Magnitude of acceleration on a ramp

[OME9A]

Why is the magnitude of acceleration on a ramp greater on the way up than it is on the way down? I can't figure it out... I was thinking maybe it has something to do with the initial force applied to the cart on the way up the ramp?

So that you know exactly what I mean: If I have a cart motionless at the bottom of the ramp, and I apply ONE force (me pushing it and then immediately letting go) so that it goes up the ramp and then comes back down, the magnitude of acceleration is greater on the way up than it is coming back down... But the only constant forces acting on it are gravity, the normal, and friction, none of which change on the way up VS the way down...

 

[russ_watters]

The magnitude of the acceleration due to gravity is exactly the same on the way up as on the way down. I'm not sure what is making you think it isn't constant.

...unless you are talking about a real-world situation with wind resistance or something like that, where the drag force depends on speed. That's a pretty small effect though for a small cart you can push with your hand.

 

[OME9A]

Not just the accel. due to gravity, but the overall acceleration... We did an experiment around 5 times, and every time, the magnitude was greater on the way up (up to right before it has a velocity of 0 and starts coming down) than on the way down (up to right before it returns to its original location at the bottom of the ramp)

EDIT: And it's not by much, for example: upwards accel. = -.65 m/s^2, downwards accel. = -.55 m/s^2. I don't think it's a coincidence because it happened every time... But if in theory they should be equal, I wonder what we could have done wrong to cause this!

 

[Raekwon]

When you apply a force to something to push it up the ramp you have kinetic energy that overtime will turn into potential energy, and it will get slower as the energy changes to potential until it is all potential energy, at which point it then turns back into kinetic energy as it comes back down... now we know a falling body accelerates uniformly, it will gain speed in equal amounts over time intervals, so as it falls it gets faster.

You also have a loss due to friction, air resistance etc as Russ said.

 

[rcgldr]

For example: upwards accel. = -.65 m/s^2, downwards accel. = -.55 m/s^2.

This implies that drag force opposes motion with the equivalent of .05 m/s^2 deceleration, and that gravity generates -.60 m/s^2 acceeleration with your cart and ramp. On the way up deceleration = -.60 m/s^2 from gravity -.05 m/s^2 from drag for a total of -.65 m/s^2. On the way down, acceleration = -.60 m/s^2 from gravity + .05 m/s^2 from drag for a total of -.55 m/s^2 of acceleration.

 

[OME9A]

This implies that drag force opposes motion with the equivalent of .05 m/s^2 deceleration, and that gravity generates -.60 m/s^2 acceeleration with your cart and ramp. On the way up deceleration = -.60 m/s^2 from gravity -.05 m/s^2 from drag for a total of -.65 m/s^2. On the way down, acceleration = -.60 m/s^2 from gravity + .05 m/s^2 from drag for a total of -.55 m/s^2 of acceleration.

We haven't learned this yet, but I did a quick google and found that it mainly exists for solids moving through liquids. Is it possible that a simple cart going up a ramp in the open (not underwater or anything...) could have a drag force equivalent to .05 m/s^2 deceleration? It seems like a lot!

 

[rcgldr]

Is it possible that a simple cart going up a ramp in the open (not underwater or anything...) could have a drag force equivalent to .05 m/s^2 deceleration? It seems like a lot!

I forgot to mention that I assumed the drag force is related to rolling resistance, not aerodynamic drag. Gravity is 9.8 m/s^2, and this drag force is only .05 m/s^2, so the drag force is only 1/196th of cart's weight, which translates into a rolling resistance factor of .0051, which isn't bad.

 

[OME9A]

And another confusing find: As the difference between upwards and downwards accel. increased, the coefficient of friction increased. This is understandable because we used the formula that Coeff. of Friction=


$$\frac{a_{up}-a_{down}}{-2gcos\theta}$$

(g is positive 9.81)

But the only problem is that the coeff. should remain constant! But I will probably chalk this one up to a significant figures problem or something...

 

[OME9A]

I assume it's related to rolling resistance, not aerodynamic drag. Gravity is 9.8 m/s^2, and this drag force is only .05 m/s^2, so the drag force is only 1/196th of cart's weight, which translates into a rolling resistance factor of .0051, which isn't bad.

Okay, just so that I have this straight:

On the way up the ramp, rolling resistance is acting in addition to friction so that both rolling resistance and friction add to the deceleration, but on the way down, rolling resistance and friction are pulling it upwards so it reduces the deceleration!

Although I probably don't have to even include friction because the coefficient varied from .002->.006 which seems negligible.

Thanks very much! I'm so glad I got past this conundrum...!

 

[rcgldr]

In this experiment, the angular momentum of the wheels will produce an upwards acceleration opposing any downwards acceleration both on the way up and the way back down, but this should be a very small effect, assuming that the mass of the wheels is much smaller than the mass of the cart.

On the way up, rolling resistance is acting in addition to gravity, but angular momentum in the wheels is acting against gravity. On the way down, both rolling resistance and angular momentum in the wheels are acting against gravity.

 

[OME9A]

In this experiment, the angular momentum of the wheels will produce an upwards acceleration opposing any downwards acceleration both on the way up and the way back down, but this should be a very small effect, assuming that the mass of the wheels is much smaller than the mass of the cart.

On the way up, rolling resistance is acting in addition to gravity, but angular momentum in the wheels is acting against gravity. On the way down, both rolling resistance and angular momentum in the wheels are acting against gravity.

Another force!? I can't believe how many forces there are in such a simple experiment... I hope we never get this far in physics; I have enough trouble dealing with just gravity, normal, and friction!

Thanks, I will keep angular momentum in mind as well!

 

[rcgldr]

Thanks, I will keep angular momentum in mind as well!

I doubt that angular momentum is significant in this case, but it will be important when considering the classic examples of a solid sphere, solid cylinder, hollow sphere, or hollow cylinder on an inclined plane.

 

[Redbelly98]

Okay, just so that I have this straight:

On the way up the ramp, rolling resistance is acting in addition to friction so that both rolling resistance and friction add to the deceleration, but on the way down, rolling resistance and friction are pulling it upwards so it reduces the deceleration!

Yes, that is the whole key here. Friction, drag, and rolling resistance forces change direction, and therefore are not the same between going up vs. going down.

 

[OME9A]

Yes, that is the whole key here. Friction, drag, and rolling resistance forces change direction, and therefore are not the same between going up vs. going down.

Thank you, it makes sense now!