A person with mass m_1 = 65 kg stands at the left end of a uniform beam with mas

Question

A person with mass m_1 = 65 kg stands at the left end of a uniform beam with mass m_2 = 97 kg and a length L = 2.6 m. Another person with mass m_3 = 69 kg stands on the far right end of the beam and holds a medicine ball with mass m_4 = 12 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? m 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? m 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?) m 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? m

Explanation / Answer

If you take the left end to be zero, then to get the center of mass you take the total moment (m*x) for all objects from here, and divide by the total mass.

The objects are person 1 mass = 65 x =0 so mx= 0

beam mass = 97 ,x = 1.3m ( its center is only half way along), mx= 126.1

person 2 mass = 69, x= 2.6, mx= 179.4

and ball mass = 12, x= 2.6, mx=31.2

Add up the total moments (336.7) and divide by the total mass (243 Kg)

1)Center of mass is 336.7/243 = 1.386 m from the left end.

2) As there is no friction the center of mass can't move no matter what. Momentum is always conserved. It remains at 1.386 m from the original position of the left end.

3) However the left end of the beam DOES move.

If we continue to measure from the left end only then the new moment from it has changed.

The medicine ball is now at the left so it has a moment of zero (the total moment has been reduced by 31.2 and is now 305.5

So the center of mass is 305.5/243 = 1.257 m from the left end.

The only way this is possible is if the beam has moved ( 1.386 - 1.257) = 0.129m to the right.

Remember the center of mass hasn't shifted at all it is still 1.386 m from where it was. It is only the left end that has moved.

4)Whether you start from the original positions or the new ones, and whether you choose to measure the moments from the floor or from the beam the result must be the same.

You can pick your origin to make the problem easy for you.

I would go back to the original information, and put the three masses together at a distance of "s1"

from the original left end.

[(65+12+69)s1 + (97x1.3)]/(65+97+69+12) = 1.386

S1 = 1.443m

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