Visualizing the Motion of Coins on a Rotating Scale

[haruspex]

From that fact I can only say 'a' is going to be negative or do you mean a substitution for N?

I get the feeling you do not know how to work with inequalities.
You have a = g-N/m and N≥0. So -N/m≤0. Adding g to each side, g-N/m≤g.

 

[Vibhor]

You have a = g-N/m and N≥0. So -N/m≤0. Adding g to each side, g-N/m≤g.

Alternatively ,

since N≥ 0 , m(g - a) ≥ 0 or a≤g

 

[Rahulrj]

I get the feeling you do not know how to work with inequalities.
You have a = g-N/m and N≥0. So -N/m≤0. Adding g to each side, g-N/m≤g.

I have not had much practice with inequalities so when I come across a problem that requires it, that idea does not strike through although I know how to work out inequalities such as this. Also how would the equation vary if I was to write it for a coin that is placed less than 2/3 L of the rod?
Specifically what in this equation, N-mg=-ma would change?

 

[haruspex]

what in this equation, N-mg=-ma would change?

Nothing in that equation would change. It would still be true that since N≥0 it must that a ≤ g.
The difference would be that the acceleration of the part of the rod it is on would be < g, implying N > 0.

 

[Rahulrj]

Nothing in that equation would change. It would still be true that since N≥0 it must that a ≤ g.
The difference would be that the acceleration of the part of the rod it is on would be < g, implying N > 0.

So a more specific value for acceleration cannot be inferred then? or is it correct to write the normal force on the coin at the part of rod with the rod's value of g at that point since they are going to be sticking onto the rod?

 

[haruspex]

is it correct to write the normal force on the coin at the part of rod with the rod's value of g at that point since they are going to be sticking onto the rod?

I think you mean the rod's value of acceleration at that point.
Yes. If at some point less than 2L/3 from the axis the normal force is zero then the coin must be accelerating faster than the rod, which is not possible. So N>0, which means the coin must be in contact with the rod.

 

[Rahulrj]

I think you mean the rod's value of acceleration at that point.
Yes. If at some point less than 2L/3 from the axis the normal force is zero then the coin must be accelerating faster than the rod, which is not possible. So N>0, which means the coin must be in contact with the rod.

I am not sure if you got what I meant, for example I am interested in finding acceleration that coin experiences at say L/3
so the equation will be, N - mg = -ma and my doubt is whether if it is correct to write a = g/2? as it is the acceleration rod has at that point

 

[haruspex]

correct to write a = g/2?

Yes, that must be right. If it were more, it would penetrate the rod; if it were less it would lose contact with the rod, making N zero, so then it would fall at rate g and immediately catch up to the rod again.

 

[zwierz]

There is a funny counterintuitive fact: if $m$ is a total mass of all coins then $$|\epsilon|\sim const \,m^{1/3},$$ as $m\to\infty$ where $\epsilon$ is angular acceleration of the rod right after it is released. Other parameters are fixed.