# Block in a rotating cylinder

## Homework Statement

A cylinder with radius R spins around its axis with an angular speed ω. On its inner surface there lies a small block; the coefficient of friction between the block and the inner sur- face of the cylinder is μ. Find the values of ω for which the block does not slip (stays still with respect to the cylinder). Consider the cases where (a) the axis of the cylinder is hori- zontal; (b) the axis is inclined by angle α with respect to the horizon. ## The Attempt at a Solution

I tried to figure out which are the forces applied to the block in the non inertial frame of the cylinder. There is ##m \vec{g} ##, the centrifugal force ##\vec{F_{c}} = m \omega^2 \vec{R}##, the normal reaction ## \vec{N} ## and the frictional force ## \vec{F_{s}}= \vec{N} \mu ##. If the block is at rest the sum ## m \vec{g} + \vec{F_{c}} + \vec{N} + \vec{F_{s}} = 0 ##.
Considering only case (a) of the problem, I don't know how to work with this equation in a three dimension, because my first solution where that ## \omega = \sqrt{ \frac{\mu g}{R}} ## but I think it's correct in a two-dimensional geometry. Any help or reference to see?

What is the centrifugal force? Is it a force on its own?

No, it's the centripetal force in the reference frame of the rotating cylinder.

haruspex
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Considering only case (a) of the problem, I don't know how to work with this equation in a three dimension, because my first solution where that ## \omega = \sqrt{ \frac{\mu g}{R}} ## but I think it's correct in a two-dimensional geometry.
That looks right. (I don't see that using the noninertial frame helps here. You get the same equation immediately.)

The 3D version is going to hurt the head.
Consider when the cylinder has rotated θ from the position where the mass is at its highest.
What are the components of the normal unit vector?

By the way, it is not true that ##\vec F_s=\vec N\mu_s##. You only know that on the point of slipping ##|\vec F_s|=|\vec N|\mu_s##.

Considering only case (a) of the problem, I don't know how to work with this equation in a three dimension, because my first solution where that ω=√μgRω=μgR \omega = \sqrt{ \frac{\mu g}{R}} but I think it's correct in a two-dimensional geometry. Any help or reference to see?
What is your work for that? I got a different solution.

• zwierz and TSny
What are the components of the normal unit vector?

I think that ## \vec{N} ##is always directed along the line that connect the mass to the centre of the cylinder, isn't it?

haruspex
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I think that ## \vec{N} ##is always directed along the line that connect the mass to the centre of the cylinder, isn't it?
Right, but you do not know at what point in the rotation is the greatest risk of slipping. So you need an expression for that vector at a general position. My preference wouldbe to stick with the usual Cartesian coordinates. If you prefer to work with cylindrical here then the challenge, instead, is to represent gravity in those.

my result is different as well • fedecolo and Isaac0427
Right, but you do not know at what point in the rotation is the greatest risk of slipping. So you need an expression for that vector at a general position. My preference wouldbe to stick with the usual Cartesian coordinates. If you prefer to work with cylindrical here then the challenge, instead, is to represent gravity in those.

As zwierz did, I have ##N=m \omega^2 R-mg sin \theta ## and ## N= \frac{mg cos \theta}{\mu}## and then ##\mu \geq \frac{g cos \theta}{\omega ^2 R-g sin \theta}## and so ## \omega^2 R \geq g sin \theta ## because the denominator must be > 1 (otherwise the fraction would be > 1 and the block would slide)

• physics_CD
my result is different as well

The final solution must be ## \omega^2 R \geq g\sqrt{1+ \mu ^-2} ##

The final solution must be ω2R≥g√1+μ−2
sure

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sure

But I don't know how to get it

But I don't know how to get it
that is after I have posted the solution ;(

that is after I have posted the solution ;(

The expressions ##\omega^2 R \geq g(cos \theta \mu^-1+ sin \theta) ## and ##\omega^2 R \geq g \sqrt{\mu^-2 +1}## aren't they?

Find a maximal value of the function ##f(\theta)=|\cos\theta|+\mu\sin\theta##

Find a maximal value of the function ##f(\theta)=|\cos\theta|+\mu\sin\theta##
Oh you are right, thank you!

The case for part (b) is really not all that different, in terms of how you go about the problem. Now, you just need to find the forces in a third direction; the direction along the length of the cylinder (often denoted by p). Step one is breaking up gravity into its r, θ and p components, in terms of m, g, θ and α. From there, it should be obvious how to continue.

The case for part (b) is really not all that different, in terms of how you go about the problem. Now, you just need to find the forces in a third direction; the direction along the length of the cylinder (often denoted by p). Step one is breaking up gravity into its r, θ and p components, in terms of m, g, θ and α. From there, it should be obvious how to continue.

How can I scompose a three-dimension vector? (I never did something like that) Do you know any reference I can study about it?

my result is different as well
View attachment 203996
Just curious, how did you get the |cosθ|+μsinθ? I got μsinθ-cosθ.

Well I have written everything I think about that in #8

Well I have written everything I think about that in #8
Right, but why did you throw in absolute value signs?

EDIT: Never mind, I just realized my mistake.

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I'm trying to decompose the 3D vector ##\vec{g}## but I don't know how to figure out it! Any help?

TSny
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I'm trying to decompose the 3D vector ##\vec{g}## but I don't know how to figure out it! Any help?
If ##\hat{u}## is a unit vector in some direction, then the component of a vector in the direction of ##\hat{u}## can be found from the dot product of the vector with ##\hat{u}##.

Is it possible that the components are ##\vec{g} (g cos \theta, g sin \theta, g cos \alpha)##?

TSny
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Is it possible that the components are ##\vec{g} (g cos \theta, g sin \theta, g cos \alpha)##?