Find the magnitude of this velocity - SHM problem

[AgNO3]

Homework Statement

Two rubber pads are affixed at the bottom of a box and the assembly thus formed is placed on a uniform slope of inclination 0.5 degrees. Coeff of friction is 0.60 between the pads and the plane.
Two electric motors installed in the box can make the pads move back and forth relative to the box simple harmonically in opposite phase along the line of fastest decent on the slope. Amplitude is 0.25 mm , and angular frequency ω= 72 rad/s. Soon after the motors were switched on simultaneously , the box acquires a constant velocity down the slope.
Find the magnitude of this velocity .
final answer is 0.4 mm/s .

Homework Equations

None .

The Attempt at a Solution

Well ..this question is more about correct approach so i spent lots of time thinking about the scenario , and this is what i got.
Initially , velocities are in opposite directions, so the friction will act in opposite directions. Hence net frictional force is zero. The box will move down coz of gravity.
But its given that velocity is constant, ...so i think that it may be bcoz the velocity of pads are slower than that of the box itself.
But i am not able to do anything with this conclusion i made.
Any hints ?

 

[jbriggs444]

But its given that velocity is constant

It may be given, but it will not be so. What will be true is that the box approaches a constant average velocity. We can reason our way to this conclusion as follows.

Assume (for a proof by contradiction) that the box slides down at a constant speed.

No matter how slow this constant speed is, there will be intervals during the cycle of harmonic motion when the two pads are near the extreme ends of their travel and are momentarily moving at a rate that is lower than the constant slide rate. During these intervals, both pads will be traveling down-slope. The force of friction on both pads will be momentarily up-slope.

With a coefficient of friction of 0.6 and a slope of only 0.5 degrees from the horizontal, the box will clearly be accelerated up-slope by friction at these times. So its velocity cannot be constant.

Since we have obtained a contradiction, the premise cannot hold. The box does not slide down at a constant speed.

What can remain true however is that the box can attain a stable average speed over the course of a pad cycle. Part of the time the pads will be traveling faster than the slide rate. Part of the time they will be traveling more slowly. Can you determine the resulting acceleration for those two situations?

 

[haruspex]

This seems to me a very tricky problem requiring some assumptions to make it amenable to solution. The difficulty may be guessing the assumptions the author is making.

Most obviously, the load would not be equally spread on the two pads. The lower pad would have the greater normal force and so the greater potential frictional force. This could lead to the lower pad remaining still while the block and higher pad do a soft-shoe shuffle up and down the slope.
So, first assumption, that the mass centre of the block is vertically above the mid point of the pads.

 

[TSny]

Most obviously, the load would not be equally spread on the two pads. The lower pad would have the greater normal force and so the greater potential frictional force. This could lead to the lower pad remaining still while the block and higher pad do a soft-shoe shuffle up and down the slope.
So, first assumption, that the mass centre of the block is vertically above the mid point of the pads.

We are given that the block does move down the slope with a constant velocity (ignoring the small, jittering accelerations mentioned by @jbriggs444). With that assumption, I don't think the answer depends on whether or not the normal force is the same for each pad. In fact, I get the same answer assuming that there is just one pad or assuming that there is any number of pads oscillating with arbitrary phase differences. (Of course, I could be making unwarranted assumptions that I'm not aware of, or I could be simply mistaken.)

 

[haruspex]

We are given that the block does move down the slope with a constant velocity (ignoring the small, jittering accelerations mentioned by @jbriggs444). With that assumption, I don't think the answer depends on whether or not the normal force is the same for each pad. In fact, I get the same answer assuming that there is just one pad or assuming that there is any number of pads oscillating with arbitrary phase differences. (Of course, I could be making unwarranted assumptions that I'm not aware of, or I could be simply mistaken.)

If the weight of the block always falls within the footprint of the lower pad there is no reason for there to be any normal force on the upper pad.
However, I was overlooking that the oscillations may be sufficiently violent that slipping on the lower pad still occurs, leading to a net downslope motion.

 

[hutchphd]

If the weight of the block always falls within the footprint of the lower pad there is no reason for there to be any normal force on the upper pad.
However, I was overlooking that the oscillations may be sufficiently violent that slipping on the lower pad still occurs, leading to a net downslope motion.

The slope is half of one degree! I think equal loading on the pads is a good enough supposition. Note that absent any motion there is no friction (or countervailing frictions rather) and at speeds slow compared to vibration speed that is still true for much of each cycle.

 

[haruspex]

The slope is half of one degree! I think equal loading on the pads is a good enough supposition.

Depends on the height and width of the block.

absent any motion there is no friction (or countervailing frictions rather)

Sorry, I do not know what you mean by that.

 

[hutchphd]

Sorry, I do not know what you mean by that.

The motion of pads at interface (and hence frictional force) are in opposite directions. They (approximately) sum to zero for (nearly) equal Normal force.

 

[haruspex]

The motion of pads at interface (and hence frictional force) are in opposite directions. They (approximately) sum to zero for (nearly) equal Normal force.

You are assuming both pads are always sliding. That is unlikely.

 

[hutchphd]

You are assuming both pads are always sliding. That is unlikely.

Not true. I am saying that what they do is exactly anti-symmetric when the block is at rest and nearly so otherwise.

 

[haruspex]

Not true. I am saying that what they do is exactly anti-symmetric when the block is at rest and nearly so otherwise.

If one pad is stationary and the other sliding, at some instant, we know what the frictional force is at the sliding pad, but how do you determine what it is at the stationary pad?

 

[TSny]

I have a problem with my solution for two pads oscillating out of phase, with each pad having the same normal force. I got an answer of 0.41 mm/s for the “constant” speed of the block down the plane, which looks good. But, I assumed that to a good approximation I could treat the speed of the block as being truly constant (even during a cycle of oscillation of the pads).

But then I noticed that during one cycle of oscillation of the pads, the friction forces of the pads cancel for almost all of the cycle. (I find that the net friction force on the block is zero for about 98.5% of the cycle). Thus, during essentially all of a cycle, the block is freely accelerating down the slope due to gravity! During this time, the block would change its speed by approximately $g \sin \theta \cdot T$, where T is the period of the oscillation of the pads. This evaluates to a change of speed of about 7.5 mm/s, which is almost 20 times the speed of 0.4 mm/s. So, this is certainly inconsistent with my assumption that I can treat the speed of the block as constant during a cycle of oscillation of the pads.

Right now, the only way I can see to have the block move at a constant speed is to assume that there are many pads, each with the same normal force, but with different phases of oscillation. Then the net friction force on the block due to all of the pads would be virtually constant.

 

[haruspex]

Then the net friction force on the block due to all of the pads would be virtually constant.

So produce a constant acceleration?

 

[TSny]

So produce a constant acceleration?

No, the constant friction force would just cancel the gravitational force along the incline.

 

[haruspex]

No, the constant friction force would just cancel the gravitational force along the incline.

Ok, so are you saying that the constant friction force is speed dependent, and will cancel gravity when at the right speed?

 

[TSny]

Ok, so are you saying that the constant friction force is speed dependent, and will cancel gravity when at the right speed?

Yes. For many pads with different phases of oscillation, I think the net friction force can have essentially a constant value that depends on the speed of the block. There will then be a particular speed of the block such that the net friction force will cancel the gravity force. I believe the speed will be found to be about 0.4 mm/s.

I'm too tired to think straight at the moment. I'll come back to it tomorrow.

 

[AgNO3]

Most obviously, the load would not be equally spread on the two pads. The lower pad would have the greater normal force and so the greater potential frictional force. This could lead to the lower pad remaining still while the block and higher pad do a soft-shoe shuffle up and down the slope.
So, first assumption, that the mass centre of the block is vertically above the mid point of the pads.

The author made the slope too small ...so i think he meant us to treat both of the pads in the same manner.

 

[AgNO3]

If one pad is stationary and the other sliding, at some instant, we know what the frictional force is at the sliding pad, but how do you determine what it is at the stationary pad?

How is that possible ?
Both of them were switched on simultaneously , so both of them will be at rest and at motion at same times.

 

[haruspex]

How is that possible ?
Both of them were switched on simultaneously , so both of them will be at rest and at motion at same times.

That is motion in relation to the block. At some moment, it may be that one pad is stationary on the slope while the block and other pad move. We know what the frictional force is on the sliding pad, but what is it on the stationary pad?

 

[AgNO3]

That is motion in relation to the block. At some moment, it may be that one pad is stationary on the slope while the block and other pad move. We know what the frictional force is on the sliding pad, but what is it on the stationary pad?

You mean that the motion of the two pads is not a mirror image of each other ??
I can't imagine that ...

 

[haruspex]

You mean that the motion of the two pads is not a mirror image of each other ??
I can't imagine that ...

Suppose that the block is about 100 times as high as it is wide. At half a degree of slope, its mass centre could be over the lower pad all the time. The upper could slide freely up and down the slope with very little friction.
With the actual parameters given that probably doesn't work because the acceleration of the block could not be sustained by the stationary pad, but you can get the idea.

 

[haruspex]

I set up a spreadsheet to model the problem as given. It's quite messy, all the different cases to consider, so it's possible I have a bug.
The model covers 1000 steps of 200μs. It starts with the pads at their closest.
It shows that at first the pads are sliding in opposite directions, so there is no net friction and the block slides down, accelerating at $g\sin(\theta)$= 85mm/s^-2.
After nearly half a cycle, the pads are near their max separation and moving slowly relative to the block. This results, briefly, in both pads sliding down the ramp, creating a large acceleration up it. This swiftly leads back to the prior state, but switches back to this state again in short order. There follows a flurry in which the state alternates between both sliding up or both sliding down. Now, that brings into question whether the granularity of the timesteps is a problem, but I played around with that and it doesn't seem to affect the result.
In summary, each half cycle it accelerates as though no friction, then is effectively brought to a halt. The average velocity is therefore $g\sin(\theta)\frac{\pi}{2\omega}=1.87mm/s$.
In the chart, the red and yellow lines are pad velocities relative to the block, and the blue line is the block's velocity.
Downslope is positive, units mm and seconds.

 

[TSny]

.

After nearly half a cycle, the pads are near their max separation and moving slowly relative to the block. This results, briefly, in both pads sliding down the ramp, creating a large acceleration up it. This swiftly leads back to the prior state, but switches back to this state again in short order. There follows a flurry in which the state alternates between both sliding up or both sliding down. Now, that brings into question whether the granularity of the timesteps is a problem, but I played around with that and it doesn't seem to affect the result.
In summary, each half cycle it accelerates as though no friction, then is effectively brought to a halt. The average velocity is therefore $g\sin(\theta)\frac{\pi}{2\omega}=1.87mm/s$.

That all looks good to me. Nice analysis!

Yes, I think that when the friction "kicks in", the blue graph will essentially just follow the sine curve down to v = 0 and the process will repeat.

So, the answer of 1.87 mm/s for the average speed of the block appears to be independent of the coefficient of friction as long as μ isn't too small. I believe things might be qualitatively different if μ is less than about $\omega^2 A /g = 0.13$. I don't know if you can run your spreadsheet for a case where μ = 0.10, say. Don't bother if it's too much trouble.

 

[TSny]

If I'm thinking correctly, then the haruspex-plots below show some of the behavior for smaller values of $\mu$ where friction is not strong enough to bring the velocity back to zero. The blue portions of the graphs are where the net friction force is zero, the red portions of the graphs are where the net friction is nonzero.

For $\mu = .05$ the graph "climbs up" on the sine wave and levels off to an average velocity of about 5 mm/s.
For $\mu = .02$ the graph is similar but "climbs up" higher until you get an average velocity of about 10 mm/s.

In the case of $\mu = .001$, we have a case where $\mu <\tan \theta$ so that gravity is stronger than friction. The red portions now slope upward. Eventually, the red portion "breaks free" of the sine bumps and the block then has a constant acceleration down the slope.

 

[haruspex]

Very helpful illustrations of this case. Remarkably, I believe this case affords a closed form solution for the average velocity, though it gets very messy.
In the original problem, the exact solution requires solving an equation with mixed linear and trig functions of the unknown.
But there is a third case.

some of the behavior for smaller values of μ where friction is not strong enough to bring the velocity back to zero.

I believe there are values of μ where the velocity does not come back to zero, but also does not match your charts. In this range, when the blue line strikes the upper curve, the acceleration afforded by the combined friction is greater in magnitude than that of the pad at that point (as in the actual problem) but less than the max acceleration of the pad. So the velocity of the pad follows the sine curve down some way, but then leaves it tangentially.

 

[TSny]

I believe there are values of μ where the velocity does not come back to zero, but also does not match your charts. In this range, when the blue line strikes the upper curve, the acceleration afforded by the combined friction is greater in magnitude than that of the pad at that point (as in the actual problem) but less than the max acceleration of the pad. So the velocity of the pad follows the sine curve down some way, but then leaves it tangentially.

Ah. So, the point of leaving the sine curve, in this case, would be where the slope of the sine curve matches the acceleration of the block when friction is acting. Is that right?

 

[haruspex]

Ah. So, the point of leaving the sine curve, in this case, would be where the slope of the sine curve matches the acceleration of the block when friction is acting. Is that right?

When max friction is acting, yes.

 

[TSny]

When max friction is acting, yes.

Thanks. Another interesting case appears to be when $μ = \tan \theta \approx .0087$, where $\theta$ is the angle of incline. Then, when the net friction force is nonzero, the gravitational force along the incline cancels the friction. So, the block has no acceleration when the net friction is nonzero. (The red segments of the graphs in post #24 would become horizontal.) If the block starts at rest, I think it will eventually reach a constant speed down the incline equal to the maximum speed of the pads relative to the block (18 mm/s).

 

[haruspex]

Thanks. Another interesting case appears to be when $μ = \tan \theta \approx .0087$, where $\theta$ is the angle of incline. Then, when the net friction force is nonzero, the gravitational force along the incline cancels the friction. So, the block has no acceleration when the net friction is nonzero. (The red segments of the graphs in post #24 would become horizontal.) If the block starts at rest, I think it will eventually reach a constant speed down the incline equal to the maximum speed of the pads relative to the block (18 mm/s).

Yes, and I think it will do so in a finite number of steps.
The top of the sine curve is near enough a parabola.
We can represent the progress within a single half period by taking the horizontal back to the left to hit the curve again. So it goes up in a zigzag, rising to the right then horizontal to the left.
As it approaches the top, the curve, magnified, looks flatter and flatter. Eventually a rightward line must hit the curve again before the peak. It will then follow the sine curve to the peak and thereafter be horizontal.

 

[TSny]

Yes, and I think it will do so in a finite number of steps.
The top of the sine curve is near enough a parabola.
We can represent the progress within a single half period by taking the horizontal back to the left to hit the curve again. So it goes up in a zigzag, rising to the right then horizontal to the left.
As it approaches the top, the curve, magnified, looks flatter and flatter. Eventually a rightward line must hit the curve again before the peak. It will then follow the sine curve to the peak and thereafter be horizontal.

Right.