Rotational Momentum

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If two balls of a different size but equal mass roll down the same incline at the same which has more energy? My calculus and physics are weak, but I know that the larger accelerates slower and has a greater moment of inertia.

However, will both register the same force on an intervening scale at the bottom? The moment of inertia may be greater for the larger mass, but the forward rate of the smaller is greater at the bottom.

I guess my question is, does one have a greater momentum or energy? And why?
 
For that matter, does the center of either ball have a greater forward rate? or acceleration?
 
If momentum is generically known for its mass times acceleration and the smaller ball accelerates faster, it does have the greater momentum/or energy even if it has the same mass as the larger ball, right?

my problem is why one requires more generating force to move but the other has a greater acceleration. they are equal in substance, equal in applied force. one requires greater generating force, yet one has greater acceleration. are they equal?
 
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If both of the balls have the same mass and start at the same height then they will have the same amount of energy. You can use the gravitational potential energy equation to see why: P = mgh. P is the gravitational potential energy and it is the force of gravity that does P amount of work on the balls in order to move them. The amount of energy, P, is deposited into both the Kinetic and Rotational Kinetic energies of the two balls.

The ball with greater inertia requires more rotational kinetic energy to roll. The Moment of inertia is a way of describing how hard it is to rotate a body about an axis. The more mass a body has around an axis and the further it is away from the axis the harder it will be to rotate it, therefore there is more energy needed.

That means when P, the amount of gravitational potential energy, is given to the larger ball, most of the energy will go into rotational kinetic energy to rotate the ball and a little bit will go into kinetic energy to give the ball a forward velocity. There is less rotational kinetic energy needed to rotate the smaller ball, so a larger portion of P will go into it's kinetic energy and it will move down the inclined plane faster.

The impact force of the smaller ball will be greater than the larger ball if they are of the same mass, because it's kinetic energy gives it a larger linear momentum and a transfer of momentum between two bodies is a way of describing the impact force.

The important thing to realize is that both of them can have the same amount of energy, but the dynamics of how the balls move depends on where that energy goes.

Here is an interesting thing to ponder.
Imagine that one of the balls were somehow electrically connected to a circuit so that if you pressed a button the motion of the ball would power a current that crossed through a large resistor.
Would the ball stop moving?
Would it slow down?
Would the resistor make any difference?
 

Doc Al

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If two balls of a different size but equal mass roll down the same incline at the same which has more energy? My calculus and physics are weak, but I know that the larger accelerates slower and has a greater moment of inertia.
What makes you think that the larger one accelerates slower?
 
the larger has a greater moment of inertia and lesser angular momentum, is this correct?
 
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the larger has a greater moment of inertia and lesser angular momentum, is this correct?
The larger has greater moment of inertia AND greater angular momentum, but lesser angular velocity.
 

Doc Al

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That means when P, the amount of gravitational potential energy, is given to the larger ball, most of the energy will go into rotational kinetic energy to rotate the ball and a little bit will go into kinetic energy to give the ball a forward velocity. There is less rotational kinetic energy needed to rotate the smaller ball, so a larger portion of P will go into it's kinetic energy and it will move down the inclined plane faster.
Both large and small balls will have the same division of rotational versus translational KE. Both have the same acceleration down the ramp.
 
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Both large and small balls will have the same division of rotational versus translational KE. Both have the same acceleration down the ramp.
Are you saying that they will have equal translational and rotational kinetic energies?
 

Doc Al

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Are you saying that they will have equal translational and rotational kinetic energies?
Yes, they will roll down the incline at the same rate. The fraction of the total energy that goes to translational vs. rotational energy is independent of radius.
 
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Yes, they will roll down the incline at the same rate. The fraction of the total energy that goes to translational vs. rotational energy is independent of radius.
I don't think that's correct. If the larger ball had twice the Inertia of the smaller ball then it would have a rotational kinetic energy that was a factor of a square root of two greater than the smaller balls rotational kinetic energy. Since both of the balls are given equal amounts of energy, the larger ball's translational kinetic energy would have to be smaller. The gravitational potential energy is not divided in two and passed as equal portions to the rotational and translational components of the balls motion.
 

Doc Al

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I don't think that's correct. If the larger ball had twice the Inertia of the smaller ball then it would have a rotational kinetic energy that was a factor of a square root of two greater than the smaller balls rotational kinetic energy.
How did you arrive at this conclusion?
Since both of the balls are given equal amounts of energy, the larger ball's translational kinetic energy would have to be smaller.
That would certainly follow from your earlier statement, but that statement isn't true.
The gravitational potential energy is not divided in two and passed as equal portions to the rotational and translational components of the balls motion.
That's true--the translational KE is greater:
The translational KE = 1/2mv²;
The rotational KE = 1/2Iω² = 1/2 (2/5mR²)(v/R)² = 1/5mv².

(Note that there's no dependence on R.)
 
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How did you arrive at this conclusion?

That would certainly follow from your earlier statement, but that statement isn't true.

That's true--the translational KE is greater:
The translational KE = 1/2mv²;
The rotational KE = 1/2Iω² = 1/2 (2/5mR²)(v/R)² = 1/5mv².

(Note that there's no dependence on R.)
Ahh yes, you are correct. Your example helped me see my mistake. The math makes sense but I am trying to understand it qualitatively. If the radius of the ball increases then it's angular momentum decreases and it's angular displacement decreases. Why then from this perspective do both of the balls move with the same speed.
 
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In other words, how can we understand this without writing an equation and making a substitution of variables.
 

Doc Al

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The math makes sense but I am trying to understand it qualitatively. If the radius of the ball increases then it's angular momentum decreases and it's angular displacement decreases. Why then from this perspective do both of the balls move with the same speed.
I'm trying to get a handle on your reasoning. For a ball rolling down an incline, the larger the radius the greater the rotational inertia (by a factor of R²), the greater the angular momentum, but the smaller the angular speed. This should make sense, since the friction force producing the angular momentum exerts a greater torque on the larger ball. Even though the angular momentum is greater for the larger ball, the rotational KE is not (since the angular displacement is smaller).
 
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I'm trying to get a handle on your reasoning. For a ball rolling down an incline, the larger the radius the greater the rotational inertia (by a factor of R²), the greater the angular momentum, but the smaller the angular speed. This should make sense, since the friction force producing the angular momentum exerts a greater torque on the larger ball. Even though the angular momentum is greater for the larger ball, the rotational KE is not (since the angular displacement is smaller).
Oops, I did not mean to say that the angular momentum decreased, I meant that the angular velocity decreases. This does make sense. But where I do not have an intuitive grasp is in understanding how the two balls can have the same speed down the ramp when one of them is rotating at a slower rate.
 

Doc Al

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But where I do not have an intuitive grasp is in understanding how the two balls can have the same speed down the ramp when one of them is rotating at a slower rate.
The translational speed is given by ωR, thus two balls with the same speed but different radii must have different rotational rates.
 
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I see. That makes sense.
 
i was out of town, thanks for the input. its my understanding Euler pointed to disks of equal mass and different size. he suggests the larger has a greater moment of inertia and the smaller has a greater angular velocity.
if we let gravity have at a marble on the incline acceleration is fast and apparently instant, but if we let it have at a handball, it is notably slower by comparison.
with two identical inclines (facing each other) that level off at the bottom, the difference in how either mass's energy works becomes more exaggerated.
the smaller mass lacks sufficient size to effectively stop the larger if they are on a direct collision course (even if does affect its trajectory, sum energy, etc.)
by the same token, the larger mass is naturally inclined to roll over the smaller mass if they collide.
so, the nature of their energy is different.
my concern is whether they move the same distance and whether or not the smaller comes to halt first. is there agreement on whether they move the same length? and if they do, will they cover this distance at the same rate?
 

GT1

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What about the following situation:
2 spheres with the same mass and radius are released on an incline, one of the spheres is hallow and the other one is solid. Which sphere would reach first to the bottom?
Which sphere would travel larger distance (after the incline) before it stops?
 

Doc Al

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What about the following situation:
2 spheres with the same mass and radius are released on an incline, one of the spheres is hallow and the other one is solid. Which sphere would reach first to the bottom?
The two spheres have different rotational inertias, of course. You can see which will roll down the incline faster by extending the reasoning I gave in post #12.
Which sphere would travel larger distance (after the incline) before it stops?
What's stopping them?
 
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For each case, the total kinetic energy of each ball, assuming equal mass and uniform density, and assuming rolling without sliding, is

E = (1/2)mv2 + (2/5)mv2 = (7/10)mv2 = mgh

The first term on the left is due to translation of the center of mass, and the second to the rotational kinetic energy. here I have used wR = v, where w is the angular velocity, moment of inertia I = (2/5)mR2, and R is the raduis.

In the above equation, m cancels out, (as does R), so the velocity of each ball is

v=sqrt(10 gh/7)

This is independed of ball radius and mass. Note that if g has units meters per sec^2 and h has units meters, then v has units meters per sec, so both balla have the same velocity, depending only on g and h.
 
thank you bob,
same velocity, same energy
this makes sense to me, such as it ought too.
 

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