A problem from a physics national competition

[Jim Alexandridis]

So we have an incline of slope θ, which is I circular motion around a pole with steady angular velocity ω. A body on the incline is at distance r frome the pole and stays at rest. Calculate the minimum friction coefficient so that it stays at rest.



So I analyzed the forces in 2 different "systems". One with the one to the center and the other perpendicular. The second one was when the one was perpendicular to the incline. From the second one I got: μ=tan θ. From the first one : fcosθ=mω²r(1) and fsinθ=mg(2) dividing (2):(1):
Tanθ=g/(ω²r)=μ. So I suppose that is the minimum coefficient. What do you all have to say about it ?

 

[erobz]

So we have an incline of slope θ, which is I circular motion around a pole with steady angular velocity ω. A body on the incline is at distance r frome the pole and stays at rest. Calculate the minimum friction coefficient so that it stays at rest.



So I analyzed the forces in 2 different "systems". One with the one to the center and the other perpendicular. The second one was when the one was perpendicular to the incline. From the second one I got: μ=tan θ. From the first one : fcosθ=mω²r(1) and fsinθ=mg(2) dividing (2):(1):
Tanθ=g/(ω²r)=μ. So I suppose that is the minimum coefficient. What do you all have to say about it ?

I'm getting something a bit more involved that. Maybe I'm mistaken, but either way would you care to show your FBD, frame of reference, equations of motion, etc...?

 

[hutchphd]

"so I suppose " is not physics as I presume you understand. Would you care to elaborate using free body diagrams and actual physical laws? Notice that in general the body can slide either down or up, depending upon the speed and
angle.
ERRATUM: I see I am reiterating......!

 

[Steve4Physics]

So we have an incline of slope θ, which is I [do you mean 'in'?] circular motion around a pole with steady angular velocity ω. A body on the incline is at distance r frome the pole and stays at rest [do you mean 'at rest with respect to the incline'?].

(My edits in square brackets.)

The setup isn't fully specified. One possible interpretation is this:

Is that what you mean? If not, you need to specify the required setup.

 

[hutchphd]

Yes. You see I assumed the slope was towards thr rotation center axis. $$ picture=~kword$$

 

[haruspex]

(My edits in square brackets.)

The setup isn't fully specified. One possible interpretation is this:

View attachment 358621

… and that is assuming the plane containing pole and object cuts the wedge along its line of greatest slope.

 

[kuruman]

What do you all have to say about it ?

I, like @erobz, am getting something more complicated. Your expression does not give the expected result in the two relevant limiting cases when the block is on the verge of sliding. These are

1. When there is no rotation, i.e. $\omega=0$, $\mu_s=\tan\!\theta.$ Your expression predicts $\mu_s \rightarrow \infty$ or $\mu_s=\tan\!\theta.$
2. When there is no incline, i.e. $\theta=0$ and static friction is the horizontal force that provides the acceleration, $\mu_s=\dfrac{\omega^2~r}{g}.$ Your expression predicts $\mu_s =0$ or $\mu_s=\dfrac{g}{\omega^2~r}.$

Clearly the equations you derived

Tanθ=g/(ω²r)=μ.

are inappropriate because they say that the same quantity, μ, can have two different values. Since you don't show the details of your work, I cannot figure out where you went wrong.

BTW, in my derivation I went by the incline shown in the picture, i.e. sloping away from the rotation axis.

 

[haruspex]

fcosθ=mω²r(1) and fsinθ=mg(2)

What is f here?
Reverse-engineering your equations, you are considering only three forces acting on the body: gravity, centrifugal force and the reaction from the surface (which therefore is the sum of the normal force and the frictional force). But in that case, $\theta$ would be the angle f makes to the horizontal, not the angle of the slope.

 

[Lnewqban]

Yes. You see I assumed the slope was towards thr rotation center axis. $$ picture=~kword$$

Isn’t this problem similar to the classic car on a banked curve?

 

[hutchphd]

Yes that is what Ithink. I recall a few icy nights in Ithaca where you needed to keep your speed juuust right on the banked curves......all's well that ends well!

 

[kuruman]

Isn’t this problem similar to the classic car on a banked curve?

Very much so. Except that there is friction and the surface is inclined away from the center of the circle. To add to @hutchphd's reminiscences, I remember a downhill turn on a country road in Vermont that became banked the wrong way due to frost heaves. Really tricky to negotiate in icy conditions until it got fixed in the summer.

 

[hutchphd]

Frost heaves are there to keep you alert.....

 

[Tom.G]

.....or terminate you.

(I grew up in Buffalo N.Y., followed by a couple of years in Burlington, VT. Definitely interesting weather in both places.)

 

[hutchphd]

Yeah I was in Orono Maine for a while, too. Four foot deep frost line.