Rotational moment of inertia for 3 balls on a pipe

In summary, the conversation revolved around a problem involving the moment of inertia of a rigid pipe rotating about the y-axis. There was confusion about which axis the rotation was actually taking place in, with some suggesting the x-axis and others suggesting the z-axis. There were also questions about the value of R in the equation 2(MR^2), with some suggesting it should be the distance of the centers of the large spheres from the axis of rotation. There were also doubts about the accuracy of the given information and assumptions made in the problem. Overall, there was a lot of uncertainty and confusion about how to approach and solve the problem.
  • #1
SakuRERE
68
5

Homework Statement



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Homework Equations

The Attempt at a Solution


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i just don't know from where to take the R which is colored with brown, in the 2(MR^2), what is the R here exactly, because i took the radius of the sphere , but i get different answer althought it's near.
 

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  • #2
Why is the moment of inertia of the rigid pipe about the y-axis ##(1/12)m L^2##? That's the moment of inertia about the x-axis passing through the mid point of the rod.
 
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  • #3
kuruman said:
Why is the moment of inertia of the rigid pipe about the y-axis ##(1/12)m L^2##? That's the moment of inertia about the x-axis passing through the mid point of the rod.
yeah the sysytem will be rotating around z ( perpendicular to the page) as we have (x,y) fixed point, so the system will be rotating in the plane of x and y but around z. so the axis of rotation z is perpendicular to the pipe and at the center of it, so that why i used 1/12.
 
  • #4
SakuRERE said:
the sysytem will be rotating around z
Why do you think that?
The text says it is about the y axis. However, that wouid mean the information about the length of the rod and the spacing of the balls is irrelevant. I notice also the dashed horizontal line through the mass centre. All of this suggests the text is wrong and that the rotation is about the x axis.
 
  • #5
I am dubious of their assumption that the two identical balls can be "treated as hoop material". I think that will introduce a significant error (I estimate a bit more 10 %).
SakuRERE said:
i just don't know from where to take the R which is colored with brown, in the 2(MR^2), what is the R here exactly, because i took the radius of the sphere , but i get different answer althought it's near.
I would think that it should be the distance of the centers of the large spheres from the axis of rotation.
 
  • #6
kuruman said:
That's the moment of inertia about the x-axis passing through the mid point of the rod.
Not quite. The rod has a given diameter.
 
  • #7
haruspex said:
Why do you think that?
The text says it is about the y axis. However, that wouid mean the information about the length of the rod and the spacing of the balls is irrelevant. I notice also the dashed horizontal line through the mass centre. All of this suggests the text is wrong and that the rotation is about the x axis.
you mean it will not move like a pinwheel?
 
  • #8
gneill said:
I am dubious of their assumption that the two identical balls can be "treated as hoop material". I think that will introduce a significant error (I estimate a bit more 10 %).

I would think that it should be the distance of the centers of the large spheres from the axis of rotation.
so the radius of the big sphere + 2.5+ the radius of the center sphere?
 
  • #9
SakuRERE said:
so the radius of the big sphere + 2.5+ the radius of the center sphere?
That would be my view, yes.
 
  • #10
SakuRERE said:
you mean it will not move like a pinwheel?
Because of the rotational symmetry about the y axis, it actually doesn't matter whether it is the x-axis or the z axis. I was just puzzled that you thought it was the z axis. As I posted, the text is probably wrong, so using x or z should be fine.
 
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  • #11
haruspex said:
Because of the symmetry, it actually doesn't matter whether it is the x-axis or the z axis. I was just puzzled that you thought it was the z axis. As I posted, the text is probably wrong, so using x or z should be fine.
yeah i understood that the text says , it will rotate in the plane of x-y. but to make sure, it will move like a pin wheel right? and what about the R so? which R should i take?
 
  • #12
SakuRERE said:
i get different answer althought it's near.
Please note post #6.
 
  • #13
haruspex said:
Please note post #6.
so how am i supposed to solve it. i am so confused really
 
  • #14
SakuRERE said:
yeah i understood that the text says , it will rotate in the plane of x-y
No, the text says it will rotate about the y axis, about the axis of the rod, so in the XZ plane. But that is at odds with several other clues, which indicate rotation in the YZ plane. Yes, like a pinwheel, but about the x axis, not the z axis. Anyway, as I posted, the answer is the same for X and Z.
 
  • #15
haruspex said:
Not quite. The rod has a given diameter.
Yes of course, I should have said "That's the moment of inertia about the x-axis passing through the mid point of a very thin rod. Thank you for the correction. The point remains that the moment of inertia used by OP is incorrect. I am also troubled by the assertion in the solution that the two outer spheres are treated as point masses with moment of inertia each ##Mr^2## instead of using the parallel axis theorem.
 
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  • #16
kuruman said:
Yes of course, I should have said "That's the moment of inertia about the x-axis passing through the mid point of a very thin rod. Thank you for the correction. The point remains that the moment of inertia used by OP is incorrect. I am also troubled by the assertion in the solution that the two outer spheres are treated as point masses with moment of inertia each ##Mr^2## instead of using the parallel axis theorem.
i am really surprised how could this be wrong, but this is in our lecture notes (college) and it's only 4 hours left for the exam. so is it hard to learn the parallel axis theorem
 
  • #17
SakuRERE said:
so how am i supposed to solve it. i am so confused really
You need another formula. It looks like you are not expected to be able to derive these formulae for yourself, so I will show you.
Consider a disc slice of the rod, radius r, thickness dy, at distance y from the x axis. Its mass is mdy/Y, where Y is the length of the rod and m its mass.
By the parallel axis theorem, its MoI is (mdy/Y)[¼r2+y2].
Integrating, ¼r2m+(1/12)mY2.
 
  • #18
kuruman said:
troubled by the assertion in the solution that the two outer spheres are treated as point masses
Yes, that is clearly wrong.
SakuRERE said:
this is in our lecture notes
I suggest you have taken something out of context. It may have been a reasonable approximation in a particular set-up, but it will not fly in general and is quite wrong here.
 
  • #19
haruspex said:
You need another formula. It looks like you are not expected to be able to derive these formulae for yourself, so I will show you.
Consider a disc slice of the rod, radius r, thickness dy, at distance y from the x axis. Its mass is mdy/Y, where Y is the length of the rod and m its mass.
By the parallel axis theorem, its MoI is (mdy/Y)[¼r2+y2].
Integrating, ¼r2m+(1/12)mY2.
wow, I don't think i am going to get this now, anyways, thanks everyone.
 
  • #20
haruspex said:
Yes, that is clearly wrong.

I suggest you have taken something out of context. It may have been a reasonable approximation in a particular set-up, but it will not fly in general and is quite wrong here.
this is really so disappointing, unfortunately
 
  • #21
SakuRERE said:
i am really surprised how could this be wrong, but this is in our lecture notes (college) and it's only 4 hours left for the exam. so is it hard to learn the parallel axis theorem
The basic premise of the Parallel Axis Theorem is that if you know the MoI for some object of mass M about a given axis, then the MoI for that same object about another axis parallel to that first axis is given by:

##MoI_{new} = MoI + Md^2##

Where d is the distance between the new axis and the original axis.

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  • #22
SakuRERE said:
this is really so disappointing, unfortunately
It's not that complicated really, but it looks like you have been poorly taught.
You have three solid spheres, and for each you can use the standard (2/5)mr2 formula. The outer two are off-axis, so you need to use the parallel axis theorem. Effectively, this says that each of these spheres can be represented by a sphere of the same mass and radius located at the axis, plus a point mass of the same mass located at the actual centres of the spheres.
Similarly, it turns out that a solid cylinder (thick rod) rotating about an axis normal to its own can be treated as a thin disc of the same mass located at the axis (as a diameter of the disc) plus a thin rod of the same mass.
 
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What is rotational moment of inertia?

Rotational moment of inertia, also known as moment of inertia or angular mass, is a measure of an object's resistance to rotational motion. It is analogous to mass in linear motion.

How is rotational moment of inertia calculated?

The rotational moment of inertia for a single object can be calculated using the formula I = mr^2, where I is the moment of inertia, m is the mass of the object, and r is the distance from the axis of rotation.

For multiple objects, the moment of inertia can be calculated by adding the individual moments of inertia together.

What is the significance of rotational moment of inertia for 3 balls on a pipe?

In the context of 3 balls on a pipe, rotational moment of inertia is important because it determines how easily the balls will rotate around the pipe. A higher moment of inertia means the balls will require more force to rotate, while a lower moment of inertia means they will rotate more easily.

How does the placement of the balls affect the rotational moment of inertia?

The placement of the balls on the pipe can affect the rotational moment of inertia. For example, if the balls are placed closer to the axis of rotation, the moment of inertia will be lower compared to if they are placed further away.

Can the rotational moment of inertia for 3 balls on a pipe be changed?

Yes, the rotational moment of inertia can be changed by altering the mass or the distribution of the mass of the balls. For example, adding a heavier ball to the pipe will increase the overall moment of inertia, while moving the balls further from the axis of rotation will also increase the moment of inertia.

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