Block in a rotating cylinder

  • Thread starter fedecolo
  • Start date
  • #1
61
1

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.

5z3xaf.jpg


Homework Equations




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?
 

Answers and Replies

  • #3
61
1
No, it's the centripetal force in the reference frame of the rotating cylinder.
 
  • #4
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
37,199
7,297
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##.
 
  • #5
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.
 
  • Like
Likes zwierz and TSny
  • #6
61
1
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?
 
  • #7
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
37,199
7,297
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.
 
  • #8
334
61
my result is different as well
63cca082c0f3.png
 
  • Like
Likes fedecolo and Isaac0427
  • #9
61
1
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)
 
  • #10
61
1
my result is different as well

The final solution must be ## \omega^2 R \geq g\sqrt{1+ \mu ^-2} ##
 
  • #11
334
61
The final solution must be ω2R≥g√1+μ−2
sure
 
Last edited:
  • #13
334
61
But I don't know how to get it
that is after I have posted the solution ;(
 
  • #14
61
1
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?
 
  • #15
334
61
Find a maximal value of the function ##f(\theta)=|\cos\theta|+\mu\sin\theta##
 
  • #16
61
1
Find a maximal value of the function ##f(\theta)=|\cos\theta|+\mu\sin\theta##
Oh you are right, thank you!
 
  • #17
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.
 
  • #18
61
1
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?
 
  • #20
334
61
Well I have written everything I think about that in #8
 
  • #21
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.
 
Last edited:
  • #22
61
1
I'm trying to decompose the 3D vector ##\vec{g}## but I don't know how to figure out it! Any help?
 
  • #23
TSny
Homework Helper
Gold Member
13,327
3,603
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}##.
 
  • #24
61
1
Is it possible that the components are ##\vec{g} (g cos \theta, g sin \theta, g cos \alpha)##?
 
  • #25
TSny
Homework Helper
Gold Member
13,327
3,603
Is it possible that the components are ##\vec{g} (g cos \theta, g sin \theta, g cos \alpha)##?
Describe your coordinate system.
 

Related Threads on Block in a rotating cylinder

  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
3
Views
4K
  • Last Post
Replies
14
Views
6K
  • Last Post
Replies
0
Views
2K
  • Last Post
3
Replies
65
Views
12K
  • Last Post
Replies
5
Views
2K
Replies
6
Views
2K
Replies
9
Views
2K
Replies
6
Views
346
Top