Barrel roll with/without slipping

[dislect]

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]

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

 

[bgq]

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

 

[dislect]

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]

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.)

 

[dislect]

Thank you for the help! much appreciated