A person with mass m1 = 67 kg stands at the left end of a uniform beam with mass

Question

A person with mass m1 = 67 kg stands at the left end of a uniform beam with mass m2 = 96 kg and a length L = 3.1 m. Another person with mass m3 = 65 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. just amswer 3 and 4 1)What is the location of the center of mass of the system? Answer is 1.60 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? Answer is 1.60 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?

Explanation / Answer

Given,

m1 = 67 kg ; m2 = 96 kg ; L = 3.1 m ; m4 = 9 kg ;

(1)The center of mass can be calculated as follows:

Xcm = (67 x 0 + 96 x 3.1/2 + 65 x 3.1 + 9 x 3.1) / (67+95+65+9) = 378.2 /236 = 1.602 m

Xcm = 1.602 m

2)The location of com will be:

Xcm' =  (67 x 0 + 96 x 3.1/2 + 65 x 3.1 + 9 x 0) / (67+95+65+9) = 378.2 /236 = 1.483 m

Xcm' = 1.483 m

3)The center of mass of the system did not move, as no component from the system has been removed or added. The medicene ball is moved to left end where x = 0 which have us the Xcom = 1.483 m. So the distance moved by the beam to teh right will be:

D = 1.602 - 1.483 = 0.119 = 0.12 m to the right

Hence, D = 0.12 m to the right.

4)When both walked to the center,

X = 3.1/2 = 1.55 m

This time the end of the beam moved:

D' = 1.55 - 1.483 = 0.067 m

So the end of the beam is now moved to:

X = 0.12 - 0.067 = 0.053 m

Hence, X = 0.053 m

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