Free cylinder on an accelerating platform

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Discussion Overview

The discussion revolves around the motion of a cylinder resting on a rough, accelerating platform. Participants explore the linear and rotational acceleration of the cylinder as the platform accelerates in the x direction, considering the implications of the no-slip condition and the effects of friction.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants propose that the linear acceleration of the cylinder is 3/2 times that of the platform, which raises questions about visualization and understanding of the motion.
  • Others argue that since there is no slipping due to the rough surface, the linear acceleration of the cylinder should match that of the platform.
  • A participant highlights the importance of the no-slip condition, suggesting that the total acceleration of the part of the cylinder in contact with the surface must equal the platform's acceleration.
  • There is a mention of rolling motion, indicating that while slipping does not occur, the cylinder still rolls as the platform accelerates.
  • One participant shares their experience of deriving the acceleration without using a pseudo force, indicating a clearer understanding of the motion.
  • Another participant discusses their preference for using an inertial reference frame to avoid confusion, emphasizing the importance of visualizing the motion correctly.

Areas of Agreement / Disagreement

Participants express differing views on the relationship between the cylinder's acceleration and that of the platform, with no consensus reached on the correct interpretation of the motion.

Contextual Notes

Participants reference the no-slip condition and the effects of friction, but the discussion does not resolve the mathematical steps or assumptions involved in deriving the cylinder's acceleration.

ash1262
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Suppose a cylinder is resting on a horizontal platform on plane x-y with its axis parallel to the y axis, and the platform accelerates in the x direction. Assume that the axis remains parallel to the y-axis and the surface is rough.
What is the motion (acceleration:linear as well as rotational) of the cylinder?
 
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hi ash1262! :wink:

tell us what you think, and why, and then we'll comment! :smile:
 
The solution says that the linear acceleration of the cylinder is 3/2 times acceleration of the platform, which I can't visualize. I feel it should be same as that of the platform as there is no slipping (the surface is rough). Again, the cylinder rolls in the opposite direction of the platform's acceleration, so how is the motion visualized.
 
call the friction force on the cylinder "F", the mass "m" and the radius "r"

what is the equation showing the effect of F on the linear motion?

what is the equation showing the effect of F on the rotational motion?
 
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a bit more advice, since this is a tricky question:
ash1262 said:
The solution says that the linear acceleration of the cylinder is 3/2 times acceleration of the platform, which I can't visualize. I feel it should be same as that of the platform as there is no slipping
That is good, you are thinking along the right lines. The 'no-slip' condition is the starting place for this problem. And as you suggest, this means the total acceleration of the bit of the cylinder next to the surface must be equal to the acceleration of the surface. But, the total acceleration of the bit of the cylinder near the surface is not equal to the linear acceleration of the cylinder. What other kind of acceleration contributes?
 
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There is no slipping but there is rolling
 
Thanks Bruce, your reply was very useful. The solution I had used a pseudo force to derive the acceleration, and I could not visualize what was really happening. Now I have derived the acceleration without using the pseudo force, and can visualize the motion!
 
hooray! nice work. yeah, pseudo forces can make things more confusing. I pretty much always use an inertial reference frame. Even when using polar coordinates, in my head, I think of
##\vec{v}= \dot{r} \hat{r} + r \dot{\theta} \hat{\theta} = \dot{x} \hat{x} + \dot{y} \hat{y}##
So I'm thinking "don't worry, underneath it all is a nice inertial reference system, with Cartesian coordinates". haha. p.s. the 'hooray' is not meant to be sarcastic. Often people think I'm being sarcastic when I'm not.
 
Thanks for the encouragement.
 

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