Reference no: EM13903069

1. Consider two objects whose position vectors are given by r1^→ and r2^→ and whose momenta are given by p1^→ = m1v1^→ and p2^→ = m2v2^→ respectively. Show that in the special case that the center of mass of the two bodies is at rest at the origin (that is P^→ = p1^→ + p2^→ = 0, and the position of the center of mass is R^→ = 0), the sum of the angular momenta of the two objects about the center of mass equals the angular momentum of a single object of mass μ = m1m2/(m1+m2), rotating in circular motion about the origin at a distance r^→ = r2^→-r1^→. The quantity μ is called the reduced mass.

           Figure 1.1

2. A Pulley system is used to lift a heavy mass. How much force must be applied to lift the object in figure 1.2 at a steady speed? Neglect friction at the axle.

           Figure 1.2

3. A man of mass 95kg is dancing with his wife, who has a mass of 68kg. Assume that each person's mass is concentrated at their respective centers of mass, which are separated by 48cm.

a. What is the gravitational attraction between them?

b. Which person has the greater gravitational attraction toward the other?