A block placed in a horizontal hollow cylinder

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SUMMARY

A block placed inside a horizontal hollow cylinder begins to slide at an angular position of 30° below the horizontal when the cylinder rotates at a constant angular speed of one revolution per second. The coefficient of friction is established at 0.6, leading to the conclusion that the radius of the cylinder must be calculated using the balance of forces just before sliding begins. The discussion clarifies that the block initially rotates with the cylinder without slipping until the critical angle is reached, at which point the transition to sliding occurs due to the balance of forces.

PREREQUISITES
  • Understanding of centrifugal force and its calculation (Centrifugal force = mv² / r)
  • Knowledge of angular speed and its relationship to tangential speed (Angular speed = Tangential speed/Radius)
  • Familiarity with friction concepts, particularly the coefficient of friction (Friction = μN)
  • Basic principles of rotational dynamics and force balance
NEXT STEPS
  • Study the derivation of forces acting on objects in rotating systems
  • Learn about the implications of angular velocity on friction and motion
  • Explore the concept of static versus kinetic friction in rotational contexts
  • Investigate the dynamics of objects in non-inertial reference frames
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Homework Statement


A block is placed inside a horizontal hollow cylinder. The cylinder is rotating with constant angular speed one revolution per second about its axis. The angular position of the block at which it begins to slide is 30° below the horizontal level passing through the center. Find the radius of the cylinder if the coefficient of friction is 0.6, What should be the minimum constant angular speed of the cylinder so that the block reach the highest point of the cylinder?

Homework Equations


Centrifugal force = mv2 / r
Angular speed = Tangential speed/Radius
Friction = μN

The Attempt at a Solution


Picturing a cross section of the hollow cylinder rotating with an angular velocity of 2π radians/sec with forces marked;
zEBZPbt_d.jpg

Equating forces as the block can be assumed to be in rest (constant slipping);
(There would be no centripetal force because the block isn't moving with the cylinder, or isn't in a circular motion)
Mgcos(30) = μ[Mgsin(30)]
=> μ= √3
Which contradicts the statement made by the question that coefficient of friction is 0.6.
Also, this isn't giving any information about the radius as the radius doesn't even come in the equations.

However, according to the book, the solution is
h5KA4Ko.jpg

The solution considers the ω of the block (equation1) to be the ω of the cylinder, when in actuality, the ω of the block is zero (it's at rest), and 5hus there shouldn't be any centrifugal forces acting on it.Please guide as to where I am wrong. I am sure 5to be missing some concept, because the question is correct. Thank you very much.
 
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I think the problem wants you to assume that the block is placed at the bottom of the rotating cylinder such that the block is initially moving with the wall of the cylinder. So, initially the block is not slipping on the wall. The block moves with the wall without any slipping until it reaches the 30 degree position at which point slippage begins.
 
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SciencyBoi said:
in actuality, the ω of the block is zero (it's at rest),
The question says
SciencyBoi said:
at which it begins to slide
That is, it was rotating with the cylinder until the given angle was reached. The transition to sliding is determined by the balance of forces just before it started sliding.
 
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haruspex said:
The question says

That is, it was rotating with the cylinder until the given angle was reached. The transition to sliding is determined by the balance of forces just before it started sliding.
I have now understood it. It was a language problem apparently. Thank you very much.
 
TSny said:
So, initially the block is not slipping on the wall. The block moves with the wall without any slipping until it reaches the 30 degree position at which point slippage begins.
Thank you very much, I seem to have understood it now.
 

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