Impacts of gradient on speed.

[SleepingBag]

We know that when a circular object is placed on a ramp , it will " roll down the ramp ".

 

How gentle or how steep the ramp is will affect the velocity of the object.

 

My question is that whats the exact explanation for how the gradient affects the speed.

 

My guess is it has to do with moments , am I right?

[timo]

Not exactly. Two different objects, say spherical ones, can roll down the same slope in a different way. That would be caused by different moments of inertia. The same object rolling down slopes with different angles in different ways is much more simple: There is a constant gravitational force pointing downwards towards the slope. This force can be uniquely written as the sum of two forces, where one of the forces is in the direction of motion and the other one is perpendicular to it. The force into the direction of motion is the one that is responsible for the motion (not too much of a surprise). The steeper the slope the larger this force, and hence the faster the object accelerates and rolls down.

[DrRocket]

Not exactly. Two different objects, say spherical ones, can roll down the same slope in a different way. That would be caused by different moments of inertia. The same object rolling down slopes with different angles in different ways is much more simple: There is a constant gravitational force pointing downwards towards the slope. This force can be uniquely written as the sum of two forces, where one of the forces is in the direction of motion and the other one is perpendicular to it. The force into the direction of motion is the one that is responsible for the motion (not too much of a surprise). The steeper the slope the larger this force, and hence the faster the object accelerates and rolls down.

 

There is one other effect to consider. That involves friction, the moment of inertia of the sphere about its center, and the distinction between rolling and moving with slippage. The sphere can slide down the hill, roll without slipping or both slip and slide. The normal force plays a role in determing this, and the final velocity is dependent on the specific scenario since some of the initial potential energy will be realized as translational speed and some as rotational speed.

[J.C.MacSwell]

There is one other effect to consider. That involves friction, the moment of inertia of the sphere about its center, and the distinction between rolling and moving with slippage. The sphere can slide down the hill, roll without slipping or both slip and slide. The normal force plays a role in determing this, and the final velocity is dependent on the specific scenario since some of the initial potential energy will be realized as translational speed and some as rotational speed.

 

...and friction losses from the slippage and/or rolling resistance you mentioned.

[DrRocket]

...and friction losses from the slippage and/or rolling resistance you mentioned.

 

Friction losses from slippage are relatively easy to calculate and come out of the model directly, if indeed here is slippage given knowledge of the coefficient of friction in a Coulomb friction scenario.

 

Rolling rresistance is a separate ingredient that may or may not be a consideration. One can formulate the problem in a sensible way with the assumption that rolling resistance is zero. This is a reasonable idealization in many circumstances. In others rolling resistance is significant. The loss mechanisms associated with rolling resistance can be very complex, as can the mechanisms for sliding friction -- which are not well understood at the most fundamental level.

 

You can make this problem as difficult as you want by boring down into all possible loss scenarios, but the major point that some energy is embodied in rigid body rotation will be quickly lost in the fog and the important physics will be obscured.

[SleepingBag]

Can you guys explain it in a more simplified manner ,

 

I don't think you are getting my point , @timo an fair experiment is conducted with gradient as independent variable. And of course , the force supplied to start the ball rolling is constant for both scenarios.

 

I was wondering if its correct to say that the steeper the gradient , the greater the effects of "free fall".

 

Another possibility is that the reaction force during point of contact of slope and ball is smaller for the steeper slope , hence it experiences a greater force due to gravity.

[J.C.MacSwell]

Friction losses from slippage are relatively easy to calculate and come out of the model directly, if indeed here is slippage given knowledge of the coefficient of friction in a Coulomb friction scenario.

 

Rolling rresistance is a separate ingredient that may or may not be a consideration. One can formulate the problem in a sensible way with the assumption that rolling resistance is zero. This is a reasonable idealization in many circumstances. In others rolling resistance is significant. The loss mechanisms associated with rolling resistance can be very complex, as can the mechanisms for sliding friction -- which are not well understood at the most fundamental level.

You can make this problem as difficult as you want by boring down into all possible loss scenarios, but the major point that some energy is embodied in rigid body rotation will be quickly lost in the fog and the important physics will be obscured.

 

Whatever. They can be idealized as appropriate to their significance. If it's based on a real experiment it might be best not to ignore them.

[timo]

I don't think you are getting my point , @timo a fair experiment is conducted with gradient as independent variable. And of course, the force supplied to start the ball rolling is constant for both scenarios.

I'm a bit surprised you apply a force to the rolling balls. Do you mean the gravitational force that pulls them down or are you applying another force?

 

I was wondering if its correct to say that the steeper the gradient , the greater the effects of "free fall".

If I understood your scenario correctly, then that is somewhat correct. It's pretty much what I meant when I said

There is a constant gravitational force pointing downwards towards the slope floor (that was a typo in my original post, sorry for that). This force can be uniquely written as the sum of two forces, where one of the forces is in the direction of motion and the other one is perpendicular to it. The force into the direction of motion is the one that is responsible for the motion (not too much of a surprise). The steeper the slope the larger this force, and hence the faster the object accelerates and rolls down.

 

 

Another possibility is that the reaction force during point of contact of slope and ball is smaller for the steeper slope , hence it experiences a greater force due to gravity.

You are mixing two things that are usually considered somewhat separately. First, "experiencing a greater force due to gravity" is the same as "greater effect of 'free fall'", since free fall is when 100% of the gravitational force goes into accelerating the object. Second, there is a force pressing the ball onto the ramp, which can determine the way the balls go down the ramp (rolling vs. sliding, and presumably mixtures of - that's also what DrRocket spoke about in his post). I was assuming your balls roll, in which case this force can usually be ignored. If you choose not to ignore it, then reduced pressure of the spheres onto the slope indeed is another reason strengthening the "goes down a steeper slope faster" effect. I believe you are best helped by trying to understand the "more freefall-like" reason first, since this is basic school knowledge (8-9th grade or so), whereas the other is already lower university level.

 

You may be interested in reading forces on an inclined plane on Wikipedia, since this shows how the gravitational force is split into a force parallel to the direction of motion, and a rest.

[SleepingBag]

@timo , thanks i get it now .