Vertical circular motion in container

[Dadface]

No. At the point of detachment the net acceleration is vertical. If you break that into radial and tangential components, the tangential component is nonzero but is missing from your equation.

This wrong equation led you to the erroneous 0.91 answer and thus to a relatively modest 2.7 for the coefficient of friction. With the correct ≈0.85 answer the coefficient of friction becomes huge (in fact, infinite).

Since you do not care about the tangential acceleration, simpler to resolve gravity into tangential and radial components: mg sin(α) = ma_rad.

I am considering the instant when contact forces reduce to zero. At an infinitesimally small time before that instant contact forces will be infinitely small and the majority of the centripetal force will be provided by the inward component of the weight. At an infinitely small time after that instant detachment will be complete and the initial free flight path of of the object will be in a direction tangential to the circle at the point of release.

 

[BvU]

the initial free flight path of of the object will be in a direction tangential to the circle at the point of release

of course not: there's something in the way: the drum.

 

[haruspex]

I assume this is intended as a response to post #25.

I am considering the instant when contact forces reduce to zero.

I understand that, but unless the coefficient of friction is effectively infinite it will have started to slide at some earlier time. That will reduce the angle at which it loses contact.

 

[Dadface]

I think I know what you mean Bvu ( post 32) but let me clarify. For something moving in a circle the instantaneous velocity at all instants is tangential to the circle. The same applies to the situation discussed here so at the instant of disconnect the instantaneous velocity is tangential to the circle. However the object will not be able follow a parabolic path because as you said the drum gets in the way. I should not have used the term "free flight path".

 

[Dadface]

I assume this is intended as a response to post #25.

I understand that, but unless the coefficient of friction is effectively infinite it will have started to slide at some earlier time. That will reduce the angle at which it loses contact.

This is one reason why I would like to see the question as it was originally presented to Alex126. My impression as it stands at present Is that it is an A level physics ( UK educational system ) question. If so certain simplifying assumptions are implied and if that's the case then I would like to know what they are. And I would like to know the other details of the question. Without the details I think that any discussion can keep going round in circles.
I'm also suspicious about the concept of infinite friction. From a practical viewpoint I can see the concept being approached by fixing the object to the drum with glue. Ensuing discussions can all get a bit sticky.

 

[PeroK]

This is one reason why I would like to see the question as it was originally presented to Alex126. My impression as it stands at present Is that it is an A level physics ( UK educational system ) question. If so certain simplifying assumptions are implied and if that's the case then I would like to know what they are. And I would like to know the other details of the question. Without the details I think that any discussion can keep going round in circles.
I'm also suspicious about the concept of infinite friction. From a practical viewpoint I can see the concept being approached by fixing the object to the drum with glue. Ensuing discussions can all get a bit sticky.

Another solution is not to have friction, but indentations that hold the object in place and prevent tangential slippage.

An adhesive force complicates the force needed to detach the object.

That said, "infinite" friction, in the sense of "large enough friction to make a negligible difference to the result" is not that different from "frictionless", in the sense of "small enough friction to make a negligible difference".

 

[haruspex]

the object will not be able follow a parabolic path

No, it will follow a parabola, not merely tangent to the drum but of second order contact. The third derivative will be different.

 

[Alex126]

Which of the two answers is supposed to be correct? Using my reasoning (see post above) I calculate f to be 0.85.

Everyone says it should be 0.85 (and I'm sufficiently convinced that's the right one), though the textbook says 0.91. Could just be an error on the textbook part, at this point that's my guess.

Alex 126. I would like to follow the discussion but I ask you please to clarify the question. For example is it the object that rotates or the drum? Do both rotate? Does the question state or imply that certain simplifying assumptions can be made? I think the best thing to do is to send the question exactly as it was presented.

I'd like to quote it exactly, but it's not in English. You can trust my translation though -- I've reported everything that was in there, and it didn't provide any additional info that I've omitted, even "subtle" things. The only "implicit" info that I was missing at first is, I think, that saying "the object falls" really means "at the point where N = 0". For the record, for what it's worth, the rotating drum is supposed to be a dryer machine.
Here's the full text, just in case:

In a dryer machine, clothes move inside the cavity of a cylinder [the machine] of radius r = 0.32m which rotates vertically. The device is built so that clothes can roll gently while they dry. This means that when a piece of clothing reaches a certain angle α above the horizontal line, it loses contact with the surface of the cylinder and falls down.
How many rounds per minute must the cylinder do so that the clothes lose contact with the surface of the cylinder when the angle α = 70° ?
+Picture:

it is an A level physics ( UK educational system ) question

It's high school level.

Parabola or not, I think it's safe to assume the problem just wanted to know when the object "detaches".

No, it will follow a parabola, not merely tangent to the drum but of second order contact

Second order of contact? If I had to guess what the parabola trajectory/equations are (pretending that the drum just disappears once the object loses contact with it), I would say (with X axis horizontal, Y axis vertical):

X component (+X to the left):
x = v*t
(with v = the tangential velocity, obtained from a = v²/r)

Y component (+Y down):
y = 1/2 * g * t²

This doesn't feel totally right though. It feels like it would be right if the object detaches at 12 o'clock, so the vector v is actually horizontal, but in this case it's not. Maybe for the X axis we would use V_x instead of v?
So V_x should be: v * cos (90-α), which should be v * sin(α) if the drawing below is right:

And likewise, on the Y axis (+Y down) it should actually be:
y = -V_y*t + 1/2 * g * t²

Gonna try and see about post #30 next.

 

[Alex126]

I think I get why μ must be infinite. With X centripetal axis, Y tangential:

X axis:
+N + W_x = m*a

Y axis:
+Friction - W_y = 0
=> N*μ = W_y
=> N = W_y/μ

Since N must be zero at the point of detachment, then N = W_y/μ => we need μ to be infinite to make that fraction -> 0.

With a finite coefficient of friction, the object must start to slip before it falls off. The object falls off when the normal force reaches zero, but it starts to slip when μF_N<gcosθ. Which must happen before F_N=0.

Just to be clear, you meant "when μF_N<mgcosθ" right?

 

[PeroK]

Just to be clear, you meant "when μF_N<mgcosθ" right?

Yes, of course. I'll correct it.

 

[Dadface]

Thanks Alex126, your post (38) clarifies the question.

 

[Dadface]

Just a brief note on a practical point ........ the drums in tumble dryers and front loading washing machines have radial projections on the walls of the drums. These help to lift the clothes. Just saying.

 

[haruspex]

Second order of contact?

Yes. For the drum, the radius of curvature is constant. For the parabola it reduces towards the apex. While the unconstrained parabolic motion of the object would have the greater radius, it remains on the drum wall. Once its radius becomes less than that of the drum it will detach.

 

[haruspex]

the drums in tumble dryers and front loading washing machines have radial projections on the walls

... and now we know why.

 

[Alex126]

... and now we know why.

lol

Yes. For the drum, the radius of curvature is constant. For the parabola it reduces towards the apex. While the unconstrained parabolic motion of the object would have the greater radius, it remains on the drum wall. Once its radius becomes less than that of the drum it will detach.

So if I'm reading this right you mean that after the object detaches it would start its parabolic motion, but it "hits its head" on the drum because the parabola trajectory "goes up" higher than the drum's circumference. Then, once the drum's circumference is "higher" than the parabola trajectory, it starts to behave like a normal parabola again. Something like this:

(The parabola is not necessarily realistic)
In the "green area" the parabola would exist, but the fact that the circle is in the way means that it obviously can't follow the green path because the black arc is in the way. But, if the drum were to disappear all of a sudden, we would have the parabola as it is, and it would have the X and Y component as I wrote them in the previous post. Is this what you meant?

 

[haruspex]

after the object detaches it would start its parabolic motion, but it "hits its head" on the drum

No, I mean that at any instant before it detaches, if you were to take away the drum then its parabola would cross the drum's circle. But at any time thereafter the drum makes no difference - the parabola would remain inside it.
At the instant of detachment, the parabola it embarks on is a smooth continuation of its previous circular path, having both the same slope and the same radius of curvature. Immediately, the radius of curvature starts to shrink, drawing the mass away from the drum.

 

[Alex126]

Alright, I think that's beyond my current understandings and knowledge. I'm going to pass on that last bit for now :D Thanks again everyone.