How Long Does It Take for a Disc to Stop Under Applied Force?

[sauri]

Help please!

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?

 

[phucnv87]

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?

The equation must be M=Fd=I\ddot{\theta}[/color]

 

[sauri]

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

 

[Hootenanny]

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.

 

[sauri]

I see, but is the working correct?

 

[arunbg]

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?

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

 

[Hootenanny]

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

 

[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!

 

[Hootenanny]

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

 

[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

 

[Hootenanny]

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?

 

[sauri]

got it.thanks

 

[Hootenanny]

No problem