A and B barrels

[devrimci_kürt]

Two similiar barrels A and B with the same radius and mass roll down the same slope starting at the same instant.Barrel A is filled with liqued water, and barrel B is filled with ice of equal mass(Barrel B slightly longer than barrel A to compensate for the slightly larger density of liquid water)

 

which barrel reaches the bottom of the slope first?

 

and why?

[Fuzzwood]

If barrel B is longer, it will endure more roll and air friction.

[npts2020]

If barrel B is longer, it will endure more roll and air friction.

 

What if the ice were in an identical barrel, either under pressure or one or both barrels not completely full? It seems to me that they would reach the bottom at the same time if both are filled to the top but I am unsure how water sloshing around would change the dynamics, especially if the barrel did not roll fast enough for centrifugal force to even out the level of water on the inside.

[Mr Skeptic]

We are assuming the water has zero rotational kinetic energy, right?

[Kyrisch]

Since the water in the barrel will not rotate nearly as much as the ice we can ignore the angular velocity of the water completely and we have the following two statements of energy conservation:

 

[math]GPE_A = \frac{1}{2}Mv_A^2 + \frac{1}{2}I_{ice}\omega^2[/math]

and

[math]GPE_B = \frac{1}{2}Mv_B^2[/math]

 

Since the energy necessary to effect the rotation of the ice inside the barrel is nonzero, the linear velocity of the barrel must be less than that of the barrel containing water. Therefore the water-filled barrel will reach the ground first.

[swansont]

http://blogs.scienceforums.net/swansont/archives/2066

[J.C.MacSwell]

Since the water in the barrel will not rotate nearly as much as the ice we can ignore the angular velocity of the water completely and we have the following two statements of energy conservation:

 

[math]GPE_A = \frac{1}{2}Mv_A^2 + \frac{1}{2}I_{ice}\omega^2[/math]

and

[math]GPE_B = \frac{1}{2}Mv_B^2[/math]

 

Since the energy necessary to effect the rotation of the ice inside the barrel is nonzero, the linear velocity of the barrel must be less than that of the barrel containing water. Therefore the water-filled barrel will reach the ground first.

 

Agree under your assumptions, however, if the slope is long enough the increased friction from the water will slow the water barrel down and the ice barrel wins!