# Homework Help: Change in angular momentum

1. Dec 28, 2009

### series111

1. The problem statement, all variables and given/known data
a steel disk 400mm in diameter and a mass of 48.98 kg accelerates from 6.28rad/s to 43.98rad/s calculate the change in angular momentum.

2. Relevant equations
final angular momentum = intial momentum

(I) final x (w) final = (i) initial x (w) intial

I= mr _2

3. The attempt at a solution

(I) initial = 48.98 x 200x10^-3_2 = 1.95 kgm_2/s

(I) final = (I) initial x (w) initial/ (w) final = 1.95 x 6.28/ 43.98 = 0.278 kg/m_2

change in angular momentum = final - initial

(I) initial x (w) initial - (I) final - (w) final

= 1.95 x 6.28 - 0.278 x 43.98 = 19.56 x 10 -3 kgm_2/s

can someone check if this correct thanx again....

Last edited: Dec 28, 2009
2. Dec 28, 2009

### EricAngle

You calculate a change in angular momentum in part 3 after assuming the angular momentum is constant in part 2.

The angular momentum is I * ω, and since the moment of inertia I is constant, the change in angular momentum is I * Δω, where

( http://en.wikipedia.org/wiki/List_of_moments_of_inertia )

I = m * r^2 / 2 = 48.98 kg * ( 0.4 m / 2 )^2 / 2

and

Δω = ( 43.98 - 6.28 ) rad/s.

3. Dec 28, 2009

### series111

so I = 3.918kgm_2/s and Δω = 37.7 rad/s

and the change in angular momentum is = I * Δω = 3.918 x 37.7 = 147.70 kgm_2/s

thanx for putting me right....

Last edited: Dec 28, 2009
4. Jan 15, 2012

### MathsRetard09

This is wrong, from the handout I have the Angular Momentum = Iw(omega)

There is no division of 2. Wikipedia though helpful, isn't the best source of evidence to use because anyone can edit it.

Where did you get 3.918 kgm^2/s from calculating this:

48.98 kg * ( 0.4 m / 2 )^2 / 2

I'd like to ask for help since I have the exact same question to do.

Using the same values I got this below:

Angular Momentum = Iw

I = mr^2 = 48.984 x (0.2)^2 = 1.95936 = 1.96 kgm^2/s

w2 - w1 = 43.98 - 6.28 = 37.7 rad/s^2

therefore: 1.96 x 37.7 = 73.892 kgm^2/s

The next question asks for the change in angular kinetic energy.

This is what I got:

Equation - Angular ke = 1/2Iw^2

1/2x(1.96 x (37.7)^2) = 1.39x10^3

Angular ke = 1.39 kJ

Is this correct??

Last edited: Jan 15, 2012
5. Jan 16, 2012

### EricAngle

The 1/2 is for the moment of inertia of a solid disk about its center perpendicular to the plane of the disk, not angular momentum. I used Wikipedia because it's a known result, and I didn't want to do this:

$$I = \int dm \ r^2 = \int \left(\frac{M}{\pi R^2} r d\theta dr\right) r^2 = \frac{M}{\pi R^2} \int_0^{2 \pi} d\theta \int_0^R dr r^3 = \frac{1}{2} M R^2$$

6. Jan 17, 2012

### series111

Yes my calculations were wrong however I did calculate them again and got the write answers through a little research :

K^2 = d^2/8 ( This is for a soild Disk i.e Steel Disk)

Where I=mk^2

Where Iw2 - Iw1 ( change in angular momentum)

As for change in angular kinetic energy you are using the correct formula just wrong values.

7. Jan 18, 2012

### MathsRetard09

@EricAngle - appolagies haha.

@series111 - I've actually figured it all out now.

My method is wrong because above I do this: (w2-w1)^2

But the correct method is (w2^2-w1^2)

Therefore: 1/2 I (w2^2-w1^2).