A person with mass m1-59 kg stands at the left end of a uniform beam with mass m

Question

A person with mass m1-59 kg stands at the left end of a uniform beam with mass m2 94 kg and a length L- 3 m. Another person with mass m3 69 kg stands on the far right end of the beam and holds a medicine ball with mass m4- 13 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? 1.65 m Submit 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? 1.65 m Submit 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 Submit

Explanation / Answer

We know that the center of mass does not change, so the system must act according to reorient itself. This means that the position of the beam is what changes.

Let's say the beam moves "d" to the right as a result of the ball throwing.

we know center of mass equation is given by

Xcm = [(m1)(x1) + (m2)(x2) + (m3)(x3) + ... + (mn)(xn)] / (m1 + m2 + m3 + ... + mn)

now Xcm = 1.65m, x1 = x2 =d, x3= 1.5+d, x4=3+d

        1.65 = [(59)*d + 94*(1.5+d) + 69*(3+d) +(13)*(0+d)] / 235

The reason why the position of the medicine ball is now 0 in the after equation is because it was thrown to the left, and caught by the person on the left who was previously at the origin
Solving this

        387.75 = 348 + 235*d

        d = 0.16914 m

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