Rotating rod of negligible mass attached to two point masses

[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?

 

[kuruman]

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.

 

[Kennedy]

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).

 

[kuruman]

The basic principle is energy conservation. It says
K_1i + U_1i + K_2i + U_2i = K_1f + U_1f + K_2f + U_2f
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.

 

[Kennedy]

The basic principle is energy conservation. It says
K_1i + U_1i + K_2i + U_2i = K_1f + U_1f + K_2f + U_2f
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?

 

[Kennedy]

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).

 

[kuruman]

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
K_1i + U_1i + K_2i + U_2i = K_1f + U_1f + K_2f + U_2f,
you can move everything to the right side and rewrite it as
- K_1i - U_1i - K_2i - U_2i + K_1f + U_1f + K_2f + U_2f = 0
or
(K_1f - K_1i) + (K_2f - K_2i) + (U_1f - U_1i) + (U_2f - U_2i) = 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 = K_1f - 0 = K_1f 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.

 

[haruspex]

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.

 

[Kennedy]

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)!

 

[Kennedy]

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!

 

[kuruman]

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.

 

[Kennedy]

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?

 

[kuruman]

What do I do with these number?

As I stated in post #7,

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?

 

[Kennedy]

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?

 

[kuruman]

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]

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]

My answer agrees with yours.

 

[Kennedy]

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.

 

[kuruman]

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.

 

[haruspex]

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".