Rolling different masses down a hill - Intertia Question

[Bob S]

It seems Bob S reaches the opposite conclusion -- the friction will affect the less massive marble more, so the more massive marble will travel further.
!

My conclusion was that rolling friction force is proportional to mass (e.g., rolling wheel), so the heavier ball has more rolling friction in direct proportion to its mass; hence there is no advantage or disadvantage in being a heavier ball. But for balls of equal density, their radius scales as the cube root of mass, while the air drag force scales only linerarly or quadratically (=frontal area) with radius, so a ball with twice the radius has 8 times the mass but only at most 4 times the air drag force.

 

[TurtleMeister]

To change gears, you surmised (from an earlier post) that the friction for the more massive marble will be greater, and thus the smaller marble will travel further.

It seems Bob S reaches the opposite conclusion -- the friction will affect the less massive marble more, so the more massive marble will travel further.

Assuming I understand both of these posts, from where does the difference of opinion arise?

I am not in total disagreement with Bob. Either marble could travel a greater distanced than the other, depending on the forces acting on them. If air drag is the only factor considered then the marble of greater mass will be affected less. If rolling resistance is the only factor considered then the lower mass marble will be affected less. It would be difficult to predict these forces. Remember my reference to the MythBusters episode? The toy car beat the much more massive Dodge Viper because the Viper had greater rolling resistance.

When dealing with forces which are difficult to predict, would it not be best to just leave out those forces and say that both marbles will travel the same distance (disregarding losses due to friction)? Of course if you disregarded friction the marbles would have to roll uphill, otherwise they would never stop. :)

 

[zwillingerj]

OK, great, I think I've got a fairly good handle on this now.

Turtlemeister, your suggestion to ignore the "hard to quantify forces" is a very good idea, and makes the concepts much easier to understand.

Thanks for all the help, everyone!

 

[rcgldr]

For a solid, uniform dense sphere, I = 2mr^2/5. This would imply that a more massive object, or an object with a larger radius (or both), will have a higher moment of inertia.

Since the force from gravity increases or decreases directly proportional to the amount of mass, the linear acceleration and linear speed on the slope are the same, because the linear acceleration and linear speed remain the same as long as I / (m r²) is the same. For a solid sphere (uniform density), I / (m r²) = 2/5. For a solid cylinder (uniform density) I / (m r²) = 1/2 so it's linear acceleration is slower. For a hollow sphere, I / (m r²) = 2/3, so it's linear acceleration is slower still. For a hollow cylinder, I / (m r²) = 1, and it's the slowest of the group.

 

[Bob S]

Since the force from gravity increases or decreases directly proportional to the amount of mass, the linear acceleration and linear speed on the slope are the same, because the linear acceleration and linear speed remain the same as long as I / (m r²) is the same. For a solid sphere (uniform density), I / (m r²) = 2/5. For a solid cylinder (uniform density) I / (m r²) = 1/2 so it's linear acceleration is slower. For a hollow sphere, I / (m r²) = 2/3, so it's linear acceleration is slower still. For a hollow cylinder, I / (m r²) = 1, and it's the slowest of the group.

Precisely. We all are assuming that the balls are rolling without slipping. This is the case when the ramp slope angle is less than theta = arctan(7C_f/2), where C_f is the coefficient of friction of the balls on the ramp. The problem is much more interesting when for example the coefficient of friction (both static and sliding) were 0.1, and theta were 30 degrees. The balls would be slipping, but they will be gaining angular momentum as well as linear momentum. Besides linear and rotational energy, which are conservative, there also has to be heat, because sliding friction = F_frict dx is work.