Textbook velocity problem

  • Thread starter sauri
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  • #1
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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?
 

Answers and Replies

  • #2
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sauri said:
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 [itex]M=Fd=I\ddot{\theta}[/itex]
 
  • #3
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The M represents what exactly? and isn't the momentum for a disk about it's center I = ½ mr2?
 
  • #4
Hootenanny
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I believe the M is turning moment or torque. Your F = Ia equation needs modifing, instead of using F, you must use torque:
[tex]\tau = I\alpha[/tex]
This is a version of Newton's second law.
 
  • #5
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I see, but is the working correct?
 
  • #6
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sauri said:
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

[tex] \omega_{f}=\omega_{i}+at [/tex]

Now solve for t putting [tex]\omega_{f}=0[/tex]

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 [tex]\tau=Ia[/tex]
and substitute [tex]\tau=rF[/tex]
 
Last edited:
  • #7
Hootenanny
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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
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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
Hootenanny
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[itex]\tau = rF[/itex] is simply torque, force([itex]F[/itex]) multiplied by distance([itex]r[/itex]) from the axis of rotation.
 
Last edited:
  • #10
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Edit:I didn't read ur sol properly.You also need to use [tex]\tau=Ia[/tex]
and substitute [tex]\tau=rF[/tex][/QUOTE]

I understand using the [tex]\tau=Ia[/tex] part where we can find [tex]\alpha[/tex](angular acceleration). But why do we need to substitue [tex]\tau=rF[/tex]?. Why can we not just find [tex]\alpha[/tex], and find t so it can be substituted into [tex] \omega_{f}=\omega_{i}+at [/tex]
 
Last edited:
  • #11
Hootenanny
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You need to use [itex]\tau = rF[/itex] and put it equal to [itex]\tau = I\alpha[/itex] to find the angular accelertatiion thus;

[tex]I\alpha = rF[/tex]

[tex]\alpha = \frac{rF}{I}[/tex]

Can you go from here?
 
  • #12
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got it.thanks
 
  • #13
Hootenanny
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No problem
 

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