Dynamics: minimum velocity to prevent slipping

[jaguar ride]

Homework Statement

A 200 lb man lies against the cushion on a Rotor ride, which is inclined at θ=10°. The coefficient of static friction is μs=0.3. Radius is 8 ft. Find minimum speed so that the man does not slip down.

Homework Equations

ΣF_n = ma_n
ΣF_z = 0
a_n = v²/r

The Attempt at a Solution

OK, so I've done several of these problems without any issues. But now I'm running into a problem finding minimum velocity.
Here's my setup:

ΣF_n = mv²/r
FnormalCos80-0.3FnormalCos10 = 6.21v²/8ΣF_z = 0
FnormalSin80+0.3FnormalSin10 = 200

The positive z axis is up, and the positive n axis is to the left.

The issue is that once I plug the normal force into my first equation, the left hand side becomes negative. This is the same approach I've taken with other problems, but with larger angles. I feel like I'm going insane. Can someone please help me out? Thanks.

 

[BvU]

Hi Jag,

so I've done several of these problems without any issues. But now I'm running into a problem finding minimum velocity

Good. My advice: when problem, make drawing. I miss that a little here... Is ##\theta## wrt the vertical ? or wrt the horizontal ?

Which way is ##\Sigma \vec F## pointing ? And ##\Sigma \vec F_n## ? If ##\vec F_n## is the normal force, then what is ##F_{\rm normal}## (just teasing: but don't use multiple symbols when referrring to one and the same variable...).

Can you explain what you do when you do ##F_n\cos 80 - 0.3 F_n \cos 10 ## ? What does 6.21 stand for ?
(advice: better to use symbols as long as feasible -- then divide out the common factors and only then calculate a value).

And almost at the end you tell me z+ is up. What is the n axis ?

No need to go insane , there's way too many of us already ...

 

[haruspex]

And ##\Sigma \vec F_n## ? If ##\vec F_n## is the normal force, then what is ##F_{\rm normal}## (just teasing: but don't use multiple symbols when referrring to one and the same variable...).

As I read it, 'n' here is indicating the horizontal axis. Confusing, I agree.

 

[haruspex]

FnormalCos80-0.3FnormalCos10 = 6.21v2/8

Careful with the signs. Is the friction helping or hindering in achieving the centripetal force needed?

 

[jaguar ride]

"n" would technically be the x-axis, but according to my book, it's to be labeled "n" in this chapter. Also according to my book, the z-axis should be "b", but my professor likes to use "z".

Anyways, here's the free body diagram. Is friction not working against centripetal force? It's preventing the man from sliding down the incline.

 

[haruspex]

Is friction not working against centripetal force? It's preventing the man from sliding down the incline.View attachment 96558

Sorry, my mistake. Got confused with max speed.
Your calculation appears correct, assuming 10 degrees is the angle to the horizontal, but bear in mind the exact definition of the coefficient of static friction. What is it?

 

[jaguar ride]

Sorry, my mistake. Got confused with max speed.
Your calculation appears correct, assuming 10 degrees is the angle to the horizontal, but bear in mind the exact definition of the coefficient of static friction. What is it?

What do you mean by exact definition? Static friction is what's keeping an object from moving. Once it moves it becomes subjected to kinetic friction.

 

[haruspex]

What do you mean by exact definition? Static friction is what's keeping an object from moving. Once it moves it becomes subjected to kinetic friction.

Yes, but I asked about the definition of the coefficient.

 

[jaguar ride]

I'm fairly certain it's the ratio of frictional forces. I'm not sure where that factors in.

 

[haruspex]

I'm fairly certain it's the ratio of frictional forces. I'm not sure where that factors in.

See section 3 of https://www.physicsforums.com/insights/frequently-made-errors-mechanics-friction/

 

[jaguar ride]

I'm sorry, I'm still not understanding where I've gone wrong.

|Fs|≤μs|FN|

So the force of static friction can be less than the normal*coefficient. How do I know what that is? Could it be zero? I don't get it.

 

[haruspex]

I'm sorry, I'm still not understanding where I've gone wrong.

|Fs|≤μs|FN|

So the force of static friction can be less than the normal*coefficient.

Yes, that's the critical point. Your calculation assumed the frictional force was at its maximum value, and that gave you a silly answer. What is your conclusion?
(By the way, you still have not confirmed that the 10 degrees is the angle to the horizontal. Taking it as 10 degrees to the vertical would not have produced a silly answer.)

 

[jaguar ride]

Yes, it's 10 degrees from the horizontal. I wasn't running into this issue with other problems since the angles had all been larger.

I still don't know what my conclusion is, I'm even more confused now. How do I figure out exactly what the frictional force is? Or how do I prove it's zero? I found the normal force by including the frictional force in my equations. If it's zero shouldn't I have left it out completely?

 

[haruspex]

Yes, it's 10 degrees from the horizontal. I wasn't running into this issue with other problems since the angles had all been larger.

I still don't know what my conclusion is, I'm even more confused now. How do I figure out exactly what the frictional force is? Or how do I prove it's zero? I found the normal force by including the frictional force in my equations. If it's zero shouldn't I have left it out completely?

No, it's not zero.
The static frictional force is whatever tangential force is required to keep the surfaces from sliding against each other, up to the maximum allowed magnitude.
If you start with the premise that it does not slide, you can deduce the frictional force for a given speed. You feel stuck because you do not know what speed to plug in.
Consider this: when you assumed static friction at its maximum value, you showed there was no speed low enough to require that. So what does that suggest the minimum speed is?

 

[jaguar ride]

Well, since there is no speed low enough to require the maximum amount of static friction, I'm inclined to believe that it could not be rotating at all and there would be no slipping. But I've been confused for this entire thread, so I'm not feeling too confident about that answer.

 

[haruspex]

Well, since there is no speed low enough to require the maximum amount of static friction, I'm inclined to believe that it could not be rotating at all and there would be no slipping.

Try that and see if it all works.
(If you are well acquainted with problems about static friction on slopes, you will know that the condition for stasis involves just the tangent of the angle and the coefficient.)

 

[jaguar ride]

Alright, so the left side of my equation must be greater than or equal to zero. If I set it equal to or greater than zero I can find out what value of frictional force will make that happen. Since the maximum frictional force wants it to be negative, it's going to take the first opportunity it has to actually work. The first possible frictional force will therefore end up making the left side equal to zero, making velocity equal zero.

Am I on the right path here, or did I just make up my own physics? I am so burnt out that this makes sense in my tired brain.

 

[jaguar ride]

Try that and see if it all works.
(If you are well acquainted with problems about static friction on slopes, you will know that the condition for stasis involves just the tangent of the angle and the coefficient.)

After talking to my professor, I discovered that they didn't even solve this problem and didn't realize they had made the angle so small. We were given credit based solely on our setup as a result, but they didn't take the time to explain what was actually going on in this instance.

Thanks for your help, haruspex.

 

[Biker]

Nvm edited. Sorry
My point was mentioned above.

 

[haruspex]

Alright, so the left side of my equation must be greater than or equal to zero. If I set it equal to or greater than zero I can find out what value of frictional force will make that happen. Since the maximum frictional force wants it to be negative, it's going to take the first opportunity it has to actually work. The first possible frictional force will therefore end up making the left side equal to zero, making velocity equal zero.

That's basically right. If the static frictional force needed to balance all the other forces is less than maximum then it will be what is needed and no more.
For a stationary slope, the standard result is that the object will slide if the tangent of the slope exceeds the coefficient of friction. As you can see, the tangent of 10 degrees is rather less than 0.3.

 

[Biker]

Alright, so the left side of my equation must be greater than or equal to zero. If I set it equal to or greater than zero I can find out what value of frictional force will make that happen. Since the maximum frictional force wants it to be negative, it's going to take the first opportunity it has to actually work. The first possible frictional force will therefore end up making the left side equal to zero, making velocity equal zero.

Am I on the right path here, or did I just make up my own physics? I am so burnt out that this makes sense in my tired brain.

Just saying that the velocity is 0 means that your are no longer in a circular motion. So It should be a bit bigger than 0.
As you said "The first possible frictional force "

 

[haruspex]

Just saying that the velocity is 0 means that your are no longer in a circular motion. So It should be a bit bigger than 0.

I don't have a problem with a minimum speed being zero.

 

[Biker]

I don't have a problem with a minimum speed being zero.

Isn't that just standing still? If the question allows that then its alright. I thought that he has to at least moving at a certain speed. I could just answered it 0 directly without trying to solve it. That doesn't make sense to me at least

 

[haruspex]

Isn't that just standing still? If the question allows that then its alright. I thought that he has to at least moving at a certain speed. I could just answered it 0 directly without trying to solve it. That doesn't make sense to me at least

As j r wrote in post #18, the question specifications were a mistake. But it is consistent, so we can still take it as a valid question. I see nothing in the wording that prohibits an answer of 0. Indeed, if it were to specify "minimum nonzero speed" then the question would not be valid since the set of values greater than zero does not include a minimum.

Although the question was a mistake, I regard it as a fruitful one. It makes the solver aware of false assumptions that may arise from only solving the usual questions. I wish teachers would set more like this.