Rolling different masses down a hill - Intertia Question (2025)

I have a very simple conceptual physics problem that I am looking to solve, and unfortunately I've been out of school a little too long to solve:

Suppose you have a hill, and at the bottom of the hill it levels off and keeps going on.

You take two marbles of different masses and roll them down the hill with the same force. Which marble will travel the farther distance? Ignore air resistance, AND assume that friction acts on them the same way (i.e., μ is adjusted so that F_f, the force of friction, is the same).

My understanding of inertia is that, although both marbles will reach the bottom of the hill with the same speed, the more massive marble will have more momentum/inertia, and will therefore take longer to stop. My sister's 4th grade teacher tried arguing the opposite -- that the smaller marble would travel further -- but conceptually this doesn't make any sense to me.

Second question: if you make the situation more realistic, by keeping μ constant, so F_f is greater for the more massive marble, how do you balance the greater inertia of the larger marble with it's high F_f? That is, will a more massive marble always travel further, or is there some mass ratio between the two marbles where the smaller marble will travel further?

Thanks,

Jacob

I understand the marble won't roll without friction. The point is not to ignore friction entirely -- then the question is meaningless -- but in the first case, to assume that the friction for the two marbles is constant, and for the second case, to closer simulate reality by allowing the friction to have different values.

In other words, I'm trying to separate the inertial properties from the friction properties to better understand how inertia works.

The force of friction slows the balls down. In the first experiment, both marbles have the same friction force working against the marbles (F_f marble 1 = F_f marble 2). In the second case, the more massive marble has a great friction force, because it it more massive, so the normal force is greater.

I am not trying to _ignore_ friction -- I am trying to isolate the principal of interia.

Re: Quiz question - why would they be traveling at different speeds? If the only force is gravity (assuming F_grav >> F_friction) acting on the marbles down the hill, won't they reach the bottom of the hill, at the same time, with the same speed?

Thanks for your questions and the help.

zwillingerj said:

The force of friction slows the balls down. In the first experiment, both marbles have the same friction force working against the marbles (F_f marble 1 = F_f marble 2). In the second case, the more massive marble has a great friction force, because it it more massive, so the normal force is greater.

I am not trying to _ignore_ friction -- I am trying to isolate the principal of interia.

Re: Quiz question - why would they be traveling at different speeds? If the only force is gravity (assuming F_grav >> F_friction) acting on the marbles down the hill, won't they reach the bottom of the hill, at the same time, with the same speed?

Thanks for your questions and the help.

Nope and nope. The balls are not like sliding blocks. The balls are rolling. So friction does not slow them down on the flat. They just roll with no retarding force forever.

Friction on the ramp helps spin the balls up. The more massive ball has a higher moment of inertia, so more energy goes into the rotational energy of the heavier ball. The heavier ball also has more potential energy at the top of the release, since the potential energy is mgH, and depends on the mass.

Actually, I should spend a bit more time checking the equations -- the point I was trying to make with my QQ was that there is more rotational energy in the heavier ball, so there would be less kinetic energy (less velocity). But since the heavier ball has a higher potential energy (PE) initially by virtue of its higher mass, maybe things cancel out. I need to look up the moment of inertia of a sphere. I'll try to do that tonight -- but if somebody else beats me to it and can help this poster out with the TE = KE + PE equations, that would be great.

berkeman said:

Actually, I should spend a bit more time checking the equations -- the point I was trying to make with my QQ was that there is more rotational energy in the heavier ball, so there would be less kinetic energy (less velocity). But since the heavier ball has a higher potential energy (PE) initially by virtue of its higher mass, maybe things cancel out. I need to look up the moment of inertia of a sphere. I'll try to do that tonight -- but if somebody else beats me to it and can help this poster out with the TE = KE + PE equations, that would be great.

The moment of inertia for both uniform density balls is I = (2/5) m r² where m=mass and r = radius of both balls. The rotational energy is (1/2) I w² (w = radians per sec), so if it is not slipping, the rotational energy is (2/10) mv², where v = linear velocity. The linear (center of gravity) kinetic energy is (1/2) mv².
So if the hill height is h, then

mgh = (2/10) mv² + (1/2) mv² = (7/10)mv² for both balls.

I apologize for the non-technical terms, for this question assume "resistance" = rolling resistance = rolling friction = rolling drag = whatever force you want to call it that actually slows spherical objects down as they roll.

This is the question:

Go to some actual hill. Place two spherical objects of the same size, but of different masses at the top of the hill. Push the marbles down the hill with the same force. Which marble, in real life, will actually go further? Ignore air resistance. How would this sum be quantified?

At the crux of the question is this: will the inertia of the larger ball cause it go further, once the balls are done accelerating at the bottom of the hill?

Thanks for all the help so far guys.

ryuunoseika said:

Neither you nor the teacher are seeing the whole picture: inertia both keeps the ball from rolling when gravity first starts pulling on the heavier object it and keeps it rolling when it reaches the bottom of the hill. Both marbles travel the same distance.

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?

flatmaster said:

Actually, if you completely ignore friction and air resistance, the marble doesn't even need to roll! The marble would simply glide down the slope without rolling. It's the frictional force that makes the marble roll.

Flatmaster is correct. If the slope tan(theta)>7/2 the balls will always slide, never roll, even if the coefficient of friction is 100%. If the coefficient of friction is C_f, then the critical slope angle is tan(theta)= 7 C_f/2. This does not depend on ball size, as long as they are uniform density. This slip vs. slide determination is a very interesting theoretical mechanics calculation.

Thanks for all the help thus far, but I don't believe anyone has answered the question yet.

I am not trying to determine if the balls will slide vs. roll.

I'm trying to understand the principal of interia

While slide vs. roll, coefficient of friction, friction vs. drag, etc. all might be helpful in other cases, I am trying to keep all other variables the same and ONLY change mass. How does this affect how far the marbles will roll?

Go to a hill. Take two marbles of different masses. Roll them down the hill. Which marble will go further?

As far as I see it, only one of three things will happen:

- The heavier marble roll farther, because it has more inertia at the bottom of the hill;
- Both marbles roll the same distance, because this case is independent of mass;
- The lighter marble roll farther, because it has less inertia, so it will be carried down the hill faster, and will have more velocity to propel it further;

Which one of these three things will happen?

The first is correct. The quantitative definition of "inertia" of each ball is its mass, which is proportional to

, as is its kinetic energy at the bottom of the hill. The air drag force may at first be the velocity-squared type (like automomiles), which is proportional to

times velocity-squared, but eventually becomes Stokes' Law drag force, which is proportional to

So even on a very smooth level surface (but requires rolling w/o sliding), the bigger ball will roll further.

TurtleMeister said:

Both balls will roll the same distance. This is the same thing as the inclined plane experiments done by Galileo. You can prove this by having the balls roll back up another inclined plane once they reach the bottom. No matter how steep or shallow the second inclined plane, they will always return to the same height at which they were released (minus losses due to friction). Which means they will both travel the same distance regardless of their difference in mass.

That ignores air resistance and rolling resistance, though. Still, that was part of the OP's statement.

zw -- do you understand that there has to be some kind of retarding force in order for the balls to stop?

TurtleMeister said:

I think the OP wanted the question answered both ways. If resistance to motion due to friction is the same for both objects then they will both travel the same distance. In real world experiment there would probably be greater friction for the more massive object.

I don't think friction is going to act to slow down a rolling object. The main retarding forces are air resistance and rolling resistance (like with a rubber tire and its deformation when rolling).

I don't think friction is going to act to slow down a rolling object. The main retarding forces are air resistance and rolling resistance (like with a rubber tire and its deformation when rolling).

I think resistance to motion due to friction includes those things.

edit:
This reminds me of the MythBusters episode where they pitted a Dodge Viper against a hot wheels toy car on a down hill slope. The hot wheels toy car beat the Viper handily over the first few hundred feet. That would be expected for the Viper because of the friction between the tires and road. In the end the Viper won because the hot wheels car kept flying off the track.

Bob S said:

The first is correct. The quantitative definition of "inertia" of each ball is its mass, which is proportional to

radius cubed

, as is its kinetic energy at the bottom of the hill. The air drag force may at first be the velocity-squared type (like automomiles), which is proportional to

radius-squared

times velocity-squared, but eventually becomes Stokes' Law drag force, which is proportional to

velocity times radius (no squares).

So even on a very smooth level surface (but requires rolling w/o sliding), the bigger ball will roll further.

Thanks Bob, I appreciate the help. I'm a bit confused: is what you said equivalent to saying that if the two balls are rolling, the heavier ball will reach the bottom with a great velocity than the lighter ball, and will thus propel it further? (And the opposite case, if the balls are slipping, then inertia doesn't play a factor, and the balls will travel the same speed?)

TurtleMeister said:

Both balls will roll the same distance. This is the same thing as the inclined plane experiments done by Galileo. You can prove this by having the balls roll back up another inclined plane once they reach the bottom. No matter how steep or shallow the second inclined plane, they will always return to the same height at which they were released (minus losses due to friction). Which means they will both travel the same distance regardless of their difference in mass.

Thanks, this conceptually makes sense. However, calculation wise, if we have

mgh = (2/10) mv2 + (1/2) mv2 = (7/10)mv2 for both balls, shouldn't this translate into the heavier ball going further? Or is this canceled out because it takes a greater force for the ball to be "pushed" up the second incline?

berkeman said:

That ignores air resistance and rolling resistance, though. Still, that was part of the OP's statement.

zw -- do you understand that there has to be some kind of retarding force in order for the balls to stop?

Yup, that part's clear to me. I only intended to ignore air resistance. In the first case, I meant for both balls to have the same retarding force, and in the second case, which I assume is a little more realistic, allow the balls to have a retarding force based on their masses. (Assuming the retarding force is proportional to mass?)

I appreciate the answers -- now I'm confused how to synthesize Bob's answer (heavier marble will go farther) with Turtlemeister's answer (both balls will travel the same, or accounting for realistic friction forces, the lighter ball will travel further).

Thanks so much!

Bob S said:

Flatmaster is correct. If the slope tan(theta)>7/2 the balls will always slide, never roll, even if the coefficient of friction is 100%.

This case was discussed in this thread, and assume that the spheres are iniitially sliding.

https://www.physicsforums.com/showthread.php?t=312948

bottom line was that a sphere continues to slide and never transitions into rolling if

tan(theta) >= (7/2) Cf.

The rate of linear acceleration is greater than rolling surface acceleration if tan(theta) > (7/2) Cf.
The rate of linear acceleration is the same as rolling surface acceleration if tan(theta) = (7/2) Cf.
The rate of linear acceleration is less than rolling surface acceleration if tan(theta) < (7/2) Cf.

If Cf = 1, then spheres continue to slide without transitioning into rolling if tan(theta) >= 7/2.
If Cf > 1, then the critical value for theta increases.

Getting back to the original post, assume that both balls start off with the same contact point height. The height to the center of mass is the starting height + radius x cos(slope angle). Once on the straight, the height of the center of mass is the radius. The smaller marble descends a greater distance and ends up with more speed.

Assume slope angle ~= 36.87 and cos(slope angle) = .8
Assume starting height = 10 inches
Assume radius of marble_1 is 1 inch
Assume radius of marble_2 is 2 inch

Initial height to center of mass of marble_1 = 10 + 1 x .8 = 10.8 inch
Final height to center of mass of marble_1 is 1 inch
Distance marble_1 descends = 10.8 inch - 1 inch = 9.8 inch

Initial height to center of mass of marble_2 = 10 + 2 x .8 = 11.6 inch
Final height to center of mass of marble_2 is 2 inch
Distance marble_2 descends= 11.6 inch - 2 inch = 9.6 inch

marble_1 descends 9.8 inches while marble_2 descends 9.6 inches, so marble_1 ends up with more speed than marble_2. The obvious case is when the radius of marble_2 is so large that it touches both the slope and the straight section in the iniital state and doesn't descend at all. As long as marble_2 is larger than marble_1, eventually you reach a time where marble_2 is touching the flat and stops descending, while marble_1 is still descending.

Inertia wasn't the issue here, it was the difference in the distance that the marbles descend.

I'm still thinking it's something simple like the size of the marbles causing the larger marble to reach the floor sooner and therefore stop accelerating sooner, such as shown in this diagram:

Jeff Reid said:

I'm still thinking it's something simple like the size of the marbles causing the larger marble to reach the floor sooner and therefore stop accelerating sooner, such as shown in this diagram:

Good point, Jeff.
If we have a 100-meter sprint, a soap box derby, or a Kentucky Derby, the entire body has to be behind the starting gate; no overhangs allowed. In this case, the smaller marble would reach the level plane first.
Bob S

Bob S said:

If we have a 100-meter sprint, a soap box derby, or a Kentucky Derby, the entire body has to be behind the starting gate; no overhangs allowed. In this case, the smaller marble would reach the level plane first.

If the stop gate is parallel to the floor, both marbles reach the floor at the same time with the same speed. If the stop gate is angled upwards, the larger marble is pushed back, the smaller marble reaches the floor first, and vice versa. For a soap box derby, getting the center of mass as far aft (above) of the stop gate would increase the GPE of the initial starting point.

Thanks, Jeff.

Let's assume now that both marbles have the same radius, but one is more massive than the other. That is, try to remove the factor of the larger marble reaching the bottom sooner because it's radius is larger, as illustrated in your graphic. Both marbles would then reach the bottom at the same time. (Right?) (For all these cases, I'm assuming the marble is rolling, not slipping/falling)

Also, can you explain why inertia isn't the governing principal here? In my mind, I was comparing the two marbles for two vehicles on a highway, both traveling 60 miles/hr, one being an 18-wheeler, and the other being a Volkswagen Beetle. In this case, if both drivers take their foot of the gas at the exact same time, won't the 18 wheeler take longer to stop, because of inertia/ having more momentum? (Assuming this is correct...) what makes the case of the marbles different?

zwillingerj said:

Also, can you explain why inertia isn't the governing principal here?

The rate of acceleration = force / inertia. As long as both marbles have the same density distribution, and since the force from gravity is relative to the amount of mass, then then the ratio of force / inertia is the same if the density distrubution is the same. If one of the marbles has the same mass but almost all of it's mass at the surface, like a heavy ping pong ball, then it's angular inertia is higher, and it's rate of linear acceleration down the slope would be less.

In the base of the small car versus the big truck if the coefficient of braking friction is the same for both, then the deceleration is the same for both.

If both balls are the same size but have different masses (uniform density), and the ramp angle is less than the critical angle (tan(theta) = 7C_f/2). they will reach the bottom of the ramp at the the same time with the same velocity:
mgh = (2/10) mv2 + (1/2) mv2 = (7/10)mv2, or v²= 10gh/7 for both balls.

, so the heavier ball will roll further. (the rolling resistance, like on railcar wheels or auto tires, is proportional to weight, so the heavier ball has no advantage.).
In the case of an 18-wheeler (70,000 pounds) vs. a beetle (2000 pounds), The air drag is proportional to frontal area and drag coefficient, which at worst is a factor of 20 larger for the truck, so the truck will coast farther.

Bob, thanks for the clear response, and Jeff, thanks for your thoroughness.

Bob S said:

If both balls are the same size but have different masses (uniform density), and the ramp angle is less than the critical angle (tan(theta) = 7C_f/2). they will reach the bottom of the ramp at the the same time with the same velocity:
mgh = (2/10) mv2 + (1/2) mv2 = (7/10)mv2, or v²= 10gh/7 for both balls.

The heavier ball has more momentum, but the air-drag forces are the same for both balls

, so the heavier ball will roll further. (the rolling resistance, like on railcar wheels or auto tires, is proportional to weight, so the heavier ball has no advantage.).
In the case of an 18-wheeler (70,000 pounds) vs. a beetle (2000 pounds), The air drag is proportional to frontal area and drag coefficient, which at worst is a factor of 20 larger for the truck, so the truck will coast farther.

I get this part, except re: truck vs. car, isn't drag a retarding force, so wouldn't the increased drag work against the truck?

Jeff, I wasn't sure how to fully parse your earlier answers - do you agree with Bob, that if we have the same radius but different masses for the marbles, the heavier marble will travel further? I'm trying to build some consensus here :)

berkeman said:

zw -- do you understand that there has to be some kind of retarding force in order for the balls to stop?

Very good point, so assuming friction only started after they reached the bottom of the slope and the force was equal on both balls, i'd say they'd both travel the same distance.

zwillingerj said:

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).

Jeff Reid said:

"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. :)

zwillingerj said:

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.

TurtleMeister said:

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!

zwillingerj said:

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