Basic Physics question

[EnginrTheFuture]

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Just going over Ch. 1 and 2 of physics before my next practice exam and asked myself a question I couldn't answer.

If you have an inclined plane with friction, it's obvious that the time up the ramp is shorter than the time back down to the original spot. Friction works with gravity on way up but against it on way down. BUT if you have no friction, is the time of the entire trip the same as the entire trip of the one with friction?

Same idea applies to a ball thrown up with air resistance. The flight time up is less than the flight time down but is the total flight time the same as with no air resistance?

I really can't reason through this mathematically or intuitively and was really curious if anyone has an intuition for this that they could share. 😎

 

[MedPR]

Just going over Ch. 1 and 2 of physics before my next practice exam and asked myself a question I couldn't answer.

If you have an inclined plane with friction, it's obvious that the time up the ramp is shorter than the time back down to the original spot. Friction works with gravity on way up but against it on way down. BUT if you have no friction, is the time of the entire trip the same as the entire trip of the one with friction?

Same idea applies to a ball thrown up with air resistance. The flight time up is less than the flight time down but is the total flight time the same as with no air resistance?

I really can't reason through this mathematically or intuitively and was really curious if anyone has an intuition for this that they could share. 😎

Why is the time up shorter than the time back down?

 

[syoung]

time up = time down, air resistance or not.
With AR it slows the ball up and it slows the ball down

 

[EnginrTheFuture]

Well you know the final speed when it reaches the ground again is going to be slower than what is was when it got shot up right? Some of that initial kinetic energy goes to heat/friction in the air. Therefore the average speed on the way up (Velocity when shot up + 0 at top)/2 is greater than the average speed on way down ( 0 at top + Velocity at bottom)/2 ... What goes up, must come down so the distance up = distance down. If the ball goes faster up than down but both cover the same distance, the ball takes longer to get down than it did to get up.

What I'm wondering is if this time saved on way up, and time added on way down are equal...😕

 

[EnginrTheFuture]

time up = time down, air resistance or not.
With AR it slows the ball up and it slows the ball down

Time up does not equal time down with air resistance. Total trip time MAY be same, which is what I'm trying to get to the bottom of, but definitely make sure you realize that the trip up is shorter with respect to time than on the way down!

 

[syoung]

Sorry.
Yes with air resistance time up is faster (shorter) than time down. Throw a balloon up and you can see the difference. It has an initial v up, slowed down by gravity and AR. but on the way down, gravity pulls it down and only AR slows it up.
With a ball the time difference will not be so exaggerated of course.

 

[EnginrTheFuture]

For those not wanting to read through everything already said... we still haven't figured out the answer 🙁

 

[syoung]

with no friction, the trip would be shorter for the ramp, given that the ball must reach the same height on the ramp in both cases

 

[EnginrTheFuture]

Yes, that I understand, but what about two cases in which they leave with the same initial V but one is on a rough surface and one is on a smooth surface?

 

[syoung]

well if we are still asking just for time of trip, and both (same mass) set off with same Vi, then given two different mu constants for friction, the one with a rough surface will reach the bottom faster than the smooth one
The rough Friction slows down the ball more and so it loses more KE than the one on the smooth surface. this is easy to see if you do a small experiment with a smooth book ramp and maybe a studded book (how would you make this???) and you can see the difference in times

 

[Buttafuoco]

This is a topic that makes me go crosseyed. But check out page 3 of this link.

http://wps.aw.com/wps/media/objects/877/898586/topics/topic01.pdf

According to this, the projectile with air resistance covers less of a range, doesn't go as high (I think we all knew that part), and it goes from ground to apex to ground again more quickly.

 

[milski]

Going up, the force is gravity plus the resistance, going down the force is gravity minus the resistance. Thus the acceleration on the way has larger magnitude than on the way up. For the same distance, the time on the way down will be less than on the way up.

For the non-resistance case, the only force is gravity. Times up and down are the same.

For the non-resistance case, the total time is sqrt(2d/g)+sqrt(2d/g)=2sqrt(2d/g)
For the resistance, it's sqrt(2d/(g+&#916😉) + sqrt(2d/(g-&#916😉).

You can easily prove that the latter is always larger than the former.

 

[hellocubed]

If you have an inclined plane with friction, it's obvious that the time up the ramp is shorter than the time back down to the original spot. Friction works with gravity on way up but against it on way down. BUT if you have no friction, is the time of the entire trip the same as the entire trip of the one with friction?

Same idea applies to a ball thrown up with air resistance. The flight time up is less than the flight time down but is the total flight time the same as with no air resistance?

hi OP, this is the misconception that you have

"Friction works with gravity on the way up, but against it on the way down." so "the time up the ramp is shorter than the time back down"
In this way, the time up a ramp will actually be LONGER than the time down. When a block is moving up a ramp, there are TWO forces hampering its motion (gravity and friction). When the block is moving down the ramp, there is only ONE force hampering its motion (friction) with gravity actually aiding motion. Hence it travels up slower, and down faster with friction.

The same applies to air resistance.



However, I know what you are thinking. If I chuck a block up a ramp, or fire a bullet into the sky, it gets to its destination faster than it falls. This is because there is something you are forgetting: the initial propulsion force.

These time differentials are not due to motion hampers. Air resistance and friction slow movement up and down to the same degree. Removing it is like subtracting 10 from both sides of the equation...

What does change is the original propulsion force becomes dissipated into heat and air displacement (example, bullet shoots up at lethal velocity, returns at below lethal terminal velocity.)

 

[milski]

hi OP, this is the misconception that you have

"Friction works with gravity on the way up, but against it on the way down." so "the time up the ramp is shorter than the time back down"
In this way, the time up a ramp will actually be LONGER than the time down. When a block is moving up a ramp, there are TWO forces hampering its motion (gravity and friction). When the block is moving down the ramp, there is only ONE force hampering its motion (friction) with gravity actually aiding motion. Hence it travels up slower, and down faster with friction.

The same applies to air resistance.



However, I know what you are thinking. If I chuck a block up a ramp, or fire a bullet into the sky, it gets to its destination faster than it falls. This is because there is something you are forgetting: the initial propulsion force.

These time differentials are not due to motion hampers. Air resistance and friction slow movement up and down to the same degree. Removing it is like subtracting 10 from both sides of the equation...

What does change is the original propulsion force becomes dissipated into heat and air displacement (example, bullet shoots up at lethal velocity, returns at below lethal terminal velocity.)

OP is correct. The time up ramp is V0/a. For larger acceleration, the time will be shorter. The distance is at^2/2=a*V0^2/(2a^2)=V0^2/2a. The distance will also be shorter for larger acceleration.

 

[EnginrTheFuture]

Going up, the force is gravity plus the resistance, going down the force is gravity minus the resistance. Thus the acceleration on the way has larger magnitude than on the way up. For the same distance, the time on the way down will be less than on the way up.

For the non-resistance case, the only force is gravity. Times up and down are the same.

For the non-resistance case, the total time is sqrt(2d/g)+sqrt(2d/g)=2sqrt(2d/g)
For the resistance, it's sqrt(2d/(g+&#916😉) + sqrt(2d/(g-&#916😉).

You can easily prove that the latter is always larger than the former.

It's looking like this kind of concept is not likely to show up but it's interesting to think through anyway. I like where this is starting to go but I think there are a few faults still left to iron out.

In the case that ∂ approaches g sqrt(2d/(g-&#8706😉) --> infinite

so I don't think that the resistance case is always larger.


Also, ∂ is velocity dependent making it time dependent. This also makes it dependent on which leg of the trip it is on since the average velocity on the way down is less than on the way up. So the function for trip time of the resistance case may not be entirely correct. It seems like a good approximation but I think there might be more to it.

I'm wondering if I'll have to bust out mathematica and some calc later this week?

 

[EnginrTheFuture]

hi OP, this is the misconception that you have

However, I know what you are thinking. If I chuck a block up a ramp, or fire a bullet into the sky, it gets to its destination faster than it falls. This is because there is something you are forgetting: the initial propulsion force.

These time differentials are not due to motion hampers. Air resistance and friction slow movement up and down to the same degree. Removing it is like subtracting 10 from both sides of the equation...

What does change is the original propulsion force becomes dissipated into heat and air displacement (example, bullet shoots up at lethal velocity, returns at below lethal terminal velocity.)

I'm starting to grasp what your saying but I'm not quite there. I don't get what you mean by initial propulsion force. If we start with a set amount of KE out of a cannon, how does this concept translate?

 

[milski]

It's looking like this kind of concept is not likely to show up but it's interesting to think through anyway. I like where this is starting to go but I think there are a few faults still left to iron out.

In the case that ∂ approaches g sqrt(2d/(g-&#8706😉) --> infinite

so I don't think that the resistance case is always larger.


Also, ∂ is velocity dependent making it time dependent. This also makes it dependent on which leg of the trip it is on since the average velocity on the way down is less than on the way up. So the function for trip time of the resistance case may not be entirely correct. It seems like a good approximation but I think there might be more to it.

I'm wondering if I'll have to bust out mathematica and some calc later this week?

δ=g is the case when there is enough friction that there is no movement. I'm not sure how that would lead to the resistance case not being larger?

It works for ramp but yes, the resistance will be dependent on the velocity for the ball thrown. I might try to work it out later but I'll be surprised if there is a case where it does not work out.

 

[hellocubed]

OP is correct. The time up ramp is V0/a. For larger acceleration, the time will be shorter. The distance is at^2/2=a*V0^2/(2a^2)=V0^2/2a. The distance will also be shorter for larger acceleration.

This was what I meant.

The OP is attributing the velocity difference to friction and air resistance, which is not true. It is acceleration that is the determinant that makes upper travel faster than the subsequent fall.

 

[hellocubed]

I'm starting to grasp what your saying but I'm not quite there. I don't get what you mean by initial propulsion force. If we start with a set amount of KE out of a cannon, how does this concept translate?

This is what I mean:
If you fire yourself out of a cannon into the sky, you will reach your max height faster than it takes you to fall to the ground. The reason you reach the sky faster is due to the initial KE that you had, which is now dissipated and you no longer have (the sound generated from your body, vibration of your cells, the sheer force it takes to displace that much volume of air that fast, etc...).

Even with absolutely no air resistance, the rate you climbed would have been ~100m/s^2 (watever) and the rate you fall would have been (9.8m/s^2 assuming you fall in a vacuum).
The reason travel up is faster is due to contributions from initial energy. Greater acceleration that you no longer have. NOT due to friction or air resistance.


If, hypothetically, you were rising and falling at the same energy, considerable air resistance would actually make your trip up slower than your trip down as
1.) trip up has 2 Forces against you: gravity and air resistance
2.) Trip down has 1 force against you: air resistance, and 1 force accelerating you: gravity.

the effect of friction and air resistance has on travel time is opposite what you originally posted.

 

[milski]

This was what I meant.

The OP is attributing the velocity difference to friction and air resistance, which is not true. It is acceleration that is the determinant that makes upper travel faster than the subsequent fall.

The difference in acceleration is due to friction/air resistance. I'm not sure what your point is. 😕

 

[EnginrTheFuture]

This sentence doesn't make much sense to me... how could the velocity difference not be attributable to the frictional force? If the frictional force isn't present, the velocity difference is non-existant isn't it?

The "acceleration" that you say contributes the velocity difference is a direct product of the two forces (weight and frictional). To say acceleration is all the contributes to the velocity difference is the same as saying that the velocity difference is due to gravity and friction.

 

[milski]

This is what I mean:
If you fire yourself out of a cannon into the sky, you will reach your max height faster than it takes you to fall to the ground. The reason you reach the sky faster is due to the initial KE that you had, which is now dissipated and you no longer have (the sound generated from your body, vibration of your cells, the sheer force it takes to displace that much volume of air that fast, etc...).

Even with absolutely no air resistance, the rate you climbed would have been ~100m/s^2 (watever) and the rate you fall would have been (9.8m/s^2 assuming you fall in a vacuum).


The reason travel up is faster is due to contributions from initial energy. Greater acceleration that you no longer have.
If, hypothetically, you were rising and falling at the same energy, considerable air resistance would actually make your trip up slower than your trip down as
1.) trip up has 2 Forces against you: gravity and air resistance
2.) Trip down has 1 force against you: air resistance, and 1 force accelerating you: gravity.

the effect of friction and air resistance has on travel time is opposite what you originally posted.

The amount of energy that you have is irrelevant to how fast that energy is being transferred to another shape of energy. For example, feel free to compare a parachute jumpers with the same weight and initial altitude/speed but one of them with open parachute and the other with close.

The only thing that can give you the times for going up and down is the second part that you wrote, describing the forces acting on the body.

He initially wrote that the time up is shorter, which is the correct answer.

 

[hellocubed]

ah sheeeeet


So wait, because there are two forces working against you on the trip up, the initial acceleration (and hence velocity) upwards has to be greater to reach the destination height?

 

[milski]

ah sheeeeet


So wait, because there are two forces working against you on the trip up, the initial acceleration (and hence velocity) upwards has to be greater to reach the destination height?

I don't think there is a single height being reached. The object rolls up with initial velocity V0 and goes as high as it gets. At least that's what I was talking about in all of my posts.

Pushing/throwing it with a velocity to reach a specific height will be another (interesting) problem.

 

[EnginrTheFuture]

I don't think there is a single height being reached. The object rolls up with initial velocity V0 and goes as high as it gets. At least that's what I was talking about in all of my posts.

Pushing/throwing it with a velocity to reach a specific height will be another (interesting) problem.

Same initial V_0, different conditions (one trial with friction and one without). They should get to different heights... yep that's what I was wondering about. 🙂

 

[milski]

This is the difference between the time with no resistance and the time with air resistance. g and m are obvious. k is some koefficient of drag in Fd=kv^2, v20 is the initial v_0 speed, should be negative since down was positive (g>0). Don't ask about the choice of positive direction and v20. 🙂 Good luck proving that this is always negative, although plugging in some number does seem to indicate that this is the case.

What makes it messy is that the differential equations for the up and down movements have different solutions (Log(Cos(...)) vs Log(Cosh(...)) so nothing cancels at the end. I did graph a few solutions with some arbitrary reasonable constants and they do seem very plausible. I could send you the Mathematica script but we are getting way off topic here.

 

[syoung]

the hell are you doing?! you were a physics major? that never showed up in my physics class...

 

[milski]

the hell are you doing?! you were a physics major? that never showed up in my physics class...

Yes, we are way outside of the MCAT now. The TL;DR version is that if there is friction, the object stops sliding up and down the ramp will stop earlier than an object which would slide up and down on a frictionless ramp.