Max Drum Rotate Speed of a Concrete Mixer Drum

[PFuser1232]

Homework Statement

In a concrete mixer, cement, gravel, and water are mixed by tumbling action in a slowly rotating drum. If the drum spins too fast the ingredients stick to the drum wall instead of mixing. Assume that the drum of a mixer has radius $R = 0.5 m$ and that it is mounted with its axle horizontal. What is the fastest the drum can rotate without the ingredients sticking to the wall all the time? Assume $g = 9.8 m/s^2$.

Homework Equations

$F_r = ma_r$

The Attempt at a Solution

If a particle of mass $m$ goes around in a vertical circle on the drum wall at angular speed $\omega$, then in polar coordinates we have:
$$-(N + mg \sin{\theta}) = -mR \omega^2$$
$$N = m(R \omega^2 - g \sin{\theta})$$
Where $\theta$ is the angle the position vector of the particle makes with the horizontal and $N$ is the magnitude of the normal reaction from the drum wall.
At the top, if we wish to find the maximum value of $\omega$ beyond which $N$ is nonzero (sticking to the wall, as described in the question), we can set $N = 0$ and solve for $\omega_{max}$:
$$\omega_{max} = \sqrt{\frac{g}{R}}$$
Is this correct?
Also, why do we choose the $N = 0$ for $\theta = \frac{\pi}{2}$? Is it because if $N = 0$ for any other point, the ingredients wouldn't go around in a circle, but would instead be in free fall?

 

[Simon Bridge]

Is this correct

Taking "N" to be the magnitude of the normal force to the sides of the drum...
If N = 0 there is certainly no sticking ... the amount of sticking would be modeled as a proportion of N rather than all of N though.

Why do we choose the $N=0$ for $\theta = \frac{\pi}{2}$ ?

You did not tell us where $\theta = 0$ is so it is difficult to answer: we don't know what position that angle corresponds to.
However, "we" did not make that choice... you did.
Why did you make that choice?

The way to understand why someone may make that choice is to look at other choices - look at them from the POV of the physics, how cement etc behaves in a drum, rather than looking at the equation.

 

[Jaya]

I don't understand one thing here , why are we considering N=0 as the critical condition? I mean the ingredients will still fall down because w points in negative direction.

 

[Simon Bridge]

I don't understand one thing here , why are we considering N=0 as the critical condition? I mean the ingredients will still fall down because w points in negative direction.

Your reasoning seems to be that a downwards net force results in falling off the loop ... and you propose that a non-zero normal force is needed to hold it up... is that correct?

Consider - when the ingredients are at the top of the drum, which direction does the normal force point?

However - if you check the question in post #1, the point is to prevent the ingredients from sticking ot the drum all the time ... this means you don't want the ingredients pressing into the wall at some stage. Now: where does the normal force come from?

 

[Jaya]

Thank you for your reply.
But I want to confirm that..
We have to find the maximum velocity... And the ingredient will fall until N <W according to me... So I was just pondering over this that why we chose N=0..
Or may be you want say that that eventhough the net force points downward but we should not have any component pressing the wall?

 

[haruspex]

why do we choose the N=0 for $\theta = \frac{\pi}{2}$? Is it because if N=0 for any other point, the ingredients wouldn't go around in a circle, but would instead be in free fall?

It's the boundary case. The normal force is least at the top. If the rotation rate is slow enough that the particles fall somewhere then the gravitational force will be greater than the centripetal force there; and vice versa.

 

[haruspex]

the ingredient will fall until N <W

At the top, the normal force and the weight both act downwards. Whether the particles fall must depend on the sum, not the difference.

 

[Simon Bridge]

It helps if you answer the questions that are put to you... they are there to help youo figure out the answer you seek for yourself, that way you learn more.

Thank you for your reply.
But I want to confirm that..
We have to find the maximum velocity... And the ingredient will fall until N <W according to me...

OK - so when I asked: "you propose that a non-zero normal force is needed to hold it up... is that correct?" your answer is, "yes".
The follow up question was: "Consider - when the ingredients are at the top of the drum, which direction does the normal force point?"
... ie, the idea was to get you to think about the direction of the vectors at different positions.
Your assumption that the normal and weight forces always point in opposite directions was false.
You can see this easily by drawing a free body diagram for a small mass at the top of the loop.

So I was just pondering over this that why we chose N=0..
Or may be you want say that that even though the net force points downward but we should not have any component pressing the wall?

You seem to be saying that you understand that the normal force and weight point in the same direction at the top of the loop.

The ingredients sticking to the wall adds a bit of a complication in that there is an additional adhesion effect ... so there would be a force pointing towards the wall due to the general stickiness of the wet concrete. That has not been included in the model... besides, it is not the "normal force", which just models the fact there is a wall in the way (or Newton's 1st law applies).

The situation in the question is more that we want the fastest the drum can rotate so that the material will still fall away from the surface. That means we don't want it pressing into the surface at all, since that just makes it stick more ... preferably we want it to pull away from the surface. How might it pull away from the surface?

It looks like the model answer is setting the N=0 condition at about a quarter turn rather than at the top ... if you watch how concrete mixers work you'll see this produces a rolling/mixing motion. It also allows for the stickiness of the concrete.[/QUOTE]

 

[haruspex]

It looks like the model answer is setting the N=0 condition at about a quarter turn rather than at the top

That's not how I read it:

θ is the angle the position vector of the particle makes with the horizontal

N=0 for $\theta = \frac{\pi}{2}$

 

[Jaya]

Ok..। understood. Thanks a lot.