What is the displacement of the beam after the medicine ball is thrown?

In summary, the position of the person at the left end of the beam is 0.167 meters after the ball is thrown from person to person. This is found by using the equation for the center of mass and solving for the displacement of the beam, which is 0.175 meters. The mistake was likely a typo in the calculator.
  • #1
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Homework Statement


A person with mass m1 = 67.0 kg stands at the left end of a uniform beam with mass m2 = 91.0 kg and a length L = 3.3 m. Another person with mass m3 = 56.0 kg stands on the far right end of the beam and holds a medicine ball with mass m4 = 12.0 kg (assume that the medicine ball is at the far right end of the beam as well). Let the origin of our coordinate system be the left end of the original position of the beam as shown in the drawing. Assume there is no friction between the beam and floor.

What is the new x-position of the person at the left end of the beam? (How far did the beam move when the ball was throw from person to person?)

I previously solved for the center of mass for the system, which came out to be about 1.65 m. This answer was correct, and I have been using it for my equations since.

Homework Equations



Xcm = [(m1)(x) + m2(x) + m3(x) + ... + mn(x)]/(m1 + m2 + m3 + ... + mn)

The Attempt at a Solution



I know that the center of mass does not change, so the system must act according to reorient itself. This means that the position of the beam is what changes. If we take the initial position of the person at the left end of the beam to be x = 0, this is the equation for the center of mass before the ball is thrown:

BEFORE:
Xcm = [(67)(0) + (91)(1.65) + (56)(3.3) + (12)(3.3)]/(67+91+56+12)
= 1.65 m

This is the center of mass. If we let x = the distance moved by the beam after the ball is thrown, we get the following equation:

AFTER:
1.65 = [(67)(x) + (91)(1.65+x) + (56)(3.3+x) + (12)(0.0 + x)]/(67+91+56+12)

And you simply solve for x to find the distance moved. The reason why the position of the medicine ball is now 0 in the after equation is because it was thrown to the left, and caught by the person on the left who was previously at the origin.

When I solve for x, I get x=0.167 m. I've double checked to make sure I'm not making any algebra mistakes or something like that. I feel like this should be the correct answer, but for some reason, it is wrong. Can anyone help me find out what I did wrong with my reasoning?
 
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  • #2
WolframAlpha gives 1.657m for the center of mass and 0.175m for the displacement, using your formulas as input. Probably just a typo in the calculator?

Cross-check: (67+91+56)kg=214kg move x to the right, 12kg move 3.3m - x to the left. Solving 214*x=12*(3.3m-x) gives x=0.175m.
 
  • #3
mfb said:
WolframAlpha gives 1.657m for the center of mass and 0.175m for the displacement, using your formulas as input. Probably just a typo in the calculator?

Cross-check: (67+91+56)kg=214kg move x to the right, 12kg move 3.3m - x to the left. Solving 214*x=12*(3.3m-x) gives x=0.175m.

That was right, thank you! I was worried I was making a mistake in my reasoning. I'm glad that wasn't it. I should double check my numbers before punching them into the calculator blindly next time. Thanks again! :D
 

1. What is the center of mass?

The center of mass is a point that represents the average location of the mass of an object. It is the point at which the total weight of the object can be considered to be concentrated.

2. How is the center of mass calculated?

The center of mass is calculated by finding the weighted average of the individual masses of an object. This is done by multiplying the mass of each part of the object by its distance from a reference point, and then dividing the sum of these values by the total mass of the object.

3. Why is the center of mass important?

The center of mass is important because it helps us understand how an object will behave when subjected to external forces. It also helps in determining the stability and balance of an object.

4. Can the center of mass be outside of an object?

Yes, the center of mass can be outside of an object. This is often the case with objects that have irregular shapes or have mass distributed unevenly.

5. How does the center of mass change when an object moves?

The center of mass remains the same regardless of the object's motion. However, the position of the center of mass may change relative to the object itself if the object is rotating or changing shape while in motion.

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