

2. Nov 18, 2014

### ehild

The moment of inertia is given, it is 2.8 kgm2. Why did you divide it by 2?

3. Nov 18, 2014

### ljucf

According to my textbook, Rotational angular momentum = (MR2/2)ω

Therefore, I divided it by two. Is that incorrect?

4. Nov 18, 2014

### ehild

MR2/2 is the moment of inertia of a homogeneous disk. The mass is not given. The moment of inertia is given as 2.8 kgm2. The angular momentum is moment of inertia times the angular speed.

5. Nov 18, 2014

### ljucf

So is the Rotational angular momentum =

6. Nov 18, 2014

### ehild

Yes, the angular momentum of a rotating body is Iω.

7. Nov 18, 2014

### ljucf

So now I take (2.8)(4) + 21.32(.63) and divide that by moment of inertia?

8. Nov 18, 2014

### ehild

Yes.

9. Nov 18, 2014