Barrel roll with/without slipping

  • Thread starter 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
 

Doc Al

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Start by drawing free body diagrams for each situation. Then you can calculate the acceleration.
 

bgq

160
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Do what Doc Al have told you, the final result is that the barrel that slips down falls faster than the one that rolls.
 
166
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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?
 

Doc Al

Mentor
44,815
1,078
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.)
 
166
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Thank you for the help! much appreciated
 

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