Rolling different masses down a hill - Intertia Question

[rcgldr]

If we have the same radius but different masses for the marbles, the heavier marble will travel further?

"Further" is dependent on the force that decelerates the marbles once they reach the floor from the slope. It would be easier to compare speed at the moment each marble reaches the floor.

If both marbles have uniform density, and the starting gate is parallel to the floor, then radius and density (as long as it's uniform) don't matter.

The only difference here is the angular moment of inertia. The higher this value, the slower the marble, cylinder, torus, ..., accelerates.

Some common examples:

http://en.wikipedia.org/wiki/List_of_moments_of_inertia

The fastest acceleration occurs when the angular inertia is zero, if all the mass is concentrated at the center of the object or if the object is sliding on a frictionless slope, I_a = 0. The slowest acceleration occurs with a hollow cylinder I_a = m r². Size and total mass don't matter (if the starting gate is parallel to the floor).

 

[zwillingerj]

"Further" is dependent on the force that decelerates the marbles once they reach the floor from the slope. It would be easier to compare speed at the moment each marble reaches the floor.

If both marbles have uniform density, and the starting gate is parallel to the floor, then radius and density (as long as it's uniform) don't matter.

The only difference here is the angular moment of inertia. The higher this value, the slower the marble, cylinder, torus, ..., accelerates.

Some common examples:

http://en.wikipedia.org/wiki/List_of_moments_of_inertia

The fastest acceleration occurs when the angular inertia is zero, if all the mass is concentrated at the center of the object or if the object is sliding on a frictionless slope, I_a = 0. The slowest acceleration occurs with a hollow cylinder I_a = m r². Size and total mass don't matter (if the starting gate is parallel to the floor).

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.

This seems to imply that a more massive marble, or a marble with a larger radius (or both), will accelerate slower, reach the flat later later, and have a slower velocity. Correct?

It seems that different people are reaching different conclusions, so I hope my confusion is understood. :)

 

[TurtleMeister]

This seems to imply that a more massive marble, or a marble with a larger radius (or both), will accelerate slower, reach the flat later later, and have a slower velocity. Correct?

It does not matter if one marble reaches the flat before the other due to distribution of mass. They will still travel the same distance (disregarding losses due to drag and friction). The kinetic energy in the marble when it reaches the flat will be divided between the forward momentum and the angular momentum. Whatever doesn't go into forward momentum will go into angular momentum and vise versa. And the energy required to stop the marble will be the sum of both.

As I've said before, you can prove all of this by setting up two inclined planes. One for acceleration and the other for deacceleration. It does not matter what angles you use for the inclined planes (except you do not want the marbles to slip). The marbles will return the to height at which they were released regardless of their mass or mass distribution.

The only factor in determining which will travel the greater distance is drag and friction.

 

[zwillingerj]

It does not matter if one marble reaches the flat before the other due to distribution of mass. They will still travel the same distance (disregarding losses due to drag and friction). The kinetic energy in the marble when it reaches the flat will be divided between the forward momentum and the angular momentum. Whatever doesn't go into forward momentum will go into angular momentum and vise versa. And the energy required to stop the marble will be the sum of both.

As I've said before, you can prove all of this by setting up two inclined planes. One for acceleration and the other for deacceleration. It does not matter what angles you use for the inclined planes (except you do not want the marbles to slip). The marbles will return the to height at which they were released regardless of their mass or mass distribution.

The only factor in determining which will travel the greater distance is drag and friction.

Thanks TurtleMeister, this is a very clear explanation.

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?

Thanks for all the help thus far!

 

[dsmith23]

I thought it was agreed that, if we assume gravity is the only force acting on the marbles as they roll down the hill, the balls will have the same speed when they reach the bottom of the hill?

what you have to take into account, assuming the balls are rolling, is that they will have different angular velocities because of the radius of their size, and because of that, they will have different linear velocities once they level out. at least that's what i think

 

[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.