A person with mass m1 = 57 kg stands at the left end of a uniform beam with mass m2 = 92 kg and a length L = 2.7 m. Another person with mass m3 = 67 kg stands on the far right end of the beam and holds a medicine ball with mass m4 = 12 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.
a. What is the location of the center of mass of the system?
b. The medicine ball is throw to the left end of the beam (and caught). What is the location of the center of mass now?
c. 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?)
d. 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?
To determine the location of the center of mass of the system, we need to calculate the weighted average position of all the masses involved.
a. To find the center of mass of the system:
Step 1: Calculate the individual weighted positions for each mass.
Position of mass m1 = 0 (since it is at the left end)
Position of mass m2 = L/2 (since it is at the midpoint of the beam)
Position of mass m3 = L (since it is at the right end)
Position of mass m4 = L (since it is at the right end)
Step 2: Calculate the total mass of the system.
Total mass (m_total) = m1 + m2 + m3 + m4
Step 3: Calculate the weighted average position.
Center of mass position (x_cm) = (m1 * 0 + m2 * (L/2) + m3 * L + m4 * L) / m_total
Substituting the given values:
x_cm = (57 * 0 + 92 * (2.7/2) + 67 * 2.7 + 12 * 2.7) / (57 + 92 + 67 + 12)
Simplifying the equation gives us the answer for part (a).
Once you have calculated the center of mass, you can proceed to part (b), (c), and (d) using the new conditions stated in the problem.