Question

A person with mass m₁ = 68 kg stands at the left end of a uniform beam with mass m₂ = 109 kg and a length L = 2.5 m. Another person with mass m₃ = 50 kg stands on the far right end of the beam and holds a medicine ball with mass m₄ = 8 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?

Answer 1

(1) Location of the center of mass of the system is given as -

Xcm = ΣMoments/ΣMasses = [109*(2.5/2) + (50+8)*2.5]/ [68+109+50+8]

= 281.25 / 235 =  1.197 m

(2) New location of the center of mass of the system -

Xcm' = [109*1.25 + 50*2.5] / 235

= 261.25 / 235 = 1.112 m from the left end of the beam.

(3) Since the CM moved (1.197 - 1.112) = 0.085 m left (relative to the beam), the beam must move the same amount to the right, as no space movement of the CM is allowed with Vi = 0.

Therefore, all points on the beam including the left end move 0.085 m to the right.

(4) Again we have -

Xcm'' = [109*1.25 + (68+8)*1.25 + 50*1.25] / 235

= [136.25 + 95 + 62.5] / 235 = 1.25 m from the left end.

The CM moved from 1.197 to 1.25 = 0.053 m right, so the beam moved from its original position 0.053 m left.

So, the people meet at x = 1.25 - 0.053 = 1.197 m

This is expected, because the beam is uniform and all the disposable mass is located at its middle. Therefore, the original location of the CM is maintained.