A problem on rotation

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Homework Statement


A horizontal platform rotates around a vertical axis
at angular velocity ω. A disk with radius R can freely rotate
and move up and down along a slippery vertical axle situated
at distance d > R from the platform’s axis. The disk is pressed
against the rotating platform due to gravity, the coefficient of
friction between them is µ. Find the angular velocity acquired
by the disk. Assume that pressure is distributed evenly over
the entire base of the disk.
d

Homework Equations


[/B]


The Attempt at a Solution


My problem is with understanding the problem in first place. The diagram given is probably not correct and if is can someone help me out to find what actually is the problem about? I feel the disk radius as given R is wrong.
ua7ic.png
 

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  • #2
kuruman
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The drawing is consistent with the inequality ##d > r##. If ##d>R## were true, the disk would be overhanging the platform. Assuming that the drawing is correct, what are your thoughts about the problem? How would you proceed to answer the question? Hint: About what axis (or axes) does the disk rotate?
 
  • #3
haruspex
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The text would match the diagram if everywhere it mentions R were changed to r. Although the diagram shows the plafform as a disk that is clearly irrelevant, so the R in the diagram is irrelevant.
I am more concerned with the rest of the question. No time is specified, so presumably were are being asked for the asymptotic state. It seems somewhat obvious for any μ>0. Maybe I am missing something.
 
  • #4
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The drawing is consistent with the inequality ##d > r##. If ##d>R## were true, the disk would be overhanging the platform. Assuming that the drawing is correct, what are your thoughts about the problem? How would you proceed to answer the question? Hint: About what axis (or axes) does the disk rotate?
So the platform is the disk with radius R and our target disk is one with radius r? And i am confused whether the axis around which smaller disc is rotating is not moving?
 
  • #5
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The text would match the diagram if everywhere it mentions R were changed to r. Although the diagram shows the plafform as a disk that is clearly irrelevant, so the R in the diagram is irrelevant.
I am more concerned with the rest of the question. No time is specified, so presumably were are being asked for the asymptotic state. It seems somewhat obvious for any μ>0. Maybe I am missing something.
That is what i was confused about. Also does it mean that the axis for smaller disk is still and not moving?
 
  • #6
kuruman
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That is what i was confused about. Also does it mean that the axis for smaller disk is still and not moving?
I believe the statement that the coefficient of friction is μ is placed there to indicate that the disk is (and remains) at rest relative to the platform. I don't think there is much to this problem except for the realization that the angular frequency of the disk is not ω but its value is ...
 
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  • #7
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I believe the statement that the coefficient of friction is μ is there to indicate that the disk is (and remains) at rest relative to the platform. I don't think there is much to this problem except for the realization that the angular frequency of the disk is not ω but its value is ...
I still dont get how to do it. Do we have to apply lamis theorum to get ω for which net frictional force is cancelled out?
 
  • #8
kuruman
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I still dont get how to do it. Do we have to apply lamis theorum to get ω for which net frictional force is cancelled out?
When the platform goes around once, how many revolutions does the disk make? This is the same situation as the Moon orbiting the Earth and showing the same face to the Earth at all times.
 
  • #9
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When the platform goes around once, how many revolutions does the disk make? This is the same situation as the Moon orbiting the Earth and showing the same face to the Earth at all times.
Thank you. If in that way its done then the answer is -w (omega).
But how did you get that the disk behaves like this? Could you elaborate what is happening in this situation?
 
  • #10
kuruman
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Thank you. If in that way its done then the answer is -w (omega).
Incorrect. This means that the disk rotates in the opposite sense from the platform. Does it look like it's doing that?
But how did you get that the disk behaves like this? Could you elaborate what is happening in this situation?
Put a coin (the disk) on a book (the platform) and use your hands to turn the platform not too fast to make sure that the coin doesn't slip. Observe how the disk behaves.
 
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When the platform goes around once, how many revolutions does the disk make? This is the same situation as the Moon orbiting the Earth and showing the same face to the Earth at all times.
Is it that the coin is rotating arouns itself once?
 
  • #12
kuruman
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Is it that the coin is rotating arouns itself once?
That's only part of the rotation. Is there another rotation about some other axis?
 
  • #13
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That's only part of the rotation. Is there another rotation about some other axis?
Around the axis of platform
 
  • #14
kuruman
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Right. There is an angular velocity associated with each of these rotations, call them ωitself and ωplatform, the subscripts indicating the appropriate axis of rotation. How are they related to the given angular velocity ω of the platform?
 
  • #15
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Right. There is an angular velocity associated with each of these rotations, call them ωitself and ωplatform, the subscripts indicating the appropriate axis of rotation. How are they related to the given angular velocity ω of the platform?
w platform =w and
w itself=w ( i doubt myconclusion)
So do we have to find the w around instantaneous axis?
The given answer is -w in the booklet.could it be wrt a different frame?
 
  • #16
kuruman
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If the answer -ω is correct and not a typo, then I have misinterpreted what's going on and I am sorry if I have misled you. However, I cannot possibly see how that can be the correct answer given the diagram and the statement of the problem. The statement "find the angular velocity acquired by the disk" indicates that the disk has (probably) zero angular velocity and then when it's "pressed" by gravity onto the platform, it acquires some angular velocity. However, I cannot see how and why this velocity is opposite to the direction of the platform's angular velocity according to the given answer. Let me think about it, but meanwhile maybe someone else can contribute a suggestion.
 
  • #17
haruspex
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I believe the statement that the coefficient of friction is μ is placed there to indicate that the disk is (and remains) at rest relative to the platform.
I misread that at first, but as a result it dawned on me that perhaps the axle is stationary. Suddenly the friction and the distance d become significant. However it makes for a very difficult problem. Certainly the asymptotic value as d tends to infinity would be zero.
the angular frequency of the disk is not ω
Umm.. with the view that the axle is fixed to the platform why would it not be ω? In your moon analogy, the moon orbits the earth once a month and rotates on its own axis, in the same sense, once a month.

I cannot think of an interpretation that leads to -ω.
 
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  • #18
kuruman
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Umm.. with the view that the axle is fixed to the platform why would it not be ω?
My thinking is that there is angular momentum of the disk about its center of mass and of its center of mass. Since the two angular momenta are collinear, the answer to the question might be the sum of the two angular velocities.
I cannot think of an interpretation that leads to -ω.
Same here.
 
  • #19
haruspex
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My thinking is that there is angular momentum of the disk about its center of mass and of its center of mass. Since the two angular momenta are collinear, the answer to the question might be the sum of the two angular velocities.

Same here.
Angular momentum is not going to be conserved, so I doubt it helps to consider that.
 
  • #20
I will attempt to help you with a solution.

First of all, I will comment, as many other users here have commented, that the problem statement and the diagram are not consistent. I will assume in this attempt of a solution that the diagram is the correct representation of the problem; the contradictions in the problem statement will be ignored.

In general, in physics and engineering you should learn to be able to recognize and classify different types of problems.
This problem here is what is called a steady-state problem; your goal is to find the equilibrium situation, which is when the system is in the same state for the rest of eternity.

The goal is to find the steady-state angular velocity [itex]\omega_f[/itex] of the smaller disk. Before the disk collides with the rotating platform, it (I assume) has an angular velocity of 0 - i.e. it is not rotating. After the disk collides with the platform, the platform exerts a net torque on the disk about the disk's central axis, which induces an angular acceleration.

After a sufficient amount of time has passed, the disk will have achieved its constant steady-state velocity [itex]\omega_f[/itex]. A constant angular velocity implies that the net torque exerted by the platform onto the disk is 0.

You might be wondering why I italicized the "net" in net torque in the preceding sentence; this is because the platform exerts a non-zero torque [itex]d\vec{T}(\vec{s})[/itex] at every point on the bottom surface of the disk, where [itex]\vec{s}[/itex] is the position vector of that point with respect to the central axis of the disk. We have the following equation for torque:
[tex]d\vec{T}(\vec{s}) = \vec{s}\times d\vec{F}(\vec{s}) [/tex]

You can make an expression for the force [itex]d\vec{F}(\vec{s})[/itex] at any given point [itex]\vec{s}[/itex] using information in the problem statement.

[itex]\omega[/itex] and [itex]\omega_f[/itex] are instrumental in determining the direction of the force at any given point. Why?

I've already done most of the problem for you. The rest is up to you.
 
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  • #21
haruspex
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I've already done most of the problem for you.
That's not how I see it. (I believe you have the same interpretation that I suggested, that the axle of the disk is fixed in the lab frame.)
The direction of the force is same as the direction of the linear velocity of the point on the platform with respect to the corresponding (touching) point on the bottom of the disk
Let me take it a bit further. Let ##\vec d## be the position vector of the axle relative to the platform's axis, ##\vec r## the position vector of some disk element relative to the disk's centre, and ##\vec s ## be the position vector of the same element relative to the platform's axis. So ##\vec s= \vec d+\vec r##.
If ##\vec \omega## and ##\vec \omega'## are the rotation vectors of the platform and disk respectively then at the element the velocities are ##\vec s\times\vec \omega## and ##\vec r\times\vec\omega'##. The relative velocity is ##\vec v=\vec r\times\vec\omega'-\vec s\times\vec \omega##, and the unit vector in that direction is ##\frac{\vec v}{|\vec v|}##. Multiplying that by a constant and taking the cross product with ##\vec r## produces the torque on the disk per unit area of the element due to that element. Finally, multiplying that by the area of the element and integrating over the disk gives the overall torque. The task is to find the value of ω' that makes that zero.
So that has taken the analysis somewhat further, yet still the hard part, the integral, lies ahead.

@jbriggs444 found the identical problem online as an Olympiad problem. It sketches the solution so vaguely that I am unconvinced their solution is valid.

Edit: wrt the proposed answer of -ω (the answer is not given at the olympiad link, so maybe that should not be trusted), it is fairly clear that for d=0 it would be +ω, as d tends to infinity it would tend to 0, and that for every other d it would be between those values.
 
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  • #22
That's not how I see it. (I believe you have the same interpretation that I suggested, that the axle of the disk is fixed in the lab frame.)

Let me take it a bit further. Let ##\vec d## be the position vector of the axle relative to the platform's axis, ##\vec r## the position vector of some disk element relative to the disk's centre, and ##\vec s ## be the position vector of the same element relative to the platform's axis. So ##\vec s= \vec d+\vec r##.
If ##\vec \omega## and ##\vec \omega'## are the rotation vectors of the platform and disk respectively then at the element the velocities are ##\vec s\times\vec \omega## and ##\vec r\times\vec\omega'##. The relative velocity is ##\vec v=\vec r\times\vec\omega'-\vec s\times\vec \omega##, and the unit vector in that direction is ##\frac{\vec v}{|\vec v|}##. Multiplying that by a constant and taking the cross product with ##\vec r## produces the torque on the disk per unit area of the element due to that element. Finally, multiplying that by the area of the element and integrating over the disk gives the overall torque. The task is to find the value of ω' that makes that zero.
So that has taken the analysis somewhat further, yet still the hard part, the integral, lies ahead.

@jbriggs444 found the identical problem online as an Olympiad problem. It sketches the solution so vaguely that I am unconvinced their solution is valid.
By "most of the problem," I meant strategy formulation, which is the most important part of solving physics problems.
 
  • #24
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That's not how I see it. (I believe you have the same interpretation that I suggested, that the axle of the disk is fixed in the lab frame.)

Let me take it a bit further. Let ##\vec d## be the position vector of the axle relative to the platform's axis, ##\vec r## the position vector of some disk element relative to the disk's centre, and ##\vec s ## be the position vector of the same element relative to the platform's axis. So ##\vec s= \vec d+\vec r##.
If ##\vec \omega## and ##\vec \omega'## are the rotation vectors of the platform and disk respectively then at the element the velocities are ##\vec s\times\vec \omega## and ##\vec r\times\vec\omega'##. The relative velocity is ##\vec v=\vec r\times\vec\omega'-\vec s\times\vec \omega##, and the unit vector in that direction is ##\frac{\vec v}{|\vec v|}##. Multiplying that by a constant and taking the cross product with ##\vec r## produces the torque on the disk per unit area of the element due to that element. Finally, multiplying that by the area of the element and integrating over the disk gives the overall torque. The task is to find the value of ω' that makes that zero.
So that has taken the analysis somewhat further, yet still the hard part, the integral, lies ahead.

@jbriggs444 found the identical problem online as an Olympiad problem. It sketches the solution so vaguely that I am unconvinced their solution is valid.

Edit: wrt the proposed answer of -ω (the answer is not given at the olympiad link, so maybe that should not be trusted), it is fairly clear that for d=0 it would be +ω, as d tends to infinity it would tend to 0, and that for every other d it would be between those values.
Thank you. I also tried doing that way but stopped because i was confused by the problem. I will work in this direction further now
 

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