Velocity of a hollow cylinder at the bottom of an incline

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Homework Statement


I was looking at the idea that a hollow cylinder will have less velocity than a solid cylinder at the bottom of an incline.

I can find the velocity of the hollow and solid cylinder from the following:

PE = Translational KE + Rotational KE

From that equation I find v and I can see that v is greater for the solid cylinder.

So, I was thinking, what about two different thick walled cylindrical tubes? One tube less hollow than the other. Which would have the greater velocity and what would be the equation for velocity for each tube?

Homework Equations



Starting again with PE = Translational KE + Rotational KE

so, mgh = 1/2mv² + 1/2 I ω²

Then the moment of inertia for a thick walled cylinder = 1/2m(r2² + r1²)
r2 being the radius of the whole cylinder
r1 being the radius of the inner hollow cylinder

Then ω = v/r (but which radius do I use???, I was thinking I should use r2, the radius of the whole cylinder)


The Attempt at a Solution



I would like a simple equation for v that includes r2 and r1 so I can see clearly how the velocity changes at the end of an incline if I had two different thick walled cylinders. I tried using the equations above, substituting into mgh = 1/2mv² + 1/2 I ω², but it's not working out for me. Any steps, guidance would be appreciated.

Cheers.
 

Answers and Replies

  • #2
gleem
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So, I was thinking, what about two different thick walled cylindrical tubes? One tube less hollow than the other. Which would have the greater velocity and what would be the equation for velocity for each tube?
PE = KEcm + KErot where KErot ∝ Mom. of Inertia Which M of I will be greater and use more of the PE.




Then ω = v/r (but which radius do I use???, I was thinking I should use r2, the radius of the whole cylinder)
Think about it. What radius determines the distance the cylinder moves down the slope per full rotation.
 
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You wrote: Then the moment of inertia for a thick walled cylinder =
Don't you mean : 1/2m(r2² - r1²)?
Also, you could solve the appropriate equations in terms of I = M k^2
where k is known as the radius of gyration.
Then you could conveniently compare solutions for various moments of inertia.
Another approach, rather than energy considerations, is to consider the torque
about the point of contact with the plane and use the moment of inertia about the point of contact
and obtain the acceleration.
Take a solid cylinder as an example:
a = g R^2 sin theta / k^2
For the solid cylinder k = 3/2 R^2 about the point of contact.
 
  • #4
gleem
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Don't you mean : 1/2m(r2² - r1²)?
I = ½m(r22 + r12) is correct
 
  • #5
haruspex
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I tried using the equations above, substituting into mgh = 1/2mv² + 1/2 I ω², but it's not working out for me.
Please post your working as far as you got and say why you find the result unsatisfactory.
 
  • #6
haruspex
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I = ½m(r22 + r12) is correct
Yes, it is a surprising result until one realises that for fixed density there is an r22-r12 term embodied in the m.
 
  • #7
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I = ½m(r22 + r12) is correct
The moment of inertia for a solid cylinder of radius R2 is 1/2 m R2^2.
If you remove a cylinder of radius R1 from the solid cylinder then you are
left with 1/2 m (R2^2 - R1^2) which is what I assumed was meant by a "thick walled cylinder".
 
  • #8
haruspex
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The moment of inertia for a solid cylinder of radius R2 is 1/2 m R2^2.
If you remove a cylinder of radius R1 from the solid cylinder then you are
left with 1/2 m (R2^2 - R1^2) which is what I assumed was meant by a "thick walled cylinder".
No, because you have also changed the mass.
Let the density be ρ.
Solid cylinder radius r2 has MoI = ½ρπLr24.
Cylinder to be removed has MoI ½ρπLr14.
Resulting MoI = ½ρπL(r24-r14)=½ρπL(r22-r12)(r22+r12)= ½M(r22+r12).
 
  • #9
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PE = KEcm + KErot where KErot ∝ Mom. of Inertia Which M of I will be greater and use more of the PE.


Think about it. What radius determines the distance the cylinder moves down the slope per full rotation.
So, I would use r2^2 as the radius when substituting v/r for ω.


Please post your working as far as you got and say why you find the result unsatisfactory.
I substituted my equations into PE = Translational KE + Rotational KE to find an expression for v that explained how the velocity changes as the internal radius of the cylinder changes.

I got this equation -

v = 2r2(gh/r2^2+r1^2)^½

So, from that equation, assuming I keep r2 the same, g remains the same and the height of the incline remains the same then the velocity will be inversely proportional to the internal radius of the cylinder? That is probably too simple for that equation. How would I describe the relationship between the velocity and the internal radius of the cylinder?

Cheers
 
  • #10
haruspex
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v = 2r2(gh/r2^2+r1^2)^½
I assume you mean ##v^2=\frac{4r_2^2gh}{r_1^2+r_2^2}##.
I get something a little different. The above doesn't give the right result as r1 tends to r2.
There is no way to describe the relationship in words simply. You could just say that as r1 increases v decreases.
 

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