Center of mass of medicine ball

In summary: Your result for #3 is 0.1632.In summary, the center of mass of the system is located at 1.6757 m from the left end of the beam. When the medicine ball is thrown to the left end of the beam, the center of mass remains at the same location. The new x-position of the person at the left end of the beam is 0.1632 m from their initial position. To return the medicine ball to the other person, both people walk to the center of the beam, which is located at 1.5125 m from the left end.
  • #1
Rifscape
41
0

Homework Statement

1.68 - ((51(0) + 109(1.5) + 66(3) + 13(0))/(51 + 66 + 109 + 13)) [/B]
A person with mass m1 = 51 kg stands at the left end of a uniform beam with mass m2 = 109 kg and a length L = 3 m. Another person with mass m3 = 66 kg stands on the far right end of the beam and holds a medicine ball with mass m4 = 13 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.

1. What is the location of the center of mass of the system?

2. The medicine ball is throw to the left end of the beam (and caught). What is the location of the center of mass now?

3.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?)

4.To return the medicine ball to the other person, both people walk to the center of the beam. At what x-position do they end up?

Homework Equations



xcm = 1/MtotΣ mi*ri

The Attempt at a Solution


I solved number one by using the formula, I did (51(0) + 109(1.5) + 66(3) + 13(3))/(51 + 109 + 66 + 13) = 1.68.

For the second part there is no external force so the center of mass stays the same at 1.68. It is only at #3 where I seem to go wrong.

I thought that you had to calculate the shift by using the same equation by adjusting it to fit the new situation. So I did 1.68 - ((51(0) + 109(1.5) + 66(3) + 13(0))/(51 + 66 + 109 + 13)) = 1.68 - 1.512 = 0.167 however this answer is wrong.

Number 4 requires number 3, so I need to figure this question out first.
Please help
Thank you

Nobody wants to help ? this is due very soon. Than you for your time though.
 
Last edited:
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  • #2
Rifscape said:

Homework Statement

1.68 - ((51(0) + 109(1.5) + 66(3) + 13(0))/(51 + 66 + 109 + 13)) [/B]
A person with mass m1 = 51 kg stands at the left end of a uniform beam with mass m2 = 109 kg and a length L = 3 m. Another person with mass m3 = 66 kg stands on the far right end of the beam and holds a medicine ball with mass m4 = 13 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.

1. What is the location of the center of mass of the system?

2. The medicine ball is throw to the left end of the beam (and caught). What is the location of the center of mass now?

3.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?)

4.To return the medicine ball to the other person, both people walk to the center of the beam. At what x-position do they end up?

Homework Equations



xcm = 1/MtotΣ mi*ri

The Attempt at a Solution


I solved number one by using the formula, I did (51(0) + 109(1.5) + 66(3) + 13(3))/(51 + 109 + 66 + 13) = 1.68.

For the second part there is no external force so the center of mass stays the same at 1.68. It is only at #3 where I seem to go wrong.

I thought that you had to calculate the shift by using the same equation by adjusting it to fit the new situation. So I did 1.68 - ((51(0) + 109(1.5) + 66(3) + 13(0))/(51 + 66 + 109 + 13)) = 1.68 - 1.512 = 0.167 however this answer is wrong.

0.167 m is the position of the new CM with respect to the left end of the beam. But it is shifted, and its new position is Xleft with respect to the origin, which is at the initial position of the left end. The CM is at Xleft+0.167. And you know, the position of the CM does not change. Determine Xleft.
 
  • #3
I don't understand is it 1.512? But isn't that the new center(not center of mass) of the system? I'm not sure how to determine x left.
 
Last edited:
  • #4
I got it all, thank you for your help!
 
  • #5
Rifscape said:
I don't understand is it 1.512? But isn't that the new center(not center of mass) of the system? I'm not sure how to determine x left.
Sorry, I misread your formula. You did it correctly, but miscalculated something or rounded off too early. The CM is at XCM=1.6757 m instead of 1.68. You should subtract 1.5125 from it.
 

1. What is the center of mass of a medicine ball?

The center of mass of a medicine ball is the point at which the weight of the ball is evenly distributed in all directions. It is the point where the ball can be balanced on a single finger without falling over.

2. Why is knowing the center of mass important for using a medicine ball?

Knowing the center of mass is important because it helps determine how the ball will move and behave when thrown or lifted. It also helps with proper form and technique when using the ball for exercises or training.

3. How is the center of mass of a medicine ball calculated?

The center of mass can be calculated by finding the geometric center of the ball, which is the point where the ball would balance on a flat surface. This can be done by measuring the diameter of the ball and dividing it by 2. The center of mass is then located at this point.

4. Does the center of mass change if the medicine ball is filled with different materials?

Yes, the center of mass will change depending on the materials inside the medicine ball. For example, if the ball is filled with sand, the center of mass will be closer to the bottom of the ball compared to a ball filled with air.

5. How does the center of mass affect the stability of a medicine ball?

The closer the center of mass is to the base of the ball, the more stable the ball will be. This is because the weight is more evenly distributed, making it less likely to tip over. However, a ball with a higher center of mass may be more challenging to use, as it requires more core strength and control to maintain stability.

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