Barrel roll with/without slipping

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Discussion Overview

The discussion revolves around the comparison of a barrel rolling without slipping versus one that slips down an inclined plane. Participants explore the mathematical implications of each scenario to determine which barrel reaches the bottom faster.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant questions which barrel will descend faster and seeks a mathematical demonstration.
  • Another participant suggests starting with free body diagrams to calculate acceleration for both scenarios.
  • It is proposed that the barrel that slips will fall faster than the one that rolls without slipping.
  • Mathematical derivations are presented for both cases, showing that for rolling without slipping, the velocity at the bottom is V=√(4/3)gh, while for slipping, it is V=√(2)gh.
  • One participant confirms the calculations and notes that the modeled barrel as a solid cylinder is acceptable.

Areas of Agreement / Disagreement

Participants generally agree that the barrel that slips reaches the bottom faster, but there is no explicit consensus on the implications of the calculations or the conditions under which they hold.

Contextual Notes

The discussion relies on the assumption that the barrels are modeled as solid cylinders, and the calculations depend on this definition. There may be limitations regarding the applicability of the results to different shapes or conditions not discussed.

Who May Find This Useful

This discussion may be useful for students or enthusiasts interested in classical mechanics, particularly in understanding the dynamics of rolling and slipping objects on inclined planes.

dislect
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Hi guys,

I'm wondering if I roll a barrel which rolls without slipping down an inclined plane vs. a barrel which only slips down the plane - who will get down faster and how can I show it mathematically ?


Thanks
 
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Start by drawing free body diagrams for each situation. Then you can calculate the acceleration.
 
Do what Doc Al have told you, the final result is that the barrel that slips down falls faster than the one that rolls.
 
For rolling without slip:
mgh=0.5Iω + 0.5mV^2, I=0.5MR^2 ->
mgh=0.25mR^2 * (V/R)^2 +0.5MV^2
gh=0.25V^2 + 0.5V^2 ->
V=√4/3 gh

For slip only:
mgh=0.5mV^2 ->
V=√2gh

and so the Velocity at the bottom of the inclined plane is higher for slip only. is that correct?
 
dislect said:
For rolling without slip:
mgh=0.5Iω + 0.5mV^2, I=0.5MR^2 ->
mgh=0.25mR^2 * (V/R)^2 +0.5MV^2
gh=0.25V^2 + 0.5V^2 ->
V=√4/3 gh

For slip only:
mgh=0.5mV^2 ->
V=√2gh

and so the Velocity at the bottom of the inclined plane is higher for slip only. is that correct?
That's correct. (I see you modeled the barrel as a solid cylinder, which is OK.)
 
Thank you for the help! much appreciated
 

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