How does torque relate to changes in angular momentum?

[fro]

I have no clue as to how to solve this. Any hints/suggestions will be helpful.

Problem: An object's angular momentum changes by 20kg*m^2/s in 4 seconds. What magnitude of torque acted on the object?

 

[Hootenanny]

Do you know of any expressions which relate a change in momentum to a torque? Or perhaps a torque to an angular acceleration?

 

[fro]

Do you know of any expressions which relate a change in momentum to a torque? Or perhaps a torque to an angular acceleration?

Maybe I could use L = I*w and T = I*angular acceleration? But I don't know what to do after that.

 

[Hootenanny]

It may be useful to note that for a constant acceleration;

$$\alpha = \frac{\Delta\omega}{\Delta t}$$

 

[fro]

It may be useful to note that for a constant acceleration;

$$\alpha = \frac{\Delta\omega}{\Delta t}$$

Sorry, I'm really confused about this.

By your equation, angular acceleration should be 5kg*m^2. Not sure how and where to plug it into get the answer.

 

[Hootenanny]

Sorry, I'm really confused about this.

By your equation, angular acceleration should be 5kg*m^2. Not sure how and where to plug it into get the answer.

Another useful observation; from your equation (I is constant);

$$\Delta L = I \Delta\omega \Leftrightarrow I = \frac{\Delta L}{\Delta \omega}$$

 

[fro]

Another useful observation; from your equation (I is constant);

$$\Delta L = I \Delta\omega \Leftrightarrow I = \frac{\Delta L}{\Delta \omega}$$

$$\Delta L = 20{kg} \cdot{m^2}$$
$$\alpha = \frac{20kg\cdotm^2}{4s} = {5kg} \cdot {m^2}$$
$$\tau = \frac{\Delta L}{\Delta \omega} \times \alpha$$
$$\tau = \frac{20kg\cdot m^2}{\Delta \omega} \times {5kg} \cdot {m^2}$$

So, how would I find $$\Delta \omega$$?

 

[Hootenanny]

From my previous posts;

$$\alpha = \frac{\Delta\omega}{\Delta t}\;\;\;\;\;\; (1)$$

$$\Delta L = I \Delta\omega \Leftrightarrow I = \frac{\Delta L}{\Delta \omega}\;\;\;\;\;\; (2)$$

Using those to facts and the formula;

$$\tau = I\alpha \stackrel{(1)\;\&\;(2)}{\Rightarrow} \tau = \frac{\Delta L}{\Delta \not{\omega}} \cdot \frac{\Delta\not{\omega}}{\Delta t}$$

$$\therefore \boxed{\tau = \frac{\Delta I}{\Delta t}}$$

This is a good formula to remember