Question

A person with mass m₁ = 58 kg stands at the left end of a uniform beam with mass m₂ = 101 kg and a length L = 2.6 m. Another person with mass m₃ = 64 kg stands on the far right end of the beam and holds a medicine ball with mass m₄ = 9 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.

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

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?)

Answer 1

center of mass = cm

cm = (m1x1 + m2x2 + m3x3 + ... ) / (m1+m2+m3+..)

here m1 = 58 kg

x1 = 0 m ( because person standing on origin )

m2 = 101 kg  

x2 = 1.3 m

m3 = 64 kg

x3 = 2.6 m ( person stand at far right end )

m4 = 9kg

x4 = 2.6 m

cm = (58*0 + 101*1.3 + 64 * 2.6 + 9 *2.6) / ( 58 + 101 + 64 + 9) = 1.38 m

when throw medicine ball there is no external force so cm will be unchanged .  

cm = 1.38 m

part b )

when ball throw from left end to right end ( left end person catch the ball ) and cm remain unchanged , assume beam move x to the right

1.38 = [58*x + 9*x + 101*(1.3+x) + 64 *(2.6+x)] / ( 58 + 9 + 101 + 64 )

1.38 *232 = 58x + 9x + 101x + 131.3 + 64x+ 166.4

320.16 = 232x + 297.7

x = 0.097 m move