3) What is the new x-position of the person at the Left end of the beam? (How fa

Question

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? A person with mass m1 = 69 kg stands at the Left end of a uniform beam with mass m2 = 105 kg and a Length L = 2.7 m. Another person with mass m3 = 62 kg stands on the far right end of the beam and holds a medicine ball with mass m4 = 11 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 angina position of the beam as shown in the drawing. Assume there is no friction between the beam and floor. 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?

Explanation / Answer

1)COM=(99*1.4+(65+8)*2.8)/(99+65+8+53)=1.524m

2)Same. Since no external force acts on the system.

3)Let the new x position on the left be x.

COM=(x*(53+8)+65*(2.8+x)+99*(1.4+x))/(99+65+8+53)=1.524

x=0.099m

4)Let the x position of the left of the beam be x now.

COM=(99*(1.4+x)+(65+53+8)*(1.4+x))/(99+65+8+53)=1.524

x=0.124

x centre=1.4+0.124=1.524m

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.