When Do Sand Particles Slip Off a Vibrating Plate?

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Homework Help Overview

The discussion revolves around a problem involving a horizontal metal plate oscillating vertically with simple harmonic motion, where dry grains of sand are placed on the plate. Participants are exploring the conditions under which the grains lose contact with the plate as the amplitude of the oscillation increases.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the necessary conditions for the grains to lose contact, with some focusing on the relationship between normal force and acceleration. Questions arise regarding how to calculate the normal force and the acceleration of the grains at the moment they lose contact.

Discussion Status

The discussion is active, with various interpretations of the problem being explored. Some participants suggest that the problem may be misinterpreted regarding the motion of the grains, while others clarify the focus on vertical oscillation. There is a mix of attempts to derive equations and conceptual understanding without reaching a consensus.

Contextual Notes

Participants note potential confusion regarding the terminology used in the problem statement, particularly the phrase "slip off" versus "lose contact." There is also mention of the gravitational acceleration and its relation to the plate's acceleration at maximum displacement.

heroslayer99
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Homework Statement
What is the max amplitude before the sand particles slip off?
Relevant Equations
a = -w^2 x
A horizontal metal plate connected to a vibration generator is oscillating vertically with simple
harmonic motion of period 0.080 s and amplitude 1.2 mm. There are dry grains of sand on the
plate. The frequency of the vibrating plate is kept constant and its amplitude is slowly increased
from zero. The grains of sand start to lose contact with the plate when the amplitude is
D. State and explain the necessary conditions when the grains of sand first lose contact
with the plate. Hence calculate the value of D.I have calculated that the acceleration of the plate is -(2pi x 1/(0.08))^2 (x) but I do not know where to go from here, any reading material / hints will be very helpful.
 
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What is the acceleration of a grain just as it loses contact?
 
haruspex said:
What is the acceleration of a grain just as it loses contact?
This seems like the typical problem that contains the condition "loses contact" which physicomathematically is being translated to "The normal force from the plate to the grain becomes zero"?
 
Delta2 said:
This seems like the typical problem that contains the condition "loses contact" which physicomathematically is being translated to "The normal force from the plate to the grain becomes zero"?
How am I meant to calculate the normal force
 
haruspex said:
What is the acceleration of a grain just as it loses contact?
I dont know :(
 
heroslayer99 said:
I dont know :(
What will be its acceleration after losing contact then?
 
heroslayer99 said:
How am I meant to calculate the normal force
You don't find some bizarre formula for Normal force, you just do newton's 2nd law for a grain and then set normal force to zero and see how the equation simplifies.
 
heroslayer99 said:
Homework Statement: What is the max amplitude before the sand particles slip off?
Relevant Equations: a = -w^2 x

A horizontal metal plate connected to a vibration generator is oscillating vertically with simple
harmonic motion of period 0.080 s and amplitude 1.2 mm. There are dry grains of sand on the
plate. The frequency of the vibrating plate is kept constant and its amplitude is slowly increased
from zero. The grains of sand start to lose contact with the plate when the amplitude is
D. State and explain the necessary conditions when the grains of sand first lose contact
with the plate. Hence calculate the value of D.I have calculated that the acceleration of the plate is -(2pi x 1/(0.08))^2 (x) but I do not know where to go from here, any reading material / hints will be very helpful.

If the plate is moving vertically, what force would cause the sand particles to move horizontally across the plate to fall off?

(or did you misread the problem?) :wink:

EDIT -- or did you mean "lose contact" with the plate, instead of "slip off"? I can use my Mentor powers to see from your location that there should be no language translation issues... :smile:
 
berkeman said:
If the plate is moving vertically, what force would cause the sand particles to move horizontally across the plate to fall off?
I assure you that they would indeed start to drift off the plate. Bouncing around will lead to some rotational, and hence horizontal, motion.
 
  • #10
heroslayer99 said:
Relevant Equations: a = -w^2 x
What is value of ##x## when the acceleration has its largest magnitude?
 
  • #11
I think the problem wants us to focus on the vertical oscillation of the grains, together with the plate, and not in some sort of Brownian horizontal motion that the grains might do and slip off the plate.
 
  • #12
Delta2 said:
I think the problem wants us to focus on the vertical oscillation of the grains, together with the plate, and not in some sort of Brownian horizontal motion that the grains might do and slip off the plate.
Sure, but the title is not inappropriate.
 
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  • #13
haruspex said:
Sure, but the title is not inappropriate.
I think he misinterpreted the statement of the problem, and in the thread title he put something according to his misinterpretation. Probably the exact wording of the problem doesn't contain the words slip off. And it doesn't seem to , if we take the exact wording of the problem to be what is in the paragraph following the relevant equations statement.
 
  • #14
Apologies for any confusion I have caused, I figured out that the acceleration of the sand particles when they lose contact is g, and hence the plate's acceleration must be greater than g, when it is at its max vertical displacement
 
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