Visualizing the Motion of Coins on a Rotating Scale

[Rahulrj]

Homework Statement

A uniform rigid scale is held horizontally with one of its end at the edge of the table and the other supported by hand. Some coins of negligible mass are kept on the meter scale.

As the hand supporting the scale is removed, the scale starts rotating about its edge on the table and the coins start moving. Which of the following would look closest to the situation if a photograph of the scale is taken soon after

2. Homework Equations
Torque = $r \times F$
$L = r \times p$
Torque = $\frac {dL} {dt}$

The Attempt at a Solution

This problem looked trivial until I tried to reason why the different options given would not be correct. This is how I thought about. Initially coins are at rest supported by reaction force provided by the hand. Now when the hand is removed, there is going to be a moment about the hinged part as a result of change in angular momentum. The torque is highest at the ends due to the length of the arm and lowest at points closer to the hinge. Also I think it is fair to assume that friction is greater than the force pulling it to the right on the coins closer to the hinge as it is about the situation immediately after the hand is removed and therefore I can conclude the coins are going to stick to the rod. However I am not able to come to conclusion regarding the coins at the right end especially trying to reason why option C and B could or could not be possible. Also I could not get my reasoning correct as to why the coins should separate from the right end as shown,

 

[zwierz]

if the coins are of negligible mass then they remain to hang in the space along straight horizontal line after the rod has fallen down

 

[Rahulrj]

if the coins are of negligible mass then they remain to hang in the air along straight horizontal line after the rod has fallen down

Why should it remain in the air? Also there are three options that show coins in the air so which one among them?

 

[zwierz]

The correct formulation of this problem should be as follows. A uniform rod of mass $M$ and length $l$ is held horizontally by its right end while the left end can rotate in a perfect hinge. There is a chain of small identical coins on the rod. The mass of the whole chain is $m$. Find acceleration of the points of the rod right after its right end has been released. The rod is smooth.

 

[Rahulrj]

The correct formulation of this problem should be as follows. A uniform rod of mass $M$ and length $l$ is held horizontally by its right end while the left end can rotate in a perfect hinge. There is a chain of small identical coins on the rod. The mass of the whole chain is $m$. Find acceleration of the points of the rod right after its right end has been released. The rod is smooth.

I do not understand how finding acceleration is relevant to the problem. Nevertheless, for the rod I can write
$Mv^2/l =Mgsin\theta$
Not quite sure how to model the equation for chain.

 

[Vibhor]

I do not understand how finding acceleration is relevant to the problem

I think , all such coins lose contact with the stick where the underlying contact surface on the stick move with an acceleration greater than 'g' ( acc. due to gravity ) after the hand is removed .

An important/interesting exercise for you would be to

1) Calculate the acceleration of the tip of the stick of mass M and length L just after hand is removed ? Note that coins are very light , so you can neglect their effect .

2) Now , find a point on the stick whose acceleration is 'g' ?

After you have solved the above two you would get length of part of the stick whose linear acceleration is greater than 'g' and part whose acceleration is less than 'g' .

This should enable you to spot the best option .

 

[zwierz]

Let $\rho=m/l$ be the density of the chain of coins; see post #4. Let $dm=\rho dx$ be infinitesimal element of this chain. By $N(x)dm$ denote the reaction force that acts on $dm$ from the rod.
By $J=Ml^2/3$ denote moment of inertia of the rod about $O$; $\epsilon$ is angular acceleration of the rod.
So we have
$$J\epsilon=-Mgl/2-\int_0^XxN(x)\rho dx,\quad a(x)dm=N(x)dm-gdm;\qquad (*)$$
here $a(x)$ is the vertical part of acceleration of the rod's point with coordinate $x,\quad a(x)=\epsilon x$; $X$ is $x-$coordinate of the point at which chain leaves the rod. The second equation of (*) holds for $0\le x\le X$.
From the second equation of (*) we have $N(x)=\epsilon x+g$ and thus $X=-g/\epsilon.$
Substituting these formulas to the first equation of (*) we have the following cubic equation on $\epsilon$
$$J\epsilon^3+\frac{1}{2}Mgl\epsilon^2+\frac{1}{6}\rho g^3=0.$$ It is easy to see that this equation has a unique real root $\epsilon_*$ and $\epsilon_*<0$.

 

[Rahulrj]

I think , all such coins lose contact with the stick where the underlying contact surface on the stick move with an acceleration greater than 'g' ( acc. due to gravity ) after the hand is removed .

An important/interesting exercise for you would be to

1) Calculate the acceleration of the tip of the stick of mass M and length L just after hand is removed ? Note that coins are very light , so you can neglect their effect .

Okay I can calculate the tangential acceleration as follows,
$a_t = r\alpha$ and torque $\tau = I\alpha$
$\alpha = \tau/I$
$\tau = r \times F = LMg/2$ (uniform rod, weight approximated to act at centre half way along)
$I = ML^2/3$
$\alpha = \frac{3g}{2L}$
$a_t = L \frac{3g}{2L} = 3/2g$

2) Now , find a point on the stick whose acceleration is 'g' ?

From the above calculation, I can say $a_t = g$ is at $\frac{2L}{3}$

After you have solved the above two you would get length of part of the stick whose linear acceleration is greater than 'g' and part whose acceleration is less than 'g'

So two third of the length of rod from the hinge has linear acceleration less than g and one third has it greater than g.
Going by that, option B should be the correct one.

However I have got a doubt, assuming that coins have negligible mass, wouldn't all the coins experience an acceleration the same as all the points on the rod? So why would some be left hanging in the air as opposed to going along with the rod?
.

 

[haruspex]

option B should be the correct one

Right.

assuming that coins have negligible mass

Does that change their acceleration in free fall?

why would some be left hanging in the air

Consider the forces on one of those coins on the right. Is it possible that these forces will cause the coins to accelerate as fast as the rod?

 

[Vibhor]

However I have got a doubt, assuming that coins have negligible mass, wouldn't all the coins experience an acceleration the same as all the points on the rod?

Why ?

So why would some be left hanging in the air as opposed to going along with the rod?
.

They are not hanging in air . Rather all those coins not in contact with stick I.e those in a horizontal line , are in free fall . They are accelerating downwards with acceleration 'g'

 

[zwierz]

The coin experiences gravity force mg. If we neglect m that is m=0 then for such a coin there is no gravity, it will hang in air. The statement of the problem is incorrect.

 

[Vibhor]

So two third of the length of rod from the hinge has linear acceleration less than g and one third has it greater than g.
Going by that, option B should be the correct one.

I agree .

 

[Vibhor]

The coin experiences gravity force mg. If we neglect m that is m=0 then for such a coin there is no gravity, it will hang in air

Coins are of negligible mass , not massless .

 

[Rahulrj]

Right.

Does that change their acceleration in free fall?

Consider the forces on one of those coins on the right. Is it possible that these forces will cause the coins to accelerate as fast as the rod?

So considering a coin at the right end,
$N-mgcos\theta = -ma$
$N = m (g cos\theta-a)$
therefore $g'=gcos\theta-a$ not sure if this is right but it tells me the coins experience a lesser acceleration and thereby I can conclude they experience a lesser force?

 

[haruspex]

So considering a coin at the right end,
$N-mgcos\theta = -ma$

As in you analysis in post #8, you only need to consider the initial acceleration, so θ=0.

g′=gcosθ−a

What is g'? You already have a for the acceleration the coins experience.

 

[zwierz]

Coins are of negligible mass , not massless .

Those are just words. Write exact equations then neglect some terms in the equations then obtain approximate equations then it will be matter for discussion

 

[Rahulrj]

As in you analysis in post #8, you only need to consider the initial acceleration, so θ=0.

Okay yes, so the equation will be N=m(g-a)

What is g'? You already have a for the acceleration the coins experience.

g' is from the normal force on the coin, (mg').

 

[haruspex]

That are just words. Write exact equations then neglect some terms in the equations then obtain approximate equations then it will be matter for discussion

There's nothing to discuss. It would have been clearer if the question had said the mass of the coins is negligible compared with that of the rod, but Vibhor's reading is obviously correct.

 

[haruspex]

the equation will be N=m(g-a)

Yes, but you are interested in the value of a, so what's a better way to write that?
What limits can you place on N?

 

[Rahulrj]

Yes, but you are interested in the value of a, so what's a better way to write that?
What limits can you place on N?

for acceleration to be g, N should be zero but that does not convince me. Putting a limit on N, we make it a case for free fall rather than it being a case of free fall as a result of the formulations.

 

[haruspex]

rather than it being a case of free fall as a result of the formulations

Not sure what you mean. Do you mean the equations? If so, that's why I asked you what limits can be placed on N.

 

[Rahulrj]

Not sure what you mean. Do you mean the equations? If so, that's why I asked you what limits can be placed on N.

I am also not quite sure what you meant by the limits on N. To me N= m(g-a) tells that coins are going to experience a lesser force and I don't know how to infer acceleration unless I take it to be a case of free fall and put N=0.

 

[haruspex]

not quite sure what you meant by the limits on N

Can it be negative?

 

[Rahulrj]

Can it be negative?

unless 'a' is greater than g, it cannot be negative.

 

[haruspex]

unless 'a' is greater than g, it cannot be negative.

You do not need that to answer the question. Can the normal force from the scale on the coins be negative? Can the normal force from the floor on your feet be negative?

 

[Rahulrj]

You do not need that to answer the question. Can the normal force from the scale on the coins be negative? Can the normal force from the floor on your feet be negative?

Normal force acts upwards in those cases and therefore it cannot be negative.

 

[haruspex]

Normal force acts upwards in those cases and therefore it cannot be negative.

Right (unless glue or nails are involved). So rearrange your equation in the form a = ... and use the fact that N≥0 to find a limit on a.

 

[Rahulrj]

Right (unless glue or nails are involved). So rearrange your equation in the form a = ... and use the fact that N≥0 to find a limit on a.

Okay now I get what you meant!
a = mg-N/m
a = g-N/m m is negligible
therefore a = g.
but if m where not negligible how would I infer 'a' in that case?

 

[haruspex]

a = g-N/m

Right.

m is negligible

No, that is not going to help. m is small, but we don't know what it is in relation to N.
Use the fact that N≥0, as I told you.

 

[Rahulrj]

Right.

No, that is not going to help. m is small, but we don't know what it is in relation to N.
Use the fact that N≥0, as I told you.

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

 

[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?