Are there two downward forces acting on the pivot of the rotating rod?

[ymnoklan]

Homework Statement

A carbon fiber rod of length L = 1.0 m and of negligible mass can rotate without
friction about an axis through its midpoint. Two small blocks of lead with masses m1 = 2.0 kg
and m2 = 3.0 kg are attached to each end of the rod. You hold the bar in a horizontal position and
release the system.

What is the speed of the blocks when the rod reaches a vertical position? You can disregard
the air resistance.
How great are the forces in the rod when it is vertical and in which directions do they act?
How big is the force on the axis, and in which direction does it act when the rod is vertical?

Relevant Equations

Ep=mgh, Ek=1/2mv^2, a=mv^2/r

I get that the speed is 3.1 m/s, the forces in the rod are 38 N downward and 58 N upward, and that the force on the axis is 49 N.

 

[kuruman]

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[ymnoklan]

I think the main problem is the last question: what do they even mean with “the force on the axis”? What is the difference between this and the forces on the rod?

 

[kuruman]

I think the main problem is the last question: what do they even mean with “the force on the axis”? What is the difference between this and the forces on the rod?

What is the difference between the force that the axis exerts on the rod and the force that the rod exerts on the axis?

 

[haruspex]

the forces in the rod are 38 N downward and 58 N upward,

Please post your working for that.

 

[ymnoklan]

Ok, so my notes are in Norwegian, but you’ll probably get the essence of it based on the equations. For the first question I basically just used conservation of mechanical energy and solved for the speed v. The second question, I used the speed from the first question and found the centripetal force on each end of the rod. Finally for the last question I just find that the only forces acting are the gravitational forces of the two blocks, but this feels kind of wrong(?) Please tell me what you think? How would you solve this kind of problem?

 

[kuruman]

Sorry, your notes are illegible, not to mention sideways. Also, "getting the essence" is not good enough because the Devil is in the details. Your strategy sounds fine for the first two parts but what about its implementation? I can't tell unless I can read your work . . . in English.

I just find that the only forces acting . . .

Acting on what system by what entity outside the system? That's an issue that it seems you have not settled in your mind. What do you think?

 

[TSny]

Welcome to Physics Forums!
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The left side of the equation is not correct for the magnitude of the change in potential energy. See if you can spot the error.

$\dfrac{m_1v^2}{r}$ gives the net force acting on $m_1$ when the rod is vertical. You want the force that the mass exerts on the rod. How many forces act on each mass? Draw a free-body diagram for each mass when the rod is vertical.

 

[haruspex]

found the centripetal force on each end of the rod.

The centripetal force on an object is a component of the net force on it. Namely, it is the component that acts orthogonally to its velocity.
Since the rod is massless, there is no net force on it.
There is a net force on each block.

 

[ymnoklan]

Sorry, your notes are illegible, not to mention sideways. Also, "getting the essence" is not good enough because the Devil is in the details. Your strategy sounds fine for the first two parts but what about its implementation? I can't tell unless I can read your work . . . in English.

Acting on what system by what entity outside the system? That's an issue that it seems you have not settled in your mind. What do you think?

Do you mean that the masses themselves are acting on the axis?

 

[ymnoklan]

Welcome to Physics Forums!
Please read the guidelines for posting homework questions. Note item 5 regarding posting images of your work.

View attachment 351606
The left side of the equation is not correct for the magnitude of the change in potential energy. See if you can spot the error.

View attachment 351607

$\dfrac{m_1v^2}{r}$ gives the net force acting on $m_1$ when the rod is vertical. You want the force that the mass exerts on the rod. How many forces act on each mass? Draw a free-body diagram for each mass when the rod is vertical.

I can’t see the error:( What is wrong? Based on the figure the change in height is half the length of the rod and the mass and g is obviously unchanged.
Are you saying that the gravitational force is acting additionally to the “centripetal force”?

 

[Steve4Physics]

I can’t see the error:( What is wrong? Based on the figure the change in height is half the length of the rod and the mass and g is obviously unchanged.
Are you saying that the gravitational force is acting additionally to the “centripetal force”?

If a mass rises, is it's change in gravitational potential energy positive or negative?
And if a mass falls...?

 

[ymnoklan]

If a mass rises, is it's change in gravitational potential energy positive or negative?
And if a mass falls...?

Oh, good point! So is the total change in potential energy (m_1+m_2)*g*L then?

 

[Steve4Physics]

Oh, good point! So is the total change in potential energy (m_1+m_2)*g*L then?

A 1kg mass rises 1m and another 1kg mass falls 1m. According to your formula, what is the total change in gravitational potential energy? Does the answer make sense?

 

[ymnoklan]

A 1kg mass rises 1m and another 1kg mass falls 1m. According to your formula, what is the total change in gravitational potential energy? Does the answer make sense?

Is it 0 then? But where does the kinetic energy come from then?

 

[Steve4Physics]

Is it 0 then?

Yes. In my example, one mass gains some gravitational potential energy (GPE). The other (equal) mass loses the same amount of GPE. The overall change is zero. So your formula is wrong!

But where does the kinetic energy come from then?

In my example, the masses were equal - but that was only to show that your formula was incorrect. In your Post #1 problem, the masses are unequal.

You need the correct formula! Can you work it out for yourself? Hint: the height-changes of $m_1$ and $m_2$ are not equal.

 

[ymnoklan]

Yes. In my example, one mass gains some gravitational potential energy (GPE). The other (equal) mass loses the same amount of GPE. The overall change is zero. So your formula is wrong!


In my example, the masses were equal - but that was only to show that your formula was incorrect. In your Post #1 problem, the masses are unequal.

You need the correct formula! Can you work it out for yourself? Hint: the height-changes of $m_1$ and $m_2$ are not equal.

Thank you! I agree. So is the midpoint of the axis not L/2 then? I have no idea how to find this though. Could you help me on the way?

 

[Steve4Physics]

Thank you! I agree. So is the midpoint of the axis not L/2 then?

The distance of each mass from the midpoint is L/2. But we are interested in the vertical displacements (which can be positive or negative).

Could you help me on the way?

a) A 2kg mass rises 0.5m. What is its change in GPE?
b) A 3kg mass falls 0.5m. What is its change in GPE?

 

[ymnoklan]

The distance of each mass from the midpoint is L/2. But we are interested in the vertical displacements (which can be positive or negative).


a) A 2kg mass rises 0.5m. What is its change in GPE?
b) A 3kg mass falls 0.5m. What is its change in GPE?

a) 2*0.5*9.81 = 9.81 J
b) -3*0.5*9.81 = 14.715 J ≈ -14.7 J
Ohh...

change in E_p = 14.715 - 9.81 = 4.905 J
E_k = 4.905 J = 1/2 * (m_1 + m_2)*v^2 => v ≈ 1.4 m/s
Is this correct?

 

[Steve4Physics]

a) 2*0.5*9.81 = 9.81 J
b) -3*0.5*9.81 = 14.715 J ≈ -14.7 J
Ohh...

change in E_p = 14.715 - 9.81 = 4.905 J
E_k = 4.905 J = 1/2 * (m_1 + m_2)*v^2 => v ≈ 1.4 m/s
Is this correct?

Yes - well done!

 

[Steve4Physics]

b) -3*0.5*9.81 = 14.715 J ≈ -14.7 J

The 3kg mass is displaced downwards so I would have written 3*(-0.5)*9.81

Can you express the total change in GPE algebraically, using the symbols $m_1, m_2, g$ and $L$?

 

[ymnoklan]

Yes - well done!

Thank you so much! What do you think of my other calculations then? Updating with the correct speed I now get that the forces in the rod are 7.81 N upwards and 11.8 N downward. For the last question I am still stuck though. Is it just the gravitational forces of the masses: (m_1*m_2)*g ≈ 49.1 N?

 

[ymnoklan]

The 3kg mass is displaced downwards so I would have written 3*(-0.5)*9.81

Can you express the total change in GPE algebraically, using the symbols $m_1, m_2, g$ and $L$?

delta_E_p = (m_1-m_2)*g*L/2 ?

 

[Steve4Physics]

Updating with the correct speed I now get that the forces in the rod are 7.81 N upwards and 11.8 N downward.

You have calculated the centripetal force ($\frac {mv^2}r$) on each mass. But these are not forces acting on (or in) the rod.

In Post #8, @TSny suggested drawing free body diagrams for each mass. That's a very good suggestion!

 

[ymnoklan]

You have calculated the centripetal force ($\frac {mv^2}r$) on each mass. But these are not forces acting on (or in) the rod.

In Post #8, @TSny suggested drawing free body diagrams for each mass. That's a very good suggestion!

I am terrible at free-body diagrams though. When I do that I only find gravity acting downward on both of the masses and that is obviously wrong.

 

[Steve4Physics]

delta_E_p = (m_1-m_2)*g*L/2 ?

Yes. I would write this:

$\Delta E_p = m_1 g \Delta h_1 + m_2 g \Delta h_2$

Since $\Delta h_1 = \frac L2$ and $\Delta h_2 = -\frac L2$

$\Delta E_p = m_1 g \frac L2 + m_2 g (-\frac L2)= (m_1 - m_2)g \frac L2$

 

[Steve4Physics]

I am terrible at free-body diagrams though.

The key is to identify all the forces that act on the body.

When I do that I only find gravity acting downward on both of the masses and that is obviously wrong.

Consider the top mass, $m_1$, for example.

Remember it is attached to the rod – so the rod exerts a force on the mass. The force on the mass could be outwards or inwards. So $m_1$ has two forces acting on it:
- its weight;
- the force exerted by the rod at the point of attachment.

It’s important to remember that $m_1$’s centripetal force (which you know) is the total* force on $m_1$, so it is the (vector) sum of the above two forces.

(*Note. We're assuming no tangential force, which is true for the top and bottom positions in this question.)

Using the above information, can you attempt free body diagrams (FBDs) for each mass?

I have some jobs to do now, so won't be able to reply for a few hours. Someone else may want to though.

 

[ymnoklan]

The key is to identify all the forces that act on the body.


Consider the top mass, $m_1$, for example.

Remember it is attached to the rod – so the rod exerts a force on the mass. The force on the mass could be outwards or inwards. So $m_1$ has two forces acting on it:
- its weight;
- the force exerted by the rod at the point of attachment.

It’s important to remember that $m_1$’s centripetal force (which you know) is the total* force on $m_1$, so it is the (vector) sum of the above two forces.

(*Note. We're assuming no tangential force, which is true for the top and bottom positions in this question.)

Using the above information, can you attempt free body diagrams (FBDs) for each mass?

I have some jobs to do now, so won't be able to reply for a few hours. Someone else may want to though.

Thank you so incredibly much for all of your help! I truly appreciate it. Using your hints I get that:
sigma_F_1 = T - G = m_1*v^2/r => T_1 ≈ 27.5 N down, G_1 ≈ 29.4 N down
sigma_F_2 = T + G = m_2*v^2/r => T_2 ≈ 41.2 N up, G_2 ≈ 29.4 down,
where T is the force(s) from the rod to the mass(es) and G is the gravitational force(s) from the mass(es).
Again I still struggle with the last question though. What is meant by the force on the axis?

 

[kuruman]

I am terrible at free-body diagrams though.

Being terrible at FBDs does not mean that you cannot do something to stop from "being terrible". Here is a tep-by-step recipe that I have used in my classes. It will see you through as long as you perform each step in order without omitting any step or doing things in your head. I have attached figures next to each step t show how the FBD develops. The method works best if you use symbols instead of numbers.

1. Identify the system and draw it.
The system is the mass $m_1=3~$ kg at the bottom of the trajectory.

2. Identify all the pieces of the Universe that interact with the system and count them.
The pieces that interact with $m_1$ are the Earth that attracts it and the rod that is attached to it. This makes a total of two pieces.

3. Draw one and only one arrow representing the force exerted by each of the items in (2). Label each arrow unambiguously.
See figure on right. The Earth exerts force $m_1g$ that is always down. The force exerted from the rod must be in a direction along the rod, i.e. vertical. It cannot be up because if it were, the mass will accelerate vertically down and not go around in a circle.

4. Choose a convenient coordinate system and draw it. Enclose your system, axes and forces in a

box with dotted lines.
Positive direction is up as indicated by the arrow labeled $y$.

5. Declare the acceleration. Draw an arrow outside the box indicating the direction of the acceleration. If you have chosen the coordinate axes wisely, the acceleration will be along one of the principal axes (x or y). Label the arrow "m"a, where "m" is the symbol you used for the mass in (1).
Mass $m_1$ is going around in a circle. Since all the force vectors in the diagram are along the vertical, the acceleration can only be along the vertical, up or down. Now the mass is going around in a circle centered directly above $m_1$. The acceleration is centripetal (towards the center), hence its direction is up.

6. Add all the forces vectorially to obtain the net force, the left side in Newton's second law. Check that the vector sum of the arrows inside the dotted line gives a resultant vector in the direction of the "mass times acceleration" vector that you drew outside the dotted line in (5).
Anything up is positive and anything down is negative, so $$F_{\text{net}}=F_{\text{rod}}-m_1g.$$7. Set the net force vector equal to the mass times acceleration vector outside the FBD. Solve for what you are asked to find.
$$F_{\text{net}}=F_{\text{rod}}-m_1g=m_1a$$ Since the acceleration is centripetal, $$F_{\text{rod}}-m_1g=m_1v^2/(L/2).$$ You have the speed $v$ from a previous part, you need to solve the force exerted by the rod on the mass. The force exerted by the mass on the rod has the same magnitude and opposite direction to that.

If you are serious about improving your free body diagram skills, apply this step-by-step method to find the force exerted by the other mass on the rod and consider using it in the future.

 

[haruspex]

Thank you so incredibly much for all of your help! I truly appreciate it. Using your hints I get that:
sigma_F_1 = T - G = m_1*v^2/r => T_1 ≈ 27.5 N down, G_1 ≈ 29.4 N down
sigma_F_2 = T + G = m_2*v^2/r => T_2 ≈ 41.2 N up, G_2 ≈ 29.4 down,
where T is the force(s) from the rod to the mass(es) and G is the gravitational force(s) from the mass(es).

$G_1$ and $G_2$ are different. Maybe that was just a typo.

As @Steve4Physics mentioned, it is much better to work algebraically. Don't plug in numbers until you have to. So far you have $v^2=\frac{(m_2-m_1)gL}{(m_2+m_1)}$.

You need to define which way is positive for your forces. It doesn’t matter what you choose as long as, for each one, you are consistent.
Say we take towards the axis as positive always. Then we have $T_1+G_1=m_1\frac{v^2}{L/2}$, $T_2-G_2=m_2\frac{v^2}{L/2}$.
What does that give you for the tensions?

 

[ymnoklan]

Being terrible at FBDs does not mean that you cannot do something to stop from "being terrible". Here is a tep-by-step recipe that I have used in my classes. It will see you through as long as you perform each step in order without omitting any step or doing things in your head. I have attached figures next to each step t show how the FBD develops. The method works best if you use symbols instead of numbers.

View attachment 3516191. Identify the system and draw it.
The system is the mass $m_1=3~$ kg at the bottom of the trajectory.

2. Identify all the pieces of the Universe that interact with the system and count them.
The pieces that interact with $m_1$ are the Earth that attracts it and the rod that is attached to it. This makes a total of two pieces.

View attachment 3516213. Draw one and only one arrow representing the force exerted by each of the items in (2). Label each arrow unambiguously.
See figure on right. The Earth exerts force $m_1g$ that is always down. The force exerted from the rod must be in a direction along the rod, i.e. vertical. It cannot be up because if it were, the mass will accelerate vertically down and not go around in a circle.

4. Choose a convenient coordinate system and draw it. Enclose your system, axes and forces in a View attachment 351623box with dotted lines.
Positive direction is up as indicated by the arrow labeled $y$.






View attachment 3516245. Declare the acceleration. Draw an arrow outside the box indicating the direction of the acceleration. If you have chosen the coordinate axes wisely, the acceleration will be along one of the principal axes (x or y). Label the arrow "m"a, where "m" is the symbol you used for the mass in (1).
Mass $m_1$ is going around in a circle. Since all the force vectors in the diagram are along the vertical, the acceleration can only be along the vertical, up or down. Now the mass is going around in a circle centered directly above $m_1$. The acceleration is centripetal (towards the center), hence its direction is up.

6. Add all the forces vectorially to obtain the net force, the left side in Newton's second law. Check that the vector sum of the arrows inside the dotted line gives a resultant vector in the direction of the "mass times acceleration" vector that you drew outside the dotted line in (5).
Anything up is positive and anything down is negative, so $$F_{\text{net}}=F_{\text{rod}}-m_1g.$$7. Set the net force vector equal to the mass times acceleration vector outside the FBD. Solve for what you are asked to find.
$$F_{\text{net}}=F_{\text{rod}}-m_1g=m_1a$$ Since the acceleration is centripetal, $$F_{\text{rod}}-m_1g=m_1v^2/(L/2).$$ You have the speed $v$ from a previous part, you need to solve the force exerted by the rod on the mass. The force exerted by the mass on the rod has the same magnitude and opposite direction to that.

If you are serious about improving your free body diagram skills, apply this step-by-step method to find the force exerted by the other mass on the rod and consider using it in the future.

Thank you so much! That is a very nice guide. Using this I get that T_1 = m_1*(2*v^2/L-g)= -11.78 N (down), G_1 = m_1*g = -19.62 N (down), T_2 = m_2*(2*v^2/L+g) = 41.19 N, G_2 = m_2*g = -29.43 N. Do you agree with this?

 

[ymnoklan]

$G_1$ and $G_2$ are different. Maybe that was just a typo.

As @Steve4Physics mentioned, it is much better to work algebraically. Don't plug in numbers until you have to. So far you have $v^2=\frac{(m_2-m_1)gL}{(m_2+m_1)}$.

You need to define which way is positive for your forces. It doesn’t matter what you choose as long as, for each one, you are consistent.
Say we take towards the axis as positive always. Then we have $T_1+G_1=m_1\frac{v^2}{L/2}$, $T_2-G_2=m_2\frac{v^2}{L/2}$.
What does that give you for the tensions?

Thank you so much! That is a very nice guide. Using your tips I get that T_1 = m_1*(2*v^2/L-g)= -11.78 N (down), G_1 = m_1*g = -19.62 N (down), T_2 = m_2*(2*v^2/L+g) = 41.19 N, G_2 = m_2*g = -29.43 N. Do you agree with this?

 

[haruspex]

I get that T_1 = m_1*(2*v^2/L-g)= -11.78 N (down),

Yes, $T_1 = -11.78 N$, but remember that we are taking towards the axis as positive. So which way is a force of -11.78N acting?

 

[ymnoklan]

Yes, $T_1 = -11.78 N$, but remember that we are taking towards the axis as positive. So which way is a force of -11.78N acting?

I get that, but that disagrees with my free-body diagram. Is T_1 really upwards? Do you agree with the other directions though?

 

[haruspex]

I get that, but that disagrees with my free-body diagram.

The free body diagram should show the forces acting in whichever direction you are choosing as positive. It doesn’t matter if the actual direction is the other way.

Is T_1 really upwards?

Yes. It is rotating quite slowly.

Do you agree with the other directions though?

Yes.