A person with mass m 1 = 52 kg stands at the left end of a uniform beam with mas

Question

A person with mass m1 = 52 kg stands at the left end of a uniform beam with mass m2 = 98 kg and a length L = 2.7 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 = 15 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

1.

location of the center of mass is,

X = m1x + m2x2 + m3x3 + m4x4 / [m1+m2+m3m+m4]

= 52*0 + 98*(2.7/2)+ 64*2.7 + 15*2.7 / [52+98+64+15]

= 1.51, from the left end

2.

in this case, the center mass location does not change since there is no fricction and momentum is conserved.

therefore, the location of the center of mas is, X = 1.51 m

3.

in this case, the medicine ball is located at the left end,

so its moment is zero. therefore, the new location of

center of mas is,

X = m1x + m2x2 + m3x3 + m4x4 / [m1+m2+m3m+m4]

= 52*0 + 98*(2.7/2)+ 64*2.7 + 15*0 / [52+98+64+15]

= 1.33 m

4.

the x-position is,

x = 1.51 m - 1.33 m = 0.18 m

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