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Questions on Rotational Kinematics/Dynamics

  • Thread starter fterh
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  • #1
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Some background information: I'm doing some reading up for PhO (so it's beyond what I'm supposed to learn), and so I'll post all my questions here (regarding both concepts and actual practice questions). Sorry if you feel there is a lack of effort on my part, but sometimes I'm really lost and don't know where to start.

Questions:

1. http://imgur.com/LEmOT

Since it's a single length of rope, shouldn't tension be constant throughout the rope? That's my only question (at the moment, that is :D).
 

Answers and Replies

  • #2
Doc Al
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Since it's a single length of rope, shouldn't tension be constant throughout the rope?
No. That would be true if the rope were massless (which it presumably is) and the pulleys were massless (which they are not) and frictionless.
 
  • #3
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Hi, then can you explain this: T2 * r - T1 * r = 0.5 * m * r^2 * (a/r)

I'm guessing that it means clockwise moment - anticlockwise moment = something?

But I'm not too sure.
 
  • #4
Doc Al
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Hi, then can you explain this: T2 * r - T1 * r = 0.5 * m * r^2 * (a/r)

I'm guessing that it means clockwise moment - anticlockwise moment = something?
It's applying Newton's 2nd law for rotation: The net torque = I*alpha.
 
  • #5
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It's applying Newton's 2nd law for rotation: The net torque = I*alpha.
Alright thanks a lot.

Back to the question on the pulley. The pulley with a mass if fixed, so the mass shouldn't affect the tension? The movable pulley is massless.
 
  • #6
Doc Al
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The pulley with a mass if fixed, so the mass shouldn't affect the tension?
It has mass and thus rotational inertia, so the tension in the string must be different on each side.
The movable pulley is massless.
For that pulley the tension is the same on either side.
 
  • #7
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Okay I get it, thanks! :D

Next question (2): A father of mass [itex]m_f[/itex] and his daughter of mass [itex]m_d[/itex] sit on opposite ends of a seesaw at equal distances from the pivot at the centre. The seesaw is modeled as a rigid rod of mass [itex]M[/itex] and length [itex]d[/itex] and is pivoted without friction. At a given moment, the system rotates in a vertical plane with an angular velocity [itex]\omega[/itex]. Obtain an expression for

a) the magnitude of the system's angular momentum

According to the answer given,

Modelling the father and daughter as particles, the moment of inertia of the system is
[itex]I = \frac{1}{12}Md^2 + \frac{1}{4}m_fd^2 + \frac{1}{4}m_dd^2 = \frac{1}{12}d^2(M + 3m_f + 3m_d)[/itex]

I get that a rod has a moment of inertia of (Md^2)/12, but how is the M.I. of the father and daughter derived? Manual derivation using calculus?
 
  • #8
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Never mind I got it :D

For anybody who needs the answer, it's because [itex]I = r^2 \times dm = (\frac{d}{2})^2 \times m_{f} = \frac{1}{4}m_fd^2[/itex],
 
  • #9
Doc Al
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Never mind I got it :D
Oops... I didn't even see your second question. Sorry about that!
 
  • #10
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Oops... I didn't even see your second question. Sorry about that!
It's okay, I feel great being able to solve a question completely out of my syllabus! :D

Okay, but now I have another question. :X

A uniform solid disk of radius R and mass M is set into rotation with an angular speed [itex]\omega_i[/itex] about an axis through its center. While still rotating at this speed, the disk is placed into contact with a horizontal surface and immediately released as shown.

What is the angular speed of the disk once pure rolling takes place?

My solution:

[itex]K_{rotational initial} = K_{rotational final} + K_{linear final}[/itex]
[itex](\frac{1}{2})(\frac{1}{2}MR^2)(\omega_i)^2 = (\frac{1}{2})(\frac{1}{2}MR^2)(\omega_f)^2 + (\frac{1}{2})(M)(R\omega_f)^2[/itex]
[itex]\frac{1}{2}\omega_i^2 = \frac{1}{2}\omega_f^2 + \omega_f^2[/itex]
[itex]\omega_f = \sqrt{\frac{1}{3}\omega_i^2}[/itex]

But the answer is [itex]\frac{1}{3}\omega_i[/itex]. I suspect a careless mistake somewhere, but I've been through my workings and I found no mistakes.
 
  • #11
Doc Al
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My solution:

[itex]K_{rotational initial} = K_{rotational final} + K_{linear final}[/itex]
Do not assume that kinetic energy is conserved. There is friction involved.
 
  • #12
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Do not assume that kinetic energy is conserved. There is friction involved.
But there is no mention of frictional forces in the question, so how do I go about compensating for energy losses due to friction?

Can you give me a clue? :D
 
  • #13
Doc Al
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But there is no mention of frictional forces in the question, so how do I go about compensating for energy losses due to friction?
There must be friction, otherwise it would just sit there rotating. Just assume that the coefficient of friction is μ and see what happens.
 
  • #14
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So,

energy lost to friction = [itex]\mu Mg \times \theta[/itex]?

Since W = F x s?

Sorry I'm wild guessing, I have no idea how to apply friction to rotational dynamics.
 
  • #15
Doc Al
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Sorry I'm wild guessing, I have no idea how to apply friction to rotational dynamics.
Sure you do. It's a force, like any other. Apply Newton's 2nd law for translation and for rotation.

Hint: The rotational speed is decreasing while the translational speed is increasing. Find the point where the condition for rolling without slipping is met.
 
  • #16
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Sure you do. It's a force, like any other. Apply Newton's 2nd law for translation and for rotation.

Hint: The rotational speed is decreasing while the translational speed is increasing. Find the point where the condition for rolling without slipping is met.
Okay... so...

[itex]\tau = I\alpha[/itex]
[itex]\mu MgR = \frac{1}{2}MR^2\alpha[/itex]
[itex]\mu g = \frac{1}{2}R\alpha[/itex]
[itex]R\alpha = 2\mu g[/itex]

[itex]F = Ma[/itex]
[itex]F = MR\alpha[/itex]
[itex]F = M2\mu g[/itex]

But since F is provided for by the friction,

[itex]\mu Mg = M2\mu g[/itex]
[itex]1 = 2[/itex]

HUH?

Condition for pure rolling is when [itex]\omega = v[/itex]. I could probably sub [itex]\alpha = \frac{d\omega}{dt}[/itex] and [itex]a = \frac{dv}{dt}[/itex], but why did I get 1 = 2? Where did I go wrong? :/
 
  • #17
Doc Al
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Okay... so...

[itex]\tau = I\alpha[/itex]
[itex]\mu MgR = \frac{1}{2}MR^2\alpha[/itex]
[itex]\mu g = \frac{1}{2}R\alpha[/itex]
[itex]R\alpha = 2\mu g[/itex]
Now write ω as a function of time.

Then write v as a function of time.

Solve for when the condition for rolling without slipping is met. (You can't assume it's met for all times; that was your error.)
 
  • #18
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Now write ω as a function of time.

Then write v as a function of time.

Solve for when the condition for rolling without slipping is met. (You can't assume it's met for all times; that was your error.)
So my formula [itex]\tau = \mu MgR[/itex] is correct?

[itex]\omega_f = \omega_i + \alpha t[/itex], [itex]v = at[/itex]

When the disk is rolling without sleeping, i.e. when [itex]\omega_f = v[/itex],

[itex]\omega_i + \alpha t = at[/itex]
[itex]\omega_i + \frac{2\mu g}{R}t = \frac{\mu Mg}{M}t[/itex]
[itex]\omega_i = \frac{\mu Mg}{M}t - \frac{2\mu g}{R}t = \mu gt (1 - \frac{2}{R})[/itex]

So from here, do I find an expression for t, then apply to [itex]\omega_f = \omega_i + \alpha t[/itex]?
 
  • #19
Doc Al
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So my formula [itex]\tau = \mu MgR[/itex] is correct?
Yes.

[itex]\omega_f = \omega_i + \alpha t[/itex], [itex]v = at[/itex]
Yes, but don't forget that alpha is negative, since the rotational speed is decreasing.

When the disk is rolling without sleeping, i.e. when [itex]\omega_f = v[/itex],
Almost: [itex]R\omega_f = v[/itex]
 
  • #20
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So my formula [itex]\tau = \mu MgR[/itex] is correct?

[itex]\omega_f = \omega_i + \alpha t[/itex], [itex]v = at[/itex]

When the disk is rolling without sleeping, i.e. when [itex]R\omega_f = v[/itex],

[itex]R\omega_i - R\alpha t = at[/itex]
[itex]R\omega_i - R\frac{2\mu g}{R}t = \frac{\mu Mg}{M}t[/itex]
[itex]R\omega_i - 2\mu gt = \mu gt[/itex]
[itex]R\omega_i = 3\mu gt[/itex]
[itex]t = \frac{R\omega_i}{3\mu g}[/itex]

[itex]\omega_f = \omega_i - \alpha t[/itex]
[itex]\omega_f = \omega_i - \frac{2\mu g}{R} \times \frac{R\omega_i}{3\mu g}[/itex]
[itex]\omega_f = \omega_i - \frac{2}{3}\omega_i[/itex]
[itex]\omega_f = \frac{1}{3}\omega_i[/itex]
Hardly a eureka moment, but still :D

Thanks a lot!
 
  • #21
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Just curious, the reason I got 1 = 2 previously is because I assumed that pure rolling occurs for all time?

And if I were to solve the equation like this:

[itex]R\omega_i = 3\mu gt[/itex]
[itex]\omega_i = \frac{3\mu gt}{R}[/itex]
[itex]\omega_f = \frac{3\mu gt}{R} - \frac{2\mu g}{R}t[/itex]
[itex]\omega_f = \frac{\mu gt}{R}[/itex]

It's not what the question wants, but is it correct as well?
 
  • #22
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Just curious, the reason I got 1 = 2 previously is because I assumed that pure rolling occurs for all time?

And if I were to solve the equation like this:

[itex]R\omega_i = 3\mu gt[/itex]
[itex]\omega_i = \frac{3\mu gt}{R}[/itex]
[itex]\omega_f = \frac{3\mu gt}{R} - \frac{2\mu g}{R}t[/itex]
[itex]\omega_f = \frac{\mu gt}{R}[/itex]

It's not what the question wants, but is it correct as well?
Oh, and in addition to the above 2 questions, part c of the question is:

Assume the coefficient of friction between the disc and surface is [itex]\mu[/itex]. What is the time interval after setting the disk down before pure rolling motion begins?

So based on this, I'm assuming that that I don't need to find t in part a? Could it be a case of "over-doing", when you do too much and too complicated for a relatively simple question?
 
  • #23
Doc Al
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Just curious, the reason I got 1 = 2 previously is because I assumed that pure rolling occurs for all time?

And if I were to solve the equation like this:

[itex]R\omega_i = 3\mu gt[/itex]
[itex]\omega_i = \frac{3\mu gt}{R}[/itex]
[itex]\omega_f = \frac{3\mu gt}{R} - \frac{2\mu g}{R}t[/itex]
[itex]\omega_f = \frac{\mu gt}{R}[/itex]

It's not what the question wants, but is it correct as well?
I don't know what problem you are trying to solve here, so I can't comment on whether it's correct or not.
 
  • #24
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I don't know what problem you are trying to solve here, so I can't comment on whether it's correct or not.
I meant is it a valid derivation? Does it hold true?
 
  • #25
Doc Al
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Oh, and in addition to the above 2 questions, part c of the question is:

Assume the coefficient of friction between the disc and surface is [itex]\mu[/itex]. What is the time interval after setting the disk down before pure rolling motion begins?

So based on this, I'm assuming that that I don't need to find t in part a? Could it be a case of "over-doing", when you do too much and too complicated for a relatively simple question?
In solving part a you wrote an expression for the speed as a function of time. You solved for the final speed, so now you can just plug in that speed and solve for the time. (In solving for the final speed you didn't need to solve for the time explicitly, but for part c you do.)
 

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