# Rotating rod of negligible mass attached to two point masses

• Kennedy
In summary, the problem involves a uniform rod with two point masses attached to either end, mounted on a horizontal axis. The rod is released from rest and the question asks for the angular velocity of the rod when the 2 kg mass passes through its low point. The initial attempt at solving the problem using rotational dynamics was not successful, and the suggestion was made to use energy conservation instead. The basic principle is to equate the initial and final potential and kinetic energies of the system. The expressions for potential and kinetic energy were found, and it was noted that the angular acceleration is not constant. The final kinetic energy of each point mass was calculated, but it is important to note that the height from the ground is not known and this may affect
Kennedy

## Homework Statement

A uniform rod (length = 2.4 m) of negligible mass has a 1.0-kg point mass attached to one end and a 2.0-kg point mass attached to the other end. The rod is mounted to rotate freely about a horizontal axis that is perpendicular to the rod and that passes through a point 1.0 m from the 2.0-kg mass. The rod is released from rest when it is horizontal. What is the angular velocity of the rod at the instant the 2-kg mass passes through its low point? (a) 1.7 rad/s (b) 2.2 rad/s (c) 2.0 rad/s (d) 1.5 rad/s (e) 3.1 rad/s

## Homework Equations

I'm thinking that using the cross product of the radius and the torque is relevant (rFsinθ), and probably the equivalent of the equations of linear motion for rotational dynamics. I would also assume that Newton's Second Law for rotational dynamics applies (τ = Iα).

## The Attempt at a Solution

First I calculated the force of gravity that would act on both point masses. So, 9.8 N for the 1.0-kg point mass, and 19.6 N for the 2.0-kg point mass. Since they're both point masses, I would assume that the torque acts at both ends of the rod. So, the torque on the 2.0-kg point mass would be 19.6 N(1 m) = 19.6 N*m, and for the 1.0-kg point mass would be 9.8 N(1.4 m) = 13.72 N*m. Because the mass of the rod is negligible, the rotational inertia is completely due to the point masses. So, I = 1(1.4^2) +2(1^2) = 3.96 kg*m^2. The net torque is 19.6 - 13.72 = 5.88 N*m. This yields the angular acceleration to be 5.88/3.96 = 1.48 rad/s^2.

Since the rod starts out horizontally, and it rotates in the direction of the 2.0-kg point mass, when the system rotates an angle of θ = pi/2, the 2.0-kg point mass is at its low point. By using the equivalent of the equations of linear motion for rotational dynamics, I solved for the final angular velocity of the 2.0-kg point mass at its low point. This gave me 2.15 rad/s, which is not the correct answer.

I was shown how to do the problem using energy conservation, but shouldn't I be able to solve the problem using this method. What's wrong with my approach?

Last edited:
Kennedy said:
By using the equivalent of the equations of linear motion for rotational dynamics ...
What are these equivalent equations? Note that the angular acceleration is not constant. You are better off using energy conservation.

kuruman said:
What are these equivalent equations? Note that the angular acceleration is not constant. You are better off using energy conservation.
I don't really understand that method at all. What's the basic principal? One gains potential energy, and the other loses potential energy, but will gain kinetic energy. I don't understand how to put this all together into one equation to eventually use KE = 1/2(I)(w^2).

The basic principle is energy conservation. It says
K1i + U1i + K2i + U2i = K1f + U1f + K2f + U2f
1. Can you find expressions for the initial and final potential energies?
2. Can you find expressions for the initial and final kinetic energies? Hint: Use K = ½ I ω2, where I is the moment of inertia of each mass relative to the pivot.

kuruman said:
The basic principle is energy conservation. It says
K1i + U1i + K2i + U2i = K1f + U1f + K2f + U2f
1. Can you find expressions for the initial and final potential energies?
2. Can you find expressions for the initial and final kinetic energies? Hint: Use K = ½ I ω2, where I is the moment of inertia of each mass relative to the pivot.
So, I believe the initial potential energy of the 2.0 kg mass is 1 m (9.8 m/s^2) (2 kg) = 19.6 J.
The initial potential energy of the 1 kg mass is (1.4 m) (9.8 m/s^2) (1 kg) = 13.72 J
Of course, both of these are assuming that the potential energy at the bottom of the circle is 0 J.

The final KE is equal to the negative of the change in potential energy, so for the 2 kg mass that would be 19.6 J, and for the 1 kg mass that would be 13.72 - (1.4 m)(9.8 m/s^2)(1 kg) = 0 J

But I feel like that's all wrong. How do I know what height they are from the ground?

The final kinetic energy of m1 is equal to (1/2)(1.96)(w^2), but isn't w going to be zero at the top of the motion, because the potential energy is a max and the kinetic energy is at a minimum? The final kinetic energy of m2 = (1/2)(2)(w^2).

Kennedy said:
But I feel like that's all wrong. How do I know what height they are from the ground?
The height from the ground is irrelevant. If you start from the equation
K1i + U1i + K2i + U2i = K1f + U1f + K2f + U2f,
you can move everything to the right side and rewrite it as
- K1i - U1i - K2i - U2i + K1f + U1f + K2f + U2f = 0
or
(K1f - K1i) + (K2f - K2i) + (U1f - U1i) + (U2f - U2i) = 0
or
ΔK1 + ΔK2 + ΔU1 + ΔU2 = 0
Note that it's the change in potential energy that counts, not its initial and final values. For that reason the zero of potential energy does not matter. Also note that since the system starts from rest, ΔK1 = K1f - 0 = K1f and likewise for ΔK2.

I recommend that you find expressions for all the Δ's in terms of symbols, not numbers and put in the numbers at the very end. Your work will be much easier to troubleshoot in case you make mistakes.

Edit: Fixed incomplete equaiton.

Last edited:
Kennedy said:
change in potential energy, ... for the 1 kg mass that would be 13.72 - (1.4 m)(9.8 m/s^2)(1 kg) = 0 J
Clearly the 1kg mass does have a change in PE, so think that through again. Your 13.72J was relative to a baseline. You must use the same baseline throughout.

ΔK1 = K1f
ΔK2 = K2f
ΔU1 = -K1f
ΔU2 = -K2F

...but that still gets me nowhere because I don’t know the change in potential energy for either mass (the change in KE is the negative of the change in PE)!

haruspex said:
Clearly the 1kg mass does have a change in PE, so think that through again. Your 13.72J was relative to a baseline. You must use the same baseline throughout.
Why wouldn’t it? It moves upwards, so should be gaining potential energy!

Kennedy said:
Why wouldn’t it? It moves upwards, so should be gaining potential energy!
Yes, the 1 kg mass gains potential energy and the 2 kg mass loses potential energy. Does that bother you? Just concentrate on finding expressions of how much potential energy is gained by the 1 kg mass and how much is lost by the 2 kg mass. To do this, consider how high one mass rises and how low the other one drops when the system goes through the lowest point of the motion.

kuruman said:
Yes, the 1 kg mass gains potential energy and the 2 kg mass loses potential energy. Does that bother you? Just concentrate on finding how much potential energy is gained by the 1 kg mass and how much is lost by the 2 kg mass. To do this, consider how high one mass rises and how low the other one drops when the system goes through the lowest point of the motion.
So, from what I understand the 1 kg has a Δh of 1.4 m and the 2 kg mass has a Δh of -1 m

Their corresponding ΔU are 1(1.4)(9.8) = 13.72 J for the 1 kg mass, and 2(-1)(9.8) = -19.6 J for the 2 kg mass.

What do I do with these number?

Kennedy said:
What do I do with these number?
As I stated in post #7,
kuruman said:
I recommend that you find expressions for all the Δ's in terms of symbols, not numbers and put in the numbers at the very end. Your work will be much easier to troubleshoot in case you make mistakes.
How many Δ's do you still need to find expressions for?

kuruman said:
As I stated in post #7,

How many Δ's do you still need to find expressions for?
Two. I still have my change in final KE for both masses, but shouldn’t those just be the negative of the change in potential energy? So, 19.6 J, and -13.72 J for the 2 kg mass and the 1 kg mass respectively?

Kennedy said:
Two. I still have my change in final KE for both masses, but shouldn’t those just be the negative of the change in potential energy? So, 19.6 J, and -13.72 J for the 2 kg mass and the 1 kg mass respectively?
Why should they? You have ΔK1 + ΔK2 = - ΔU1 - ΔU2. This doesn't necessarily mean that ΔK1 = - ΔU1 and ΔK2 = - ΔU2. Although it is true that 3 + 2 = 3 + 2, it is also true that 1 + 4 = 3 + 2.

Kennedy said:
Two. I still have my change in final KE for both masses, but shouldn’t those just be the negative of the change in potential energy? So, 19.6 J, and -13.72 J for the 2 kg mass and the 1 kg mass respectively?
OKAY! I think I got it. I added up the total change in KE (19.6 - 13.72 = 5.88), and then used 1/2(I)(w^2) to solve for w. I took 3.96 to be the rotational inertia (1(1.4^2) + 2(1^2) = 3.96), and then I solved for w, which yields 1.7 rads/sec! Thank you!

kuruman said:
Why should they? You have ΔK1 + ΔK2 = - ΔU1 - ΔU2. This doesn't necessarily mean that ΔK1 = - ΔU1 and ΔK2 = - ΔU2. Although it is true that 3 + 2 = 3 + 2, it is also true that 1 + 4 = 3 + 2.
...and I thought that was just a rule. The amount of PE lost is always converted into KE due to energy conservation.

Kennedy said:
The amount of PE lost is always converted into KE due to energy conservation.
Yes, it is a rule. In this case PE is the sum of two individual PE's and the same for the KE's. It's the total mechanical energy of the two-particle system that is conserved not the individual mechanical energies of the two particles making up the system. That's what I tried to indicate in post #15.

Kennedy
Kennedy said:
Why wouldn’t it? It moves upwards, so should be gaining potential energy!
In post #5, and as I quoted in post #8, you wrote that the change in PE for the 1kg mass was zero. You got that because you took the gain in PE from the level position to the top position, then subtracted the PE that it had in the level position from your baseline of the bottom position.
In post #8, I wrote that it does have a gain in PE, so I do not understand your asking "why wouldn't it".

## 1. What is a rotating rod of negligible mass attached to two point masses?

A rotating rod of negligible mass attached to two point masses is a physical system where two point masses are connected by a thin, rigid rod that has a very small mass compared to the point masses. The rod is able to rotate around its axis, allowing the point masses to move in a circular motion.

## 2. What are the applications of a rotating rod of negligible mass attached to two point masses?

This system is commonly used in physics experiments to study rotational dynamics and to demonstrate concepts such as angular momentum and torque. It can also be found in mechanical systems, such as balancing scales and rotating devices.

## 3. How does the mass of the rod affect the motion of the point masses?

The mass of the rod has a negligible effect on the motion of the point masses. This is because the rod is considered to be massless in comparison to the point masses, and therefore its contribution to the overall rotational motion is insignificant.

## 4. Can the point masses move independently of each other in this system?

Yes, the point masses can move independently of each other in this system. The movement of one point mass does not affect the other, as they are connected by a rigid rod and do not exert any forces on each other.

## 5. How is the angular velocity of the rod related to the angular velocity of the point masses?

The angular velocity of the rod is equal to the average of the angular velocities of the two point masses. This means that if one point mass is rotating faster than the other, the rod will rotate at an angular velocity that is between the two.

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