Homework Help: Textbook velocity problem

1. Mar 12, 2006

sauri

I found this question in my textbook but the solution is not stated. A force of 2 N vertically down is applied to a vertical disc at a distance 2 cm to the right (horizontally) of the centre of the disc. The disc has a mass of 1.2 kg and a radius of 15 cm. Its initial angular velocity is 2 rad.s-1 anticlockwise.

How long does it take to come to a (momentary) stop?

I took a crack at it but is it correct?.
I=(mr^2)/2, I=[1.2x(15/100)^2]2. Then the value for I is substituted to F=Ia, where 2=value for Ixa, where a is found, and then substitute for a=w/t. So a is substituted=2/t and t is found.

Is this o.k?

2. Mar 12, 2006

phucnv87

The equation must be $M=Fd=I\ddot{\theta}$

3. Mar 12, 2006

sauri

The M represents what exactly? and isn't the momentum for a disk about it's center I = ½ mr2?

4. Mar 12, 2006

Hootenanny

Staff Emeritus
I believe the M is turning moment or torque. Your F = Ia equation needs modifing, instead of using F, you must use torque:
$$\tau = I\alpha$$
This is a version of Newton's second law.

5. Mar 12, 2006

sauri

I see, but is the working correct?

6. Mar 12, 2006

arunbg

The first part is allright. After finding a(angular acceleration) you need to
use the eqn

$$\omega_{f}=\omega_{i}+at$$

Now solve for t putting $$\omega_{f}=0$$

I recommend you go through the eqns of motion for angular variables.

Edit:I didn't read ur sol properly.You also need to use $$\tau=Ia$$
and substitute $$\tau=rF$$

Last edited: Mar 12, 2006
7. Mar 12, 2006

Hootenanny

Staff Emeritus
You are quite right arundg, it would be alot simpler using equations of motion, I didn't stop to consider the question, I was simply looking for errors.

8. Mar 12, 2006

sauri

I thank you all for your help. But could you tell me what the T in the equation T=rF stands for. Thanks again!

9. Mar 12, 2006

Hootenanny

Staff Emeritus
$\tau = rF$ is simply torque, force($F$) multiplied by distance($r$) from the axis of rotation.

Last edited: Mar 12, 2006
10. Mar 13, 2006

sauri

Edit:I didn't read ur sol properly.You also need to use $$\tau=Ia$$
and substitute $$\tau=rF$$[/QUOTE]

I understand using the $$\tau=Ia$$ part where we can find $$\alpha$$(angular acceleration). But why do we need to substitue $$\tau=rF$$?. Why can we not just find $$\alpha$$, and find t so it can be substituted into $$\omega_{f}=\omega_{i}+at$$

Last edited: Mar 13, 2006
11. Mar 13, 2006

Hootenanny

Staff Emeritus
You need to use $\tau = rF$ and put it equal to $\tau = I\alpha$ to find the angular accelertatiion thus;

$$I\alpha = rF$$

$$\alpha = \frac{rF}{I}$$

Can you go from here?

12. Mar 13, 2006

sauri

got it.thanks

13. Mar 13, 2006

Hootenanny

Staff Emeritus
No problem