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Cylinders on an incline

  • Thread starter fruitl00p
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  • #1
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Homework Statement



The five masses below all have the same radius and a cylindrically symmetric mass distribution. They start to roll down an inclined plane, starting from rest, at the same time and from the same height. Give theri order of arrival at the bottom (i.e., ABCDE, DCBAE)

A. Icm= 728gcm^2, M =53 g
B. Icm= 669gcm^2, M= 47 g
C. Icm= 686 gcm^2 M= 41 g
D. Icm= 830 gcm^2 M= 41 g
E. Icm= 728 gcm^2 M= 50 g


Homework Equations



I= 1/2 MR^2

The Attempt at a Solution



Though this is a homework question, I simply want to understand why a larger mass may reach the bottom of the incline first while a smaller one trails behind, etc. I understand that the larger the mass, the larger the moment of inertia and therefore the longer it takes to stop and start motion. I understand the mass distribution effects the moment of inertia ( as in, if the mass is distributed towards the center of object, the moment of inertia is smaller and therefore takes less time to move). Yet, when I look at a scenario like the above question, I do not understand how to comprehend the order of arrival at the bottom.

I would think that the first one to arrive at the bottom of the incline would be B because the moment of inertia is the smallest; then C because even though the mass of C and D is the same, the moment of inertia indicates that the mass in distributed differently; A would be next because even though the mass is larger than E, they have the same moment of inertia which makes me think that the mass of A is distributed in a way that makes the amount of time it takes to reach the bottom shorter; after that E and lastly D since the moment of inertia is significantly larger than all of the masses.

However, I've been told that is the incorrect answer. Though I want to get the correct answer, I rather understand what components I do not understand about moments of inertia of an object.

I hope my information is sufficient and coherent.:blushing:
 

Answers and Replies

  • #2
682
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A nice way of tackling these problems is to look at the total energy (it's nice becuase the change is gravitational energy must be the same for all of them!)

[tex]KE=\frac{1}{2}I\omega^2+\frac{1}{2}mv^2[/tex]

assume rolling without slipping (very important... you must make sure that you can make such assumption otherwise the question wouldn't make sense at all)

rolling without slipping implies
[tex]r\omega=v[/tex]

put v in terms of energys, what kind of expression do you get?

basically you should get something involving I/r^2 and m. how does v change with these two variables? what can you say if an object has a greater v compared to others (when the height is the same)? notice that it is not just aboue the rotational inertia, it is a combination of rotational inertia and the mass.
 
Last edited:
  • #3
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If you write down the Lagrangian for this setup,

L = Kinetic energy - pot. energy you get -

[tex]L = \frac{1}{2}mr^2\dot{\theta^2} + \frac{1}{2}m\dot{x^2} - mgh[/tex]

Notice that m cancels out but there is still a dependence on r ( radius), so forget the masses, just consider the radii.

[edit] Ouch - I left an m out[edit]
 
Last edited:
  • #4
Doc Al
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Newton's 2nd law

Notice that m cancels out but there is still a dependence on r ( radius), so forget the masses, just consider the radii.
Don't forget the "rolling without slipping" constraint.

In addition to examining energy, per tim_lou's suggestion, why not just apply Newton's 2nd law and solve for the translational acceleration? It's easy! (Hint: Apply it twice; once for rotation, once for translation.)
 
  • #5
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Oh I understand now!! :smile: Thank you very much.
 
  • #6
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I dont understand still. Im working on the same type of problem...
 

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