A ball climbing a ramp while "rolling the wrong way"

In summary, the conversation is about a problem involving a ball rolling down a ramp with friction (red part) and then climbing up a ramp without friction (blue part). The final height the ball reaches on the blue slope is a fraction of the initial height due to energy conservation. The question arises about whether the ball can reach the same height on the return trip, which the book claims to be possible, but the speakers believe is incorrect due to the rotational kinetic energy of the ball. They discuss the potential errors and inconsistencies in the problem and its solution, and conclude that the average student would have difficulty solving it.
  • #1
FranzDiCoccio
342
41
Homework Statement
a ramp with rolling friction faces another ramp without friction. A ball rolls down the first ramp and climbs up the second, sliding without rolling.
The problem asks for the height reached by the ball on the second ramp, and then the height it reaches when it climbs again up the ramp it originally came from
Relevant Equations
conservation of mechanical energy, made up of gravitational potential energy and kinetic energy, both translational and rotational (at least for the first question).
ramp.png

This image represents the ramp.
The first part is pretty easy.

The red part has friction, and the ball rolls down it. The blue part has no friction, and the ball climbs it only owing to the translational kinetic energy that it gained at the bottom of the red ramp, which is only a fraction of its initial potential energy. Thus the height it reaches on the blue slope is necessarily lower than the initial one. Specifically, the final height on the blue slope is [itex]h'=5/7 h[/itex], where [itex]h[/itex] is the initial height on the red ramp (here I assume a solid ball).
The angular velocity (and hence rotational kinetic energy) does not vary while the ball is on the blue slope.

The second question seems a bit harder. The book the problem came from claims that the ball gets back to the initial height, due to energy conservation.
This seems at least really counterintuitive to me. I would dare to say that it is probably plainly wrong.

At the bottom of the red ramp the ball is rolling clockwise. If the blue ramp had friction, the rotational kinetic energy would "help" the ball climb up to (ideally) the same height it fell from.
On the other hand the ball is rolling "the wrong way" at the start of the red slope. In order to climb the red slope, it should roll counterclockwise.
It seems to me that its rotational kinetic energy should hinder its progress on the red slope, and not favour it.
I'm not really sure what would happen when the ball reaches the red part falling down from the blue ramp while "rolling backwards".
My guess is that it cannot possibly roll without sliding, and if it slides it would dissipate some energy due to friction.
Hence the height on the return on the red slope cannot be the same as the initial one. How lower than that? It probably depends on the friction coefficient, which however is not specified in the problem.

It seems to me that the claim of the book is similar to claiming that an object would reach an height [itex]h = v^2/2g[/itex] irrespective of the angle formed by its initial velocity and the horizontal direction.

Am I missing anything? Could the solution proposed by the book be possibly right?

Thanks a lot for any insight
Franz
 
  • Like
Likes Delta2
Physics news on Phys.org
  • #2
FranzDiCoccio said:
My guess is that it cannot possibly roll without sliding, and if it slides it would dissipate some energy due to friction.
I agree. What book is this from?
 
  • #3
This is from the Italian "version" of "Physics, 11th edition" by Cutnell and Johnson. I have 9th edition, and I could not find that problem there.

The Italian authors used a lot of material from the original version, but added some parts. I have to say that they did not appear to have thought these parts really through. Sometimes they just change perfectly good original parts for no apparent reason, and often end up making them less clear or even wrong.

Also, the authors of the theory parts and problems seem to be different people. Lots of inconsistencies and errors in the problems and in their solutions.
 
  • #4
If they provide their "solution", it might provide some answers.
 
  • #5
Their explanation consists in a single sentence that goes like this "When the marble gets back to the rough ramp, its rotation slows down and it reaches its starting point".
But this is what would happen for a marble rolling counterclockwise. This one isn't.
The final height on the red ramp cannot be the same as the initial one, as they claim.

As I say, their solution only takes into account energy, ignoring the direction. It's like expecting that an object reaches the height [itex]v^2/(2g)[/itex] above the initial point when thrown downward with velocity [itex]v[/itex]. The energy balance works in this case too, but the object would never go up.

My feeling is that authors add new problems getting inspiration elsewhere, and hire freshly graduated (or undergraduate students) to solve them. I get this impression from the fact that the solutions are often wrong or imprecise. Other times they rely on techniques that are never discussed in the book, because they are too advanced for its readers. Here we're talking about 3rd year high school students (16-17 yrs old).

Maybe there is something I'm missing, but it seems to me that finding the maximum height of the marble on the red ramp is a pretty messy problem. I guess that the ball might even be still rotating clockwise at the point of maximum height. That likely depends on the friction coefficient and on the exact shape of the ramp.
The average student targeted by this book has no chance of solving this problem.
 
  • #6
FranzDiCoccio said:
Maybe there is something I'm missing, but it seems to me that finding the maximum height of the marble on the red ramp is a pretty messy problem. I guess that the ball might even be still rotating clockwise at the point of maximum height. That likely depends on the friction coefficient and on the exact shape of the ramp.
The average student targeted by this book has no chance of solving this problem.
You are correct. The authors messed up with this problem.
 
  • #7
FranzDiCoccio said:
Their explanation consists in a single sentence that goes like this "When the marble gets back to the rough ramp, its rotation slows down and it reaches its starting point".
Yeah, it slows down real quickly with a "chirp" like when an airplane tire touches down on a runway during landing... :wink:
 
  • Like
Likes Keith_McClary
  • #8
FranzDiCoccio said:
The average student targeted by this book has no chance of solving this problem
The average student perhaps. But it is not really that difficult a problem. The solution dynamics of the "chirp" as backward spinning ball re-engages with the rolling (static) friction to spin it the other way involves only the assumption that it is very quick. For the advanced student it might be fun
Of course the book is really wrong. My intuition says it will be a small fraction of the original height and independent of the radius of the ball. I will work it out tomorrow if I get the new wheel strung on my bicycle.
 
  • Like
Likes Doc Al and berkeman
  • #9
FranzDiCoccio said:
Hence the height on the return on the red slope cannot be the same as the initial one. How lower than that? It probably depends on the friction coefficient
berkeman said:
Yeah, it slows down real quickly with a "chirp" like when an airplane tire touches down on a runway during landing... :wink:
hutchphd said:
The solution dynamics of the "chirp" as backward spinning ball re-engages with the rolling (static) friction to spin it the other way involves only the assumption that it is very quick.
It would take a very high coefficient of friction to revert to rolling so quickly. The normal force is just mg - there is no vertical impact to lead to a very high normal force, so no "impulsive friction". And for a landing plane, the plane's weight is enormous compared to the 'inertia' (angular momentum/radius) of the tyres.

But if we don’t assume it's quick it looks intractable, so assume it is. That reduces it to the classic one about a bowling ball's speed when it achieves rolling contact, which is solved instantly by taking angular momentum about a well chosen axis.
 
  • #10
haruspex said:
But if we don’t assume it's quick it looks intractable, so assume it is. That reduces it to the classic one about a bowling ball's speed when it achieves rolling contact, which is solved instantly by taking angular momentum about a well chosen axis.

Hi,
I'm not sure I understand your reply. It's probably my English, but it seems to me that in your first sentence above you're saying that this is an intractable problem, and in the second that this is solved easily.

I can see that the problem of a bowling ball "rolling the wrong way" has a tractable solution, at least when it rolls on a horizontal plane.

The surface in this problem is curved, though, which entails that the friction force depends on the position. Off the top of my head I cannot think of a quick way of solving the problem. When I say "quick" I mean a way that there has at least a hope of being understood by the average 17 yrs old student.
From the figure I attached, one can assume that the ramp is a circular arc, which possibly simplifies the problem a bit, at least when compared to an arbitrary shape.

The difficulty I see is that the dissipated energy depends on the friction force (which varies along the way) and on the length of the marble's "trajectory".

Am I missing something?
 
  • #11
FranzDiCoccio said:
in your first sentence above you're saying that this is an intractable problem, and in the second that this is solved easily.
No, I meant that if we assume the skidding is all over very quickly, so effectively while on a flat plane, then it's easy. If we try not to make that assumption it looks intractable.
FranzDiCoccio said:
the friction force depends on the position
Yes, though that isn't necessarily a problem. For the bowling ball case, the velocity when rolling is established can be determined even if the coefficient of friction varies along the rink in an unknown way.
 
  • #12
I'm not entirely clear on the details of the problem you have in mind.
Is it something like the following?

A bowling ball slides with some initial velocity [itex]v[/itex] on a flat surface with unknown, possibly position-dependent dynamic friction, while initially spinning with an angular velocity [itex]\omega\neq v/r[/itex] about its center. Find the velocity of the ball when it starts rolling.

I guess that this can be solved by observing that the frictional force does all the work. However I see a few different cases.

  1. the ball is not spinning or it spins the same way it will in the end, but with a slower angular velocity. Here the friction slows down the translational motion and accelerates the rotational motion. The two works are the same, assuming that friction does not depend on speed/acceleration (it usually does not).
  2. the ball spins the same way it will spin in the end, but with a larger speed. Here the friction slows down the rotational motion and accelerates the translational motion. The two works are the same.
  3. the ball spins "the wrong way". Here the friction slows down both motions, at leas initially. I guess one should compare the work needed to stop either motion and consider the smallest one. The ball could stop moving while still spinning, or stop spinning while still moving. From there on, the problem falls in one of the two previous cases.
The only point I still have to get around is the meaning of "the frictional force does all the work". I expect at least some energy is dissipated due to friction, so that the final overall kinetic energy is less than the initial one. Energy is not conserved.

This is interesting. However I think that the original problem is probably tougher, in that it requires the final height on the red ramp.
I think that a problem with a small cube simply sliding on the ramp with friction would be tough enough, because the amount of energy dissipated by friction is hard to calculate. A marble rolling backwards is even harder.
 
  • #13
FranzDiCoccio said:
I'm not entirely clear on the details of the problem you have in mind.
Is it something like the following?

A bowling ball slides with some initial velocity [itex]v[/itex] on a flat surface with unknown, possibly position-dependent dynamic friction, while initially spinning with an angular velocity [itex]\omega\neq v/r[/itex] about its center. Find the velocity of the ball when it starts rolling.

I guess that this can be solved by observing that the frictional force does all the work. However I see a few different cases.

  1. the ball is not spinning or it spins the same way it will in the end, but with a slower angular velocity. Here the friction slows down the translational motion and accelerates the rotational motion. The two works are the same, assuming that friction does not depend on speed/acceleration (it usually does not).
  2. the ball spins the same way it will spin in the end, but with a larger speed. Here the friction slows down the rotational motion and accelerates the translational motion. The two works are the same.
  3. the ball spins "the wrong way". Here the friction slows down both motions, at leas initially. I guess one should compare the work needed to stop either motion and consider the smallest one. The ball could stop moving while still spinning, or stop spinning while still moving. From there on, the problem falls in one of the two previous cases.
The only point I still have to get around is the meaning of "the frictional force does all the work". I expect at least some energy is dissipated due to friction, so that the final overall kinetic energy is less than the initial one. Energy is not conserved.

This is interesting. However I think that the original problem is probably tougher, in that it requires the final height on the red ramp.
I think that a problem with a small cube simply sliding on the ramp with friction would be tough enough, because the amount of energy dissipated by friction is hard to calculate. A marble rolling backwards is even harder.
The act of bowling does not impart much spin, so assume there is none initially.
The simplest method is to consider angular momentum about a point on the ground in the path of the ball.
 
  • #14
ok.. this way is surely simpler.
I had a quick go at it and came up with a final velocity [itex]v_1 = 5/7 v_0[/itex], for a solid ball. If this is correct, friction dissipates [itex]2/5 [/itex] of the initial kinetic energy.
This is for a uniform friction coefficient on a horizontal plane.
I guess the result would be the same for a non uniform friction coefficient. It should suffice to work with the average friction force and torque.

I have not tried yet, but working with energy is probably trickier, because one cannot simply use the displacement of the ball on the plane. Since the ball is gaining an angular velocity, the displacement associated to the work done by friction is smaller than the actual displacement of the ball on the plane.
Interesting.
 
  • #15
FranzDiCoccio said:
It should suffice to work with the average friction force
No, you don't need that either. The greater the friction force the less the distance traveled before the transition to rolling. The work lost comes out the same.
With angular momentum taken about a point on the ground, the friction exerts no torque, so the angular momentum is conserved.
 
  • #16
Hmm, I see...

[tex]L_1 = I \omega_1+ m v_1 r =7/5 \, m v_1 r = L_0 = m v_0 r [/tex]

I have a little problem of interpretation though. At least at first glance an increase in the angular momentum of a ball seems odd without a torque. Of course we're not saying there is no torque, we're only saying that the torque about a point on the ground is zero.

This reminds me of a problem where a person of mass [itex]m_1[/itex] stands on a platform of mass [itex]m_2[/itex] that can slide without friction on a horizontal plane. As soon as the person starts walking on the platform in some direction, the latter starts moving in the opposite direction with respect to the ground.
The relation between the velocities of the objects is obtained via the conservation of linear momentum (which in this specific case is zero).
This comes about from the fact that the sum of external forces is zero, and everything happens due to internal forces. Here internal and external refer to the system formed by the person and the platform.

I was wondering: can the problem of the bowling ball be seen in the same way? If so, I'm not clear about what "the system" is here.
Or else, is this something different? After all in this case the "external torque" is zero even though the sum of external forces is nonzero, due to a clever choice of the center of rotation.
 
  • #17
FranzDiCoccio said:
an increase in the angular momentum of a ball seems odd without a torque
Angular momentum and torque are always in relation to a chosen axis. With the axis on the ground, there is no torque and no change in angular momentum; with the centre of the ball as axis, the friction exerts torque and the angular momentum changes.
Not sure if this could be framed as somehow analogous to a choice of system... don't see how. Choice of system already applies in both linear and rotational aspects.
Note that linear motion can be thought of as the limit of rotational motion as the axis tends to infinitely far away. E.g. the y component of momentum of an object is the limit of (its angular momentum about an arbitrary point (x,0))/(x) as x tends to infinity. (Or minus that, depending on sign convention.)
 
  • #18
haruspex said:
Angular momentum and torque are always in relation to a chosen axis.
Of course.

I'm still thinking about this problem, because I found out that it is present in the same book as the problem in the OP. In this case the book gives the correct solution, which is worked out exploiting the conservation of angular momentum.

However, students have little hope of solving this problem, because the book never mentions, let alone prove, the formula for the angular momentum wrt an arbitrary axis parallel to the one going through the center of mass.
It is not hard to prove it, but one cannot reasonably expect that the average student works it out on his/her own, without any input (again, we're talking about 17 yrs old).

So I double checked that the same result can be obtained with kinematics. At least naively, the answer seems to be affirmative.

Here is my possibly naive "kinematic approach".

The (kinetic) friction is responsible for both the angular acceleration and the linear deceleration. Since this force is constant, the accelerations are uniform, and the final angular and linear velocities are obtained through the "time-velocity relation" of uniformly accelerated motion. It is important to observe that in this case the two accelerations are proportional, but not in the same way as for a ball rolling without slipping.
Anyway, the final result is the same as in the approach based on angular momentum conservation.

Thus it appears that the angular velocity increases linearly and the translation velocity decreases linearly, until their relation is [itex] v = \omega r[/itex]. When this happens the contact point with the surface is "instantaneously" not moving wrt the plane, and the ball is rolling without slipping. In this situation the friction is not kinetic but static, and it does not produce work. Hence the accelerations vanish.

Everything seems sort of consistent, but is this correct?

I'm wondering because this would mean that the velocity is not differentiable at the "initial" and "final" times of their variation. This seems kind of strange. I would expect the velocity to vary smoothly.

What is the point here? Perhaps the "naive" kinematic approach applies only for averaged quantities?
 
  • #19
FranzDiCoccio said:
I'm still thinking about this problem
Here is the problem to consider:
I have a frictionless bowling alley. Halfway down the alley I nail down some perfectly high friction very thin carpet. I roll the ball with speed v and "equal backspin" (meaning the bottom of the ball slips with speed 2v forward).
Find speed of ball when it hits the pins ? (I get ##\frac 3 7 v## how bout you?)
The rest of the original question is then energy conservation.
 
  • #20
hutchphd said:
Here is the problem to consider:
I have a frictionless bowling alley. Halfway down the alley I nail down some perfectly high friction very thin carpet. I roll the ball with speed v and "equal backspin" (meaning the bottom of the ball slips with speed 2v forward).
Find speed of ball when it hits the pins ? (I get ##\frac 3 7 v## how bout you?)
I get [itex]\frac{3}{7}v[/itex] too, from both approaches (angular momentum conservation and "kinetic").
I'm not sure what you're trying to say here.
This does not seem to answer my last question.

hutchphd said:
The rest of the original question is then energy conservation.

What do you mean?
The part in the OP where energy conservation can be used is trivial. My question concerned the second part, when the ball climbs up the slope where friction is present. Mechanical energy is not conserved in that case.
 
  • #21
FranzDiCoccio said:
Mechanical energy is not conserved in that case.
Am I looking in correct place?

So long as the wheel does not slip relative to the ground it is static friction and mechanical energy is conserved. Rolling up the hill is the same physics as rolling down the hill.
The only non-energy conserving event is the "chirp"

.
 
  • #22
Uhm are you saying that the friction is so strong that the ball has lost all the energy it is supposed to loose while it's still at the lowest point of the ramps?
In that case of course the final height on the red ramp can be calculated via energy conservation.

But that is an approximation that follows from a rather strong additional assumption, isnt'it?

If that is not the case, one should still calculate the height at which the ball starts rolling without slipping.
 
  • #23
How do you know it doesn't slip on the downward leg? The problem says it has "rolling friction" on the red. Absent any other information I take the author at her word: no slipping on red.
You may define as many complications as you wish.
 
  • #24
I'm not sure I follow.
Are you saying that "rolling friction" means that the object cannot slide on the slope?

By the way, the text of the problem is my translation from Italian. What I meant to say is that the red slope is "rough enough" that the ball rolls on it without slipping, as soon as its starts going downward.
Apologies if that was not clear.

Maybe I'm wrong, but I'm assuming that a ball rolls without sliding if the static friction is large enough that it keeps the contact point instantaneously at rest. This does not mean that it should be extremely large.
In real life it is pretty common that a ball rolls without slipping on a surface, if the initial conditions are right. This does not mean that it would immediately roll without sliding when its angular velocity does not "agree" with its translational velocity. The friction should be really huge for that to happen, which I believe to be a strong assumption.

If I'm getting that right, it was pointed out above that this happens with airplane wheels because the weight they sustain is huge. Hard to imagine it happens in the problem at hand.

In my opinion yours is an additional assumption that should be clearly stated in the problem. Something like "the friction is so large the ball does not slide at all when it is on the red ramp". There is nothing like that in the text of the problem, though.

Anyway I agree that this further assumption makes the problem tractable.
 
  • #25
No matter what you do you have to make an assumption about the nature of the friction. A further assumption is required no matter how you solve it. It is therefore not "additional"
 
  • #26
not sure I agree...no assumption on the friction is needed to calculate the final velocity.
For instance, in your problem on the horizontal surface you do not need to assume that the friction it is extremely large, but only that the ball rolls long enough to attain the final velocity.
How long, it depends on the amount of friction.
As pointed out above by haruspex, you do not even need to assume that the friction is uniform, but only that the ball eventually starts rolling without sliding.

Personally, I do not like problems where one has to make assumptions like the above.
I feel like they are problems created by people who create problems for other people who create and solve problems.
If a 17 years old student is given this problem, the text should be more precise.

ADDENDUM: your comment gives me an idea though. In order to make the problem tractable one could slightly change the ramps, adding a horizontal plane between them, so that its left half has friction, and the right half has no friction.
Then one could add this sentence to the text "assume that the horizontal stretch with friction is long enough for the ball to resume rolling without slipping before climbing the red ramp again.". Then the problem would be tractable with no further assumption on the strength of the friction force.
 
Last edited:
  • #27
hutchphd said:
No matter what you do you have to make an assumption about the nature of the friction. A further assumption is required no matter how you solve it. It is therefore not "additional"
Since the problem setter did not understand the physics (thought work was conserved), it is not possible to defend the problem statement. The discussion can then move to how to turn it into a suitable question, and since the answer depends on how long the transition to rolling takes, that is part of the necessary information.

Since the target is a high school student, it had better say the transition is very quick.

Wrt the initial descent, i cannot know whether the original Italian makes it clear that there is no sliding there. If not, we have the same problem. The diagram shows it starting vertical, so no normal force, so no frictional force.
FranzDiCoccio said:
Here is my possibly naive "kinematic approach".
Since your argument involves forces, it is not kinematics. In particular, you will need to use the moment of inertia of the ball to get the answer.
But you are right that the student does not need to use the choice of axis trick; writing two equations involving the frictional force and thus eliminating it works, just takes a bit longer.
 
  • Like
Likes FranzDiCoccio and hutchphd
  • #28
If there were just a "long enough" flat section with friction between the semicircles the problem wouldn't require the extremely short impulse. It would be a better problem .

The point about 17 yrs old (with a some exceptions) is well taken and care is required but requiring all the information to be spoon fed can be a trap. The real world is seldom that accommodating
 
  • Like
Likes FranzDiCoccio and haruspex
  • #29
thanks everyone for helping me gaining more insight in this problem.

About "kinematics", what I meant is that the solution is obtained by solving a system of two kinematic equations giving the linear and angular velocities of uniformly (linear and rotational) accelerated motion as functions of time. Of course, the relevant (linear and rotational) accelerations are obtained from forces and torques, via Newton's second law, as usual.
In that case one can work out the time required for the onset of rolling motion, or the space. This is not readily obtained from the law of conservation of angular momentum.

On the other hand, the problem with nonuniform friction is considerably harder in the dynamics/kinematics approach.

The original problem is not really well thought. I agree that the figure introduces the further problem of the absence of normal force at the initial point. However I interpreted that figure as a simple sketch of the situation, since the text does not refer in any way to the exact shape of the ramp.
This is reasonable because, in the mind of the author, the problem should be solved using conservation laws, for which the details of the ramps should be irrelevant.
Of course, as observed, this has some exceptions.
I have the feeling that the author found the text of the problem somewhere and added the figure almost as an afterthought. I'll try and search for the image. Maybe I'll find its original source, if any.

Also, I agree that the level of detail in the text should be fine-tuned. If it is too much it becomes counterproductive.

Do you guys believe that it is realistic that the linear and angular velocities of the ball in the transient stage are non differentiable? For instance
[tex]
v(t) =
\left\{
\begin{array}{ll}
v_0 & t < t_0 \\
v_0 - a t & t_0 \leq t < t_1 \\
v_0 - a t_1 & t \geq t_1
\end{array}
\right.
[/tex]

Of course, if this is the case, I expect that "in real life" the "corners" would be smoothed out anyway.

Thanks again for your help.
 
  • Like
Likes hutchphd
  • #30
FranzDiCoccio said:
the problem with nonuniform friction is considerably harder in the dynamics/kinematics approach.
You can get around that with calculus.
##\Delta L=I\Delta\omega=r\int F.dt##
##\Delta p=m\Delta v=\int F.dt##.
Eliminate ##\int F.dt##.

FranzDiCoccio said:
Do you guys believe that it is realistic that the linear and angular velocities of the ball in the transient stage are non differentiable?
It would be very rare to worry about acceleration being discontinuous.
 
  • Like
Likes hutchphd
  • #31
haruspex said:
You can get around that with calculus.
##\Delta L=I\Delta\omega=r\int F.dt##
##\Delta p=m\Delta v=\int F.dt##.
Eliminate ##\int F.dt##.

Sure, I know. But 17 yr old students won't know integrals for a couple of years or so. I meant harder for them.

On second thoughts, they do have the concept of average. So for them we say: even if friction changes along the way, by definition its overall effect can be described through an average friction
##\Delta L=I\Delta\omega=r \bar F \Delta t##
##\Delta p=m\Delta v=\bar F \Delta t##.
Eliminate ##\bar F \Delta t##

This is actually what I was referring to when I mentioned "average friction force and torque" a few posts back [post-6492781].

haruspex said:
It would be very rare to worry about acceleration being discontinuous.

Well, at the end of the frictionless surface the discontinuity is perhaps to be expected. But I find the other one, when the ball starts rolling, kind of intriguing.
For some reason I expect the speed to vary smoothly between [itex]v_0[/itex] and [itex]v_1[/itex].
 
Last edited:
  • #32
FranzDiCoccio said:
Well, at the end of the frictionless surface the discontinuity is perhaps to be expected. But I find the other one, when the ball starts rolling, kind of intriguing.
For some reason I expect the speed to vary smoothly between #v_0# and #v_1#.
How about a sliding object coming to a stop? That's usually taken as a sudden transition from a constant-ish nonzero acceleration to a zero one. Of course, one could allow that no object is perfectly rigid, so it doesn’t all stop at once, etc., ultimately getting down to oscillations of interacting dipoles. But does the standard simplification keep you awake at night?
 
  • Like
Likes FranzDiCoccio
  • #33
Ok, I see... The point is that acceleration is caused by kinetic friction, which vanishes as soon as the object stops.

But it works even if the object does not stop, e.g. because the horizontal plane it is sliding on has a "rough patch" of finite length, that is not sufficient to stop it.

Here too the speed decreases linearly, but won't vanish.

Something similar happens with the sliding ball, although it's a bit more subtle.
When the object attains the "rolling without slipping" speed, the kinetic friction vanishes even though the surface has no discontinuity.
The relevant friction is static, which does not produce work, and hence it does not slow down the object further.Cool. Thanks
 

Related to A ball climbing a ramp while "rolling the wrong way"

1. How is it possible for a ball to climb a ramp while rolling the wrong way?

This phenomenon is possible due to the conservation of energy. When the ball rolls up the ramp, it gains potential energy. As it continues to roll in the opposite direction, this potential energy is converted into kinetic energy, allowing the ball to continue climbing the ramp.

2. What factors affect the height and speed of the ball while climbing the ramp?

The height and speed of the ball depend on the angle of the ramp, the mass and shape of the ball, and the surface of the ramp. A steeper ramp will result in a higher and faster climb, while a heavier or more aerodynamic ball will also affect the height and speed.

3. Why does the ball eventually stop climbing the ramp?

The ball will eventually stop climbing the ramp due to friction. As the ball rolls, it rubs against the surface of the ramp, converting its kinetic energy into heat. Eventually, the ball will lose all of its energy and come to a stop.

4. Can the ball climb the ramp while rolling the wrong way indefinitely?

No, the ball cannot climb the ramp indefinitely. As mentioned before, the ball will eventually come to a stop due to friction. Additionally, the ramp will eventually reach a maximum height, at which point the ball will roll back down.

5. How does the height of the ramp affect the ball's motion?

The height of the ramp affects the ball's motion by determining the amount of potential energy it gains as it rolls up the ramp. A higher ramp will result in a greater potential energy, allowing the ball to climb higher before coming to a stop.

Similar threads

  • Introductory Physics Homework Help
Replies
14
Views
2K
  • Introductory Physics Homework Help
2
Replies
60
Views
478
  • Introductory Physics Homework Help
Replies
18
Views
2K
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
21
Views
2K
  • Introductory Physics Homework Help
Replies
17
Views
1K
Replies
10
Views
516
Replies
12
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
33
Views
7K
Back
Top