Explaining Why Sand Loses Contact with Cone During Oscillation

[mystreet123]

Homework Statement

Some sand is sprinkled onto the cone. The sand oscillates vertically with the frequency of the cone. The amplitude of oscillation of the cone is increased.
At a particular amplitude of oscillation the sand begins to lose contact with the cone.
By considering the forces acting on a grain of sand, explain why this happens.

Homework Equations

From mark scheme it says "Weight - contact force = m(w^2)x So as x increases, contact force decreases, sand loses contact with cone when contact force = 0"

But I don't understand why it is m(w^2)x, not -m(w^2)x? as I thought they take upward as negative and losing contact only happens at maximum upward displacement position.

The Attempt at a Solution

 

[drvrm]

From mark scheme it says "Weight - contact force = m(w^2)x So as x increases, contact force decreases, sand loses contact with cone when contact force = 0"

But I don't understand why it is m(w^2)x, not -m(w^2)x? as I thought they take upward as negative and losing contact only happens at maximum upward displacement position.

pl. make out a diagram of oscillations and then mark out the forces acting on sand particle ;perhaps then you can gigure out the contact forces and required centrepetal/centrifugal force .

 

[mystreet123]

pl. make out a diagram of oscillations and then mark out the forces acting on sand particle ;perhaps then you can gigure out the contact forces and required centrepetal/centrifugal force .

Yes I did!

 

[drvrm]

Yes I did!

then show it as attachment so that we can get to your problem

 

[mystreet123]

then show it as attachment so that we can get to your problem

 

[drvrm]

@mystreet123
well the attachment fails to open -perhaps a template will be better so that it can easily open.

 

[mystreet123]

@mystreet123
well the attachment fails to open -perhaps a template will be better so that it can easily open.

Can you open it now?

 

[drvrm]

Can you open it now?

yes, thanks

 

[drvrm]

at some amplitude of oscillations of the cone the sand particles leaves the contact with surface

At a particular amplitude of oscillation the sand begins to lose contact with the cone.

From mark scheme it says "Weight - contact force = m(w^2)x S

suppose N is the contact force so N= mg - mw^2.x at any displacement x so when contact force is zero the wt will be balanced by the term on the right hand ;
as i feel mg is down ward so mw^2 .x shoukd be upward.

or think in terms of F(harmonic osc.) = F(Weight) -N
what is wrong in the above picture of forces?

 

[mystreet123]

at some amplitude of oscillations of the cone the sand particles leaves the contact with surfacesuppose N is the contact force so N= mg - mw^2.x at any displacement x so when contact force is zero the wt will be balanced by the term on the right hand ;
as i feel mg is down ward so mw^2 .x shoukd be upward.
what is wrong in the above picture of forces?

But why I found N-mg = mw^2.x at highest position? if we take upward positive

Because when I use this formula, as x increases, N also increases, instead of becoming zero

 

[drvrm]

mw^2.x at highest position?

in a harmonic oscillation the force which is always a restoring force acts opposite to the displacement

F(osc)= -constantx displacement
if displacement is +ve - force will be negative and just vice-versa
if displacement from mean position is negative then force will be in opposite direction.
will this clarify your picture?

 

[mystreet123]

in a harmonic oscillation the force which is always a restoring force acts opposite to the displacement

F(osc)= -constantx displacement
if displacement is +ve - force will be negative and just vice-versa
if displacement from mean position is negative then force will be in opposite direction.
will this clarify your picture?

Yes understood thanks!

 

[drvrm]

Yes understood thanks!

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