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

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.

 

[kuruman]

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?

I do not agree. Check your algebra.
I thought you were referring to the larger mass as $m_1$ as defined in my FBD in post #29.

 

[haruspex]

The acceleration is up towards the center.

$T_1$ corresponds to $m_1$, the lesser mass. Its acceleration will be downward.

 

[kuruman]

$T_1$ corresponds to $m_1$, the lesser mass. Its acceleration will be downward.

Yes, I was referring to my $m_1$ in he FBD in post #29 which is the larger mass and I thought that OP did too. I deleted the earlier post to avoid confusion. I think I will remove myself from cooking this broth any further.

 

[ymnoklan]

Thank you all so much for your help! I am still stuck with the hardest one though… the force on the axis? What do they ask for here? What is different from this and the previous question? And where do I even begin?

 

[Steve4Physics]

Thank you all so much for your help! I am still stuck with the hardest one though… the force on the axis? What do they ask for here? What is different from this and the previous question? And where do I even begin?

Can you solve some simpler problems first? These should lead you in the right direction.

An object, weight 10N, is attached to one end of a rod of negligible weight. The rod+mass rotates in a vertical circle, about pivot P at the other end of the rod.

Q1 When the mass is at the top, suppose |centripetal force| = 15N.
a) What is the force (size and direction) exerted by the rod on the mass?
b) What is the tension or compression in the rod?
c) What is the force (size and direction) exerted by the rod on the pivot?

Q2. Same again but this time the speed is slower so |centripetal force| = 7N.

Q3. Same again with mass at bottom and |centripetal force| = 15N.

Q4. Same again with mass at bottom and |centripetal force| = 7N.

When you’ve sorted these out, go back to the original problem, Pretend there are two rods – one rod (length L//2) connects $m_1$ to the pivot; the other rod (also length L/2) connects $m_2$ to the pivot. The force on the pivot is then the (vector) sum of the forces exerted by each rod on the pivot.

(I’m off to bed now so can’t contribute for several hours!)

 

[Lnewqban]

I am still stuck with the hardest one though… the force on the axis? What do they ask for here?

Copied from the problem statement: "You hold the bar in a horizontal position and release the system."

At the time that you release it, the system would free fall, unless a force at the axis is preventing that from happening.

As the center of mass of the system is located off that axis, a rotation, and related centrifugal forces, are induced.

While rotation happens, the centrifugal forces are directed radially, while the weight forces are directly downwards.

The question seems to be directed to the point at which the system has reached a vertical position (top mass-axis-bottom mass all aligned vertically).

 

[ymnoklan]

Can you solve some simpler problems first? These should lead you in the right direction.

An object, weight 10N, is attached to one end of a rod of negligible weight. The rod+mass rotates in a vertical circle, about pivot P at the other end of the rod.

Q1 When the mass is at the top, suppose |centripetal force| = 15N.
a) What is the force (size and direction) exerted by the rod on the mass?
b) What is the tension or compression in the rod?
c) What is the force (size and direction) exerted by the rod on the pivot?

Q2. Same again but this time the speed is slower so |centripetal force| = 7N.

Q3. Same again with mass at bottom and |centripetal force| = 15N.

Q4. Same again with mass at bottom and |centripetal force| = 7N.

When you’ve sorted these out, go back to the original problem, Pretend there are two rods – one rod (length L//2) connects $m_1$ to the pivot; the other rod (also length L/2) connects $m_2$ to the pivot. The force on the pivot is then the (vector) sum of the forces exerted by each rod on the pivot.

(I’m off to bed now so can’t contribute for several hours!)

Thank you for the examples! I am not sure if I understand them correctly though. Trying to solve them I get:
Q1: a) 5 N down, b) 5 N (equal in magnitude to a), c) 5 N up (opposite to a)
Q2: a) 3 N up, b) 3 N, c) 3 N down
Q3: a) 25 N up, b) 25 N, c) 25 N down
Q4: a) 17 N up, b) 17 N, c) 17 N down
Based on this I would guess that the force on the axis is then sqrt(T_1^2 + T_2^2) = sqrt(11.78^2 + 41.18^2) ≈ 42.84 N downwards. What do you think of this?

 

[ymnoklan]

Copied from the problem statement: "You hold the bar in a horizontal position and release the system."

At the time that you release it, the system would free fall, unless a force at the axis is preventing that from happening.

As the center of mass of the system is located off that axis, a rotation, and related centrifugal forces, are induced.

While rotation happens, the centrifugal forces are directed radially, while the weight forces are directly downwards.

The question seems to be directed to the point at which the system has reached a vertical position (top mass-axis-bottom mass all aligned vertically).

Hm okay, do you mean that the force on the axis will be the resultant force of the radial forces and the vertical forces, where the radial forces are the centrifugal forces (m_1+m_1)*v^2/r and the vertical forces are the gravitational forces?
Then I would get this:
(m_1+m_2)*g => sqrt((m_1+m_1)*v^2/r)^2+((m_1+m_1)*g)
=> sqrt((2+3)*1.4^2/0.5)^2+((2+3)*9.81)^2) ≈ 52.82 N

 

[Steve4Physics]

You're getting there!

Thank you for the examples! I am not sure if I understand them correctly though. Trying to solve them I get:
Q1: a) 5 N down, b) 5 N (equal in magnitude to a), c) 5 N up (opposite to a)

Agreed. Because the direction of the force of the rod on the mass (answer a)) is downwards and the the mass is at the top, this tells us that the rod is in tension, So the answer to part b) is really 'Tension = 5N'.

Q2: a) 3 N up, b) 3 N, c) 3 N down

Agreed. But the answer to part b) is really 'Compression = 3N'.

Q3: a) 25 N up, b) 25 N, c) 25 N down

Agreed. But the answer to part b) is really 'Tension= 25N'.

Q4: a) 17 N up, b) 17 N, c) 17 N down

Agreed. But the answer to part b) is really 'Tension = 17N'.

(Note that the rod must always be in tension when the mass is at the bottom position, but it can be in tension or compression when the mass is at the top position. Worth thinking about if not clear. )

Based on this I would guess that the force on the axis is then sqrt(T_1^2 + T_2^2) = sqrt(11.78^2 + 41.18^2) ≈ 42.84 N downwards. What do you think of this?

Disagree. Guessing is dangerous and no substitute for careful thinking!

When we add two forces by vector addition the answer depends on the angle between their directions. Look at these examples:

2N to the right+ 5N to the right: vector addition gives 7N to the right.
2N to the right + 5N to the left: vector addition gives 3N to the left.
2N to the right +5N up : note angle between the forces is 90$^0$ so we use Pythagoras and vector addition gives $\sqrt{2^2 +5^2}$ at an angle of $\tan^{-1} ( \frac 52)$ to the horizontal.

 

[ymnoklan]

You're getting there!


Agreed. Because the direction of the force of the rod on the mass (answer a)) is downwards and the the mass is at the top, this tells us that the rod is in tension, So the answer to part b) is really 'Tension = 5N'.


Agreed. But the answer to part b) is really 'Compression = 3N'.


Agreed. But the answer to part b) is really 'Tension= 25N'.


Agreed. But the answer to part b) is really 'Tension = 17N'.

(Note that the rod must always be in tension when the mass is at the bottom position, but it can be in tension or compression when the mass is at the top position. Worth thinking about if not clear. )


Disagree. Guessing is dangerous and no substitute for careful thinking!

When we add two forces by vector addition the answer depends on the angle between their directions. Look at these examples:

2N to the right+ 5N to the right: vector addition gives 7N to the right.
2N to the right + 5N to the left: vector addition gives 3N to the left.
2N to the right +5N up : note angle between the forces is 90$^0$ so we use Pythagoras and vector addition gives $\sqrt{2^2 +5^2}$ at an angle of $\tan^{-1} ( \frac 52)$ to the horizontal

2N to the right+ 5N to the right: vector addition gives 7N to the right.
2N to the right + 5N to the left: vector addition gives 3N to the left.
2N to the right +5N up : note angle between the forces is 900 so we use Pythagoras and vector addition gives 22+52 at an angle of tan−1⁡(52) to the horizontal.

I get that, but in my case I thought the only forces working are the radial forces and the vertical forces, where the radial forces are the centrifugal forces (m_1+m_1)*v^2/r and the vertical forces are the gravitational forces?
This would result in:
(m_1+m_2)*g => sqrt((m_1+m_1)*v^2/r)^2+((m_1+m_1)*g) at angle tan^-1(((m_1+m_2)*g)/((m_1+m_1)*v^2/r))
=> sqrt((2+3)*1.4^2/0.5)^2+((2+3)*9.81)^2) ≈ 52.82 N at angle theta ≈ 68.22 degrees
What do you think of this then?

 

[Steve4Physics]

I get that, but in my case I thought the only forces working are the radial forces and the vertical forces, where the radial forces are the centrifugal forces (m_1+m_1)*v^2/r and the vertical forces are the gravitational forces?
This would result in:
(m_1+m_2)*g => sqrt((m_1+m_1)*v^2/r)^2+((m_1+m_1)*g) at angle tan^-1(((m_1+m_2)*g)/((m_1+m_1)*v^2/r))
=> sqrt((2+3)*1.4^2/0.5)^2+((2+3)*9.81)^2) ≈ 52.82 N at angle theta ≈ 68.22 degrees
What do you think of this then?

It is unreadable (you need to learn to use LaTeX) and clearly wrong because it uses a tangent!

I think you have misunderstood. In Post #42 you were wrong to use Pythagoras because the 2 forces you were adding were not perpendicular.

At the end of Post #44 I was trying to explain your mistake by using some simple examples.

You need to add the two forces on the pivot. What are the magnitude and direction of each force on the pivot?

 

[ymnoklan]

It is unreadable (you need to learn to use LaTeX) and clearly wrong because it uses a tangent!

I think you have misunderstood. In Post #42 you were wrong to use Pythagoras because the 2 forces you were adding were not perpendicular.

At the end of Post #44 I was trying to explain your mistake by using some simple examples.

You need to add the two forces on the pivot. What are the magnitude and direction of each force on the pivot?

I find this very difficult to get my head around, so thank you a lot for your help. On the pivot I only really find T_1 = 11.78 N upwards and T_2 = 41.18 N upwards. Aren't these kind of "on top of each other" when the rod is vertical? I think this seems very odd though.

 

[Steve4Physics]

On the pivot I only really find T_1 = 11.78 N upwards and T_2 = 41.18 N upwards. Aren't these kind of "on top of each other" when the rod is vertical? I think this seems very odd though.

Your values are OK although I get 11.772N and 41.202N using a spreadsheet, so it seems you have introduced rounding errors at some point(s). But one (or both) of the direction(s) is (are) wrong.

Call the upper half of the rod A; this part connects the pivot to $m_1$.
Call the lower half of the rod B; this part connects the pivot to $m_2$.
You need to treat A and B as if they are separate rods connected to the same pivot.

What is the force (size and direction) exerted by A on the pivot?
What is the force (size and direction) exerted by B on the pivot?
(These questions are just like Q1-Q4 in Post #40 - which you correctly answered).

The total force on the pivot is then the sum of these 2 forces.

 

[ymnoklan]

Your values are OK although I get 11.772N and 41.202N using a spreadsheet, so it seems you have introduced rounding errors at some point(s). But one (or both) of the direction(s) is (are) wrong.

Call the upper half of the rod A; this part connects the pivot to $m_1$.
Call the lower half of the rod B; this part connects the pivot to $m_2$.
You need to treat A and B as if they are separate rods connected to the same pivot.

What is the force (size and direction) exerted by A on the pivot?
What is the force (size and direction) exerted by B on the pivot?
(These questions are just like Q1-Q4 in Post #40 - which you correctly answered).

The total force on the pivot is then the sum of these 2 forces.

I mean, obviously the sum of the forces must be towards the center of the "circle", which would mean that the total force by A would be downwards and the force by B would be upwards. Do you mean that the total force on the pivot and on the axis is 41.2 N - 11.8 N ≈ 29.4 N upwards (rounded) ?
If so, what are your thoughts on the directions on the forces in the rod then? I thought this would be T_1 ≈ 11.8 N up, G_1 ≈ 19.6 N down, T_2 ≈ 41.2 N up and G_2 ≈ 29.4 N down.
What do you think of this?

 

[Steve4Physics]

I mean, obviously the sum of the forces must be towards the center of the "circle"

No. The pivot IS the centre of the circle. The direction of the total force on the pivot can't be towards itself; that wouldn't make sense.

The centripetal force on a object is always towards the centre of the circular motion. But the force on the pivot is not the centripetal force. Please try these questions (similar to the Post #40 questions):
______

Q1. An object weighs 19.62N. It is attached to one end of a rod (length 0.5m, though you won't need that) of negligible weight. The rod+mass rotates in a vertical circle, about pivot P at the other end of the rod.

When the mass is at the top, |centripetal force on mass| = 7.848N.

a) What is the direction (up/down) of the centripetal force on the mass?
b) What is the force (size and direction (up/down)) exerted by the rod on the mass?
c) Is the rod under compression or tension?
d) What is the compressive or tensile force in the rod?
e) What is the force (size and direction (up/down)) exerted by the rod on the pivot?
______________

Q2. Another object weighs 29.43N. It is attached to one end of another rod (length 0.5m, though you won't need that) of negligible weight. The rod+mass rotates in a vertical circle, about pivot P at the other end of the rod.

When the mass is at the bottom, |centripetal force on mass| = 11.772N.

a) What is the direction (up/down) of the centripetal force on the mass?
b) What is the force (size and direction(up/down)) exerted by the rod on the mass?
c) Is the rod under compression or tension?
d) What is the compressive or tensile force in the rod?
e) What is the force (size and direction) exerted by the rod on the pivot?
___________

Q3 If the 2 systems (in Q1 and Q2) are on the same pivot, what is the total force on the pivot?