Vertical circular motion in container

[Alex126]

Homework Statement

We have an object rotating inside a circular container. The rotation is vertical (see picture below).
The radius r of the container is given. Find the frequency (which then ties back to finding the centripetal acceleration) necessary, so that the objects rotating inside the container will fall down once they form a fixed/given angle α with the horizontal plane (the angle is 70°).

Homework Equations

Force (centripetal) = m*acceleration (centripetal)
Centripetal acceleration = v²/r

The Attempt at a Solution

I tried drawing the free-body diagram, and this is my best attempt:

F = centripetal force
W = weight force
R = reaction force

At first I actually thought of putting R (reaction force) perpendicular to the surface, i.e. in the same direction as F, but things just didn't add up. I watched several videos on this topic, and most of them just talked about "convenient/easy" positions inside the circle (at the very bottom, at the very top, or at "3/9 o'clock"). Just one video mentioned intermediate positions, which is the case with this problem.

That video said, and that's the way I've set up the problem, that you need to "deduce" the position of R (so R is NOT perpendicular to the tangent to the circle in the point where the object is), knowing that the sum of all forces must be equal to F. So I tried my best doing an accurate drawing, using the geometric vector sum rule (the parallelogram thing), summing F and -W to "deduce" R in the drawing.

I was then going to proceed summing the X and Y components to happily solve for a, and then I got stuck because I can't find any way to obtain the green angle in the figure, which seems necessary to calculate the Rx and Ry components.

The axis I was going to choose were +X axis in the same direction of F, and then +Y perpendicular to it and pointing "northwest" (up and left). That would have given me:

Y axis:
Ry - Wy = 0

X axis:
Rx + Wx = m*a

But, again, I can't seem to find the green angle to use for Ry and Rx.

 

[BvU]

Hello Alex,

At first I had trouble forming a picture of what is happening. Looks like a washing machine or something. The 'objects' are stuck against the inner wall of the cylinder, and if the whole thing doesn't rotate fast enough they will get unstuck. The objects indeed rotate, i.e. follow a circular trajectory (the black circle in your picture). You worry about R but can you save yourself the troubl: As soon as W can't "deliver" the vertical component of F, the objects come loose from the wall.

 

[PeroK]

@Alex126 you may need to rethink the problem. Whatever videos you have watched perhaps haven't helped.

Think about why the object stays moving in a circle or falls off. What is the critical factor that determines when it falls?

 

[Alex126]

Looks like a washing machine or something

Yea, that's the idea.

Think about why the object stays moving in a circle or falls off. What is the critical factor that determines when it falls?

The weight, I guess. Like, at some point the weight component is too strong, so the centripetal force "can't keep up" (or rather, the centrifugal force, but I'm trying to avoid using that one). But there should be some R influence there too, no?

You worry about R but can you save yourself the trouble: As soon as W can't "deliver" the vertical component of F, the objects come loose from the wall.

So...this?

Putting the Y axis vertical, X axis horizontal.

W = F_y
W = m*a*sin (α)

I tried following up, with the numbers, and the numerical solution I get is quite close to the number the textbook says, but it's a bit off. I kept rounding to a minimum, so the discrepancy (+0.06) in the final result shouldn't be there (0.91 vs 0.85).

Regardless, why does R not matter? The only way I can think R wouldn't matter is if R is exactly horizontal, and therefore it has no vertical component. But if R does have a vertical component, then it should influence F_y just like W does.

Before all that though, is R even the one I drew in the diagram before, or was that wrong?

 

[Bandersnatch]

Before all that though, is R even the one I drew in the diagram before, or was that wrong?

It's wrong in the FB diagram. Your initial hunch was correct - it's always perpendicular to the surface of the rim.

 

[PeroK]

Yea, that's the idea.
The weight, I guess. Like, at some point the weight component is too strong, so the centripetal force "can't keep up" (or rather, the centrifugal force, but I'm trying to avoid using that one). But there should be some R influence there too, no?

You are on the right track. Think about what "can't keep up" really means.

Think also about how the components of the gravitational force and the speed of the object change as it moves up the cylinder.

 

[Bandersnatch]

and the speed of the object change

The speed shouldn't change, though. It's a constant rotation, right?

 

[PeroK]

It's wrong in the FB diagram. Your initial hunch was correct - it's always perpendicular to the surface of the rim.

It doesn't say that the cylinder is frictionless.

However, what is not clear is what "frequency" is supposed to be found.

 

[PeroK]

The speed shouldn't change, though. It's a constant rotation, right?

Gravity has a bad habit of slowing things down as they move upwards!

 

[Alex126]

It doesn't mention anything about friction, which was my very first problem (as I'm about to type).
It is constant rotation, yes.
By "frequency" it means rounds per second [Hz], so as I mentioned earlier I presume it wants me to find the acceleration, and from the acceleration I can then find the velocity, and from there the period/frequency.

So, new diagram with R perpendicular.

When I originally attempted this, I set the equations and they were:

X axis:
R + Wx = m*a

Y axis:
-Wy = ?

As anticipated earlier in this post, I thought there must have been something to add to Wy so that the sum would be = 0, and I speculated it could have been friction. But there is no mention of friction (or lack thereof) in the problem, so even if I assumed there was friction I would still have the issue of having three unknown data but only two equations (R, a, and the additional coefficient of friction).

Gravity has a bad habit of slowing things down as they move upwards!

Now that you mention it, yea...is it not uniform circular motion anymore?

 

[Bandersnatch]

@PeroK I think I'm reading the problem rather different than you do. I see the cylinder as having sufficient friction to neglect sliding (no friction coefficient is given), and the rotation as constant (it's what I think the frequency in the question indicates - there wouldn't be one frequency to find if the angular velocity were changing).

 

[PeroK]

@PeroK I think I'm reading the problem rather different than you do. I see the cylinder as having sufficient friction to neglect sliding (no friction coefficient is given), and the rotation as constant (it's what I think the frequency in the question indicates - there wouldn't be one frequency to find if the angular velocity were changing).

Yes, I think you are right, it's the cylinder that is rotating.

I missed @BvU post. His must have been posted when I was writing mine. It is like a washing machine!

 

[PeroK]

@Alex126 sorry, this is all a bit muddled now. But, the reaction force cannot be perpendicular, because there is a tangential component of gravity. So, there must be friction to keep the ball in place.

That said, the key is to think about the critical factor when the ball falls.

 

[Bandersnatch]

Admittedly, I'm not as on top of the problem as I'd like to be. But let's plod along.

Y axis:
-Wy = ?

I think you can ignore this part. Whatever the Y component, it can be assumed to be counteracted by friction.

So, normally, these questions use the cardinal directions (i.e. 6, 3, 12 'o clock). At 12 'o clock one can write the condition for unsticking from the rotating drum, that is: (edit: cough, maybe I shouldn't give it away like that!)

Here, I think it's just a matter of using the X (i.e. radial, as per your diagram) component of weight instead.

 

[Alex126]

But, the reaction force cannot be perpendicular, because there is a tangential component of gravity.

Ehr...tangential component of gravity?
If R is not perpendicular, then where is it?

That said, the key is to think about the critical factor when the ball falls.

Centrifugal force is not enough anymore to counterbalance the Weight (in the second drawing I made, the W_x). Or, in other words, the W_x cannot provide a big-enough centripetal force.

Here, I think it's just a matter of using the X (i.e. radial, as per your diagram) component of weight instead.

Ok, but how? I attempted it earlier in response to the first reply. I'll quote it down below:

So...this?

Putting the Y axis vertical, X axis horizontal.

W = F_y
W = m*a*sin (α)

I tried following up, with the numbers, and the numerical solution I get is quite close to the number the textbook says, but it's a bit off. I kept rounding to a minimum, so the discrepancy (+0.06) in the final result shouldn't be there (0.91 vs 0.85).

So this process gives me an acceleration that is a bit too high to lead to the correct solution.

r = 0.32, for the record.

By the same process, but using the second diagram, it would be:
W_x = F
mg*sin(α) = ma

Which would be the same as with the other diagram anyway, since R is the difference between the two, but R isn't used anyway.

In any case, also as I mentioned later in that post, I don't understand why R doesn't matter in the calculation. The only way that can be, at least that I can see, is if R is exactly perpendicular to F, and therefore gives no contribution to F itself. However, in the first diagram R has some component along the " F axis ", whereas in the second diagram R is "radial", so entirely on the " F axis ". Is R supposed to be "tangential", so that it doesn't have any impact on F?

So, normally, these questions use the cardinal directions (i.e. 6, 3, 12 'o clock). At 12 'o clock one can write the condition for unsticking from the rotating drum, that is: the weight must be greater than the required centripetal force.

So, for instance, at 6 o'clock, what would we have? I thought we would have +R - W = m*a
Likewise, at 12 o'clock, we would have +R +W = m*a
At 3 and 9 o'clock we would have R = m*a; in this case W is perpendicular to the centripetal force, so it gives no contribution. And if that's true, then in this case (3 and 9 o'clock) W has no impact at all, whereas R is the one that does all the job. So surely R can't be ignored, no? That is, unless the Weight force turns out not to be straight down at all times (is that what "tangential component of gravity" meant?).

 

[PeroK]

@Alex126 as the normal force of gravity increases, the normal force from the cylinder decreases. In other words, gravity provides more and more of the required centripetal force. Unless the cylinder is rotating quickly enough,then the normal force from the cylinder will reduce to zero. The normal force of gravity from then on is more than is needed for circular motion at that radius, so pulls the object off the cylinder.

That's the insight you have been missing.

 

[BvU]

So...this?

Putting the Y axis vertical, X axis horizontal.

W = F_y
W = m*a*sin (α)

Your free body diagram for the vertical component should look as follows: One arrow down, modulus $mg$ plus another arrow down, modulus $F_{N,y}$. That is two forces, no more. The resultant has to provide the vertical component of the centripetal force. Once $F_{N,y}$ goes below zero the object detaches from the cylinder.

But I realize I am now repeating PK post #16.

 

[Alex126]

@Alex126 as the normal force of gravity increases, the normal force from the cylinder decreases. In other words, gravity provides more and more of the required centripetal force. Unless the cylinder is rotating quickly enough,then the normal force from the cylinder will reduce to zero. The normal force of gravity from then on is more than is needed for circular motion at that radius, so pulls the object off the cylinder.

That's the insight you have been missing.

Uhm...no, sorry, I just don't get it. I've always thought that the weight force is the same, so W can't possibly have any other value than m*g. The thing that would change is, with different angles, how much W_x and W_y we have, but W is the same.

So how can gravity "provide more and more" force? The only guess I have is that the "gravity force" here is not synonym with "weight force", i.e. gravity acts through the W force and something else. I have no clue what this "something else" is...unless it's the Normal force.

Then you say that "unless the cylinder is rotating quickly enough, then N will be reduced to zero". I'm interpreting this as the key point of the problem; that is to say, the object will fall at the angle 70° because at that point N will be reduced to zero, and therefore won't provide the centripetal force necessary anymore (which I think is another way of saying what you wrote: if it's not fast enough, i.e. not enough centripetal force, then N is zero; just with backwards cause-consequence relation).

I still don't understand what I'm supposed to do though.

Your free body diagram for the vertical component should look as follows: One arrow down, modulus $mg$ plus another arrow down, modulus $F_{N,y}$. That is two forces, no more. The resultant has to provide the vertical component of the centripetal force. Once $F_{N,y}$ goes below zero the object detaches from the cylinder.

But I realize I am now repeating PK post #16.

Isn't this what I did in post #10 of the topic? With "radial" Normal force, and the "centripetal axis" as +X, it was:
+N +W_x = m*a

Or, with vertical +Y axis instead:
+mg +N_y = F_y

So "once N_y goes below zero the object detaches" should mean:
mg +0 = F_y
=> mg = m*a * sin(α)
=> g = a * sin(α)
=> a = g / sin (α)

Which ties back to all the previous attempts I made, and ends up with frequency of 0.91 instead of 0.85.

Can I ask something else instead? IF the problem had given me the value of μ (friction coefficient), would you still interpret the problem the same way, or would you do the X and Y equations separately? So for instance, with "centripetal axis" as +X:

X axis:
N + W_x = m*a

Y axis (positive up and left):
+Friction - W_y = 0
=> N*μ - W_y = 0

Would you still ignore N, friction, and just put it as mg + 0 = F_y ? (If that's even what you meant :p)

And if I'm misinterpreting what you guys are saying, could you at least tell me if by doing it your way you end up with 0.85 rounds/second ? :D (radius = 0.32m ; angle = 70°)

 

[Dadface]

The speed needed can be found by considering the instant when the object first exhibits any sign of falling and this will be at the instant when the object loses contact with the cylinder. At that instant both R and F become zero and the only force maintaining that (momentary) circular path will be the component of the weight (W sin alpha) towards the centre.

 

[PeroK]

@Alex126 of course the total force of gravity is constant, but its components in the directions normal to and tangential to the particle's motion change as the particle moves.

At 3 o'clock gravity is all tangential and provides no centripetal contribution at all. At 12 o'clock gravity is all centripetal. At this point, unless the particle is moving fast enough it will fall.

The same is true at 70 degrees, except the centripetal component is less.

In this problem you must assume the coefficient of friction is infinite. I.e. Don't worry about the object slipping. Just assume friction counters the tangential component of gravity.

 

[Bandersnatch]

Isn't this what I did in post #10 of the topic? With "radial" Normal force, and the "centripetal axis" as +X, it was:
+N +W_x = m*a

Yes, this was the correct condition. Only reaction force = 0.

could you at least tell me if by doing it your way you end up with 0.85 rounds/second ? :D (radius = 0.32m ; angle = 70°)

I've just ran the numbers, comes out to 0.85.

 

[Alex126]

@Alex126 of course the total force of gravity is constant, but its components in the directions normal to and tangential to the particle's motion change as the particle moves.

Oh, so you mean W_x and W_y. Yea.

At 3 o'clock gravity is all tangential and provides no centripetal contribution at all. At 12 o'clock gravity is all centripetal. At this point, unless the particle is moving fast enough it will fall.

So "tangential" here means "perpendicular to the centripetal axis". Yea.

Yes, this was the correct condition. Only reaction force = 0.I've just ran the numbers, comes out to 0.85.

Ok, thanks (to you and everyone else).

The same is true at 70 degrees, except the centripetal component is less.

In this problem you must assume the coefficient of friction is infinite. I.e. Don't worry about the object slipping. Just assume friction counters the tangential component of gravity.

Alright. I still am not entirely convinced though; it kinda feels like we are making the most out of what we have, but we are missing a necessary piece of information (μ) which would be needed for a proper solution. As I mentioned earlier, IF we had the friction coefficient (maybe in a different problem) would we be proceeding differently?

Below is my previous post (updated a bit), my "final" body diagram thingy, and as-complete-as-I-can-fathom last considerations. If someone could read through and tell me if I made any mistakes down below, I'd appreciate it:

IF the problem had given me the value of μ (friction coefficient), would you still interpret the problem the same way, or would you do the X and Y equations separately? So for instance, with "centripetal axis" as +X:

X axis:
N + W_x = m*a
N + mg*sin(α) = m*a

Y axis (positive up and left):
+Friction - W_y = 0
=> N*μ - W_y = 0
=> N*μ - mg*cos(α) = 0

Full drawing:

From this setup, I would do:

X axis:
N = m*(a - g*sin(α) )

Y axis:
(plug the N from X axis in)
m*(a - g*sin(α) )*μ - mg*cos(α) = 0
=>(a - g*sin(α) )*μ = g*cos(α)
=> $a*μ = g*sin(α) + g*cos(α) / μ$

So the value of $a$ depends on a fixed part ($g*sin(α)$) and a μ-dependent part ($g*cos(α) / μ$) which becomes smaller and less significant the greater the value of $μ$ is.

If all of what I just said is true, by proceeding "backwards" from the theoretical solution of 0.91 rounds/second, I've tried to deduce $μ$:

$v = 2 π r * f$ (f = frequency) = 1.83 m/s
$a = v^2/r$ = 10.46 m/s²

From the X axis equation:
N = m*(a - g*sin(α) )

Plugged into the Y axis equation:
N*μ = m*g*cos(α)
=> m*(a - g*sin(α) )*μ = m*g*cos(α)
=> μ = [g*cos(α)] / [a - g*sin(α)]
=> μ = 2.7

Which, thinking about it, actually sounds close to "assume it's infinite" (because as far as I know, coefficients of friction are usually around or below 1, so a 2.7 is relatively "infinite"). With a number as "big" as 2.7, the variable part $+ g*cos(α) / μ$ is 1.243, which is rather small compared to the fixed part $g*sin(α)$ (= 9.218). Still, there would be a difference if $μ$ was bigger (10, 100, towards +infinity), although I reckon that between "virtually infinite" and the actual value of 2.7 gives an end result that is "just" 0.056 off. Which really bugs me, but oh well.

 

[PeroK]

Alright. I still am not entirely convinced though; it kinda feels like we are making the most out of what we have, but we are missing a necessary piece of information (μ) which would be needed for a proper solution. As I mentioned earlier, IF we had the friction coefficient (maybe in a different problem) would we be proceeding differently?

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 $\mu F_N < mg \cos \theta$. Which must happen before $F_N = 0$.

That's why you have to assume infinite friction. Or, less dramatically, than friction is great enough that the slipping occurs only a short distance before it falls off and makes a negligible difference to the object's velocity: i.e. the object continutes to move approximately with the cylinder until it falls.

Your calculations for $\mu$ are not valid, as $F=0$ at the point at which you appear to be calculating it!

In general, I would say there is a lesson to learn from this:

An FBD is valuable, but you must be prepared to think about the problem before/after you draw one. You made this complicated not because you drew the FBD first, but because you didn't simplify it (without the normal force and friction) once you had sussed out the key that $F_N =0$.

You still don't seem happy to solve the problem using the insight into its solution, but insist in retaining variables that can be dispensed with. This is not all bad, but you should learn from this that if a problem can be simplified, then simplify it.

 

[Dadface]

As I see it contact forces such as friction and reaction can be ignored in this problem. See post 19. Falling begins at the instant contact is broken and so the contact forces become equal to zero at that instant. The only force maintaining momentary circular motion at that instant is the component of W towards the centre. In other words the centripetal acceleration is given by gsin70.

 

[haruspex]

As I see it contact forces such as friction and reaction can be ignored in this problem.

Yes, but only after supposing the frictional coefficient is effectively infinite, as PeroK points out.

 

[haruspex]

mg = m*a * sin(α)

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.

 

[Alex126]

You still don't seem happy to solve the problem using the insight into its solution, but insist in retaining variables that can be dispensed with. This is not all bad, but you should learn from this that if a problem can be simplified, then simplify it.

Yea, but I'm better at solving things when I understand them than when I have to accept an "insight" that falls out of the schemes I'm familiar with. I guess I try too hard to make things fit into a pre-made picture that I have of the problem, and thus find it hard to "adjust" to different scenarios. It's not that I don't "trust" what you guys are saying of course (quite the opposite, I'm trying to learn), it's more that in my mind this problem seems to be missing something and I'm trying to prepare for when I encounter a different problem which has all the variables that (in my mind) are needed, and then apply it to this particular case. Hence my attempt to figure the coefficient "backwards" from 0.91. Which I understand is wrong because of this:

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 μFN<gcosθμFN<gcos⁡θ\mu F_N < g \cos \theta. Which must happen before FN=0FN=0F_N = 0.

So basically the problem with what I said in my previous post is not in the calculation itself, but in the fact that while the calculation is correct "at the end" it has a problem "before" the end (the point when the object falls) because a finite 2.7 coefficient will cause slippage before falling has a chance to even happen.

I'm curious what the coefficient would be (if there is a finite answer) if I take in consideration that fact too though.

Anyway. If I encounter this problem again I'll follow this kind of reasoning then:
1. Do the diagram.
2. If I notice that something seems to be missing (in this case, the tangential component of Weight is not counterbalanced by anything, although Friction could be implied) I will assume that there is an implicit "something" (and hopefully "Friction" can fit the role) that does the job, silently in the background. I then assume that this "something" is "big enough" that it will do its job regardless of its actual value.
3. Given that the tangential part is automatically taken care of, the only part that is still needed will be the radial component.

The only problem I have with this whole thing is understanding the part where you all say "N = 0" (which then leads to: W_radial = m*a). Even after this whole topic, I would still put this equation instead: W_radial + N = m*a. Of course if N = 0 then I get what you are all saying the equation should be, but I don't get how we deduce that N = 0 from the fact that the object falls down. I sort of understand it if I think about an object hanging on the ceiling, in which case if N = 0 (or generally N < mg) then the object would fall down to the ground.

So here, in a similar fashion, once N = 0 then we can find a critical value of a so that the object falls off, because basically before that point N "helps" W_radial in forming m*a, but after that point W_radial is all alone and it's not enough anymore. For instance, with an angle of 60° we should have (radial axis):

+N + W_rad = m*a

With the a found previously (a = 9.128), and W_rad being m*g*sin(α), the N here would be:

N = m*(a-g*sin(α) ) = m*0.632

With α of 50° N = m*1.612, and so on, with N increasing as the angle becomes smaller, i.e. W_rad provides less and less contribution (to the centripetal force) as the angle gets smaller, whereas N provides more and more contribution as the same thing happens. On the opposite end of the spectrum, as the angle increases we have bigger W_rad and smaller N. The biggest value of W_rad is m*g (at 12 o'clock), but that comes after the smallest value of N (in this particular problem, and with the acceleration found before, that is), so the point of detachment happens when W_rad reaches a certain value that alone is not enough to provide centripetal force anymore.

So, basically, as we move from angle 0° to 90° we have these two components, W_rad and N, the sum of which is constant and is always equal to m*a. As the angle increases from 0° to 90°, the contribution from W_rad increases while the one from N decreases, so we have a situation like [contribution from W_rad + contribution from N = m*a] 0+10 = 10, then 1+9 = 10, then 2+8 = 10, then 3+7 = 10, [...], 9+1 = 10, 10+0 = 10. Past this point the W_rad COULD still increase, but the fact that N has reached 0 means the object falls off. If we had a higher "sum result" (higher m*a, so higher a), then the situation would be like this instead: 0+12 = 12, 1+11 = 12, [...], 12+0 = 12. So in this second case what happens is still the same, but it takes a bigger angle before we get to the critical point where N = 0, so the object falls off later (or doesn't fall off at all if we make it past the 12 o'clock mark, which is the "hardest" point).

Therefore, asking if the object can complete the whole lap would be like asking "can we get to an angle of 90° before N reaches zero?".

So the core idea here is not really about the centripetal force, but the fact that the object falls really depends on "when do we get to N = 0". It's basically a race "against the angle", a matter of how far we can go before our "timer" (the ever-decreasing value of N) reaches 0. So that's the critical point that matters, that's the real question: "knowing that the object will fall when N = 0, and this happens at an angle of 70° in our case, what's the value of W_rad [since it's the only other component in the "sum"] at the point when N = 0?".

If this way of interpreting the problem is correct, I think I get it now :D

 

[Dadface]

I tried following up, with the numbers, and the numerical solution I get is quite close to the number the textbook says, but it's a bit off. I kept rounding to a minimum, so the discrepancy (+0.06) in the final result shouldn't be there (0.91 vs 0.85).

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

 

[Dadface]

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.

 

[PeroK]

The only problem I have with this whole thing is understanding the part where you all say "N = 0"

You could change the problem and imagine a horizontal rotation. The cylinder is pushing the object to keep it in the circular path. All the centripetal force is provided by the cylinder.

Now imagine you have a string attached to the object and you start to pull the object towards the centre. As you pull harder, you provide more of the centripetal force and the normal force with the cylinder reduces. If you pull at just the right amount, then the normal force with the cylinder reduces to zero (as you are providing all the centripetal force). At this point, the cylinder could disappear and the object would continue to move in the circle.

If you pull harder stil, then you pull the object off the cylinder.

For the vertical problem, imagine that (the normal/radial component of) gravity is pulling the object away from the cylinder. You have the same scenario. This component of gravity increases until it pulls the object away from the cylinder.

 

[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.