A person with mass m1 60 kg stands at the left end of a uni

A person with mass m_1 = 60 kg stands at the left end of a uniform beam with mass m_2 = 103 kg and a length L = 2.6 m. Another person with mass 1TI3 = 67 kg stands on the far right end of the beam and holds a medicine ball with mass 1114 = 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?)

Solution

Mass m1= 60 kg

Mass m2 = 103 kg

Mass m3 = 67 kg

Mass of ball m4 = 8 kg

Length L = 2.6 m

Position of m1 is x1 = 0

Position of center of beam x2 = L/2

                                            = 1.3 m

Position of mass m3 is x3 = L = 2.6 m

(a). Location fo center of mass of the system X =(m1x1+m2x2+m3x3 +m4x3)/(m1+m2+m3+m4)

Since the ball is also at the end of the beam.

Substitute values you get , X=[(60x0)+(103x1.3)+(67x2.6)+(8x2.6)]/(60+103+67+8)

                                          = 328.9 / 238

                                          = 1.381 m

(2).Location fo center of mass of the system when ball is throw X \' =(m1x1+m2x2+m3x3+m4x1 )/(m1+m2+m3+m4)

Substitute values you get , X \' =[(60x0)+(103x1.3)+(67x2.6)+(8x0)]/(60+103+67+8)

                                          = 308.1 / 238

                                          = 1.2945 m

(3).

Required distance S = X - X \'

                              = 1.381 m - 1.2945 m

                              = 0.0865 m