# Falling rod with a mass on either end

• clope023
In summary, the conversation discusses the solution to a problem involving a falling rod with a mass on either end. The system is released from rest with the rod horizontal and the goal is to determine the speeds of the two masses as the rod swings through a vertical position. After some calculations, it is determined that the correct equation should be magR - mbgR = 1/2mav^2 + 1/2mbv^2 and the final answer is v = \sqrt{2gr((ma-mb)/(ma+mb))}.
clope023
[SOLVED] Falling rod with a mass on either end

## Homework Statement

A wooden rod of negligible mass and length 85.0cm is pivoted about a horizontal axis through its center. A white rat with mass 0.520kg clings to one end of the stick, and a mouse with mass 0.250kg clings to the other end. The system is released from rest with the rod horizontal.

If the animals can manage to hold on, what are their speeds as the rod swings through a vertical position?

## Homework Equations

(K1a + K1b) + (U1a + U1b) = (K2a + K2b) + (U2a + U2b)

U = mgR, R = 1/2d of the rod
K1 = 0
U2 = 0

## The Attempt at a Solution

(magR + mbgR) = 1/2v^2(ma + mb)

gR (ma+mb) = 1/2v^2 (ma+mb)

2gr = v^2, v = $$\sqrt{2gr}$$

v = 2.9m/s wrong

not sure what I'm doing wrong here.

any help is appreciated.

Last edited:
The rat goes down, losing potential energy and the mouse goes up gaining potential energy. The potential energy changes don't add. They subtract. They have different signs.

hmmm, so you're saying it would be:

magR - mbgR = 1/2mav^2 + 1/2mbv^2?

and from here would'nt it be:

gR(ma-mb) = 1/2v^2(ma+mb)

$$\sqrt{2gr((ma-mb)/(ma+mb))}$$ = v?

I think so. In fact, I know so. Does that work?

Last edited:
Dick said:
I think so. In fact, I know so. Does that work?

yes it did! wow I was stuck on that one for hours and I was actually thinking of changing the signs, but when I did it I changed the signs of both the potential and kinetic energies so they (the masses) both canceled out anyways, thanks for pointing out the difference; thanks again!

Last edited:

## 1. How does the mass at the ends of the rod affect its falling motion?

The mass at the ends of the rod affects its falling motion by increasing its inertia. This means that the rod will resist changes in its motion and will take longer to fall compared to a rod with no mass attached.

## 2. Does the length of the rod impact its falling speed?

Yes, the length of the rod does impact its falling speed. A longer rod will have a greater distance to fall, thus taking longer to reach the ground compared to a shorter rod.

## 3. Is the mass of the rod itself a factor in its falling speed?

No, the mass of the rod itself does not affect its falling speed. The mass of the rod only becomes a factor when additional masses are attached to the ends of the rod.

## 4. What is the relationship between the mass at the ends of the rod and its falling acceleration?

The mass at the ends of the rod has a direct relationship with its falling acceleration. The greater the mass at the ends, the slower the rod will accelerate towards the ground. This is due to the increased inertia and resistance to changes in motion.

## 5. Are there any external factors that can influence the falling motion of the rod with masses on either end?

Yes, there are external factors that can influence the falling motion of the rod. These include air resistance, wind, and surface friction. These factors can affect the speed and trajectory of the falling rod.

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