Ch. 9

A stalactite in a cave has drops of water falling from it to the cave floor below. The drops are equally spaced in time and come in rapid succession, so that at any given moment there are many drops in midair. Is the center of mass of the midair drops higher than, lower than, or equal to the halfway distance between the tip of the stalactite and the cave floor?

Higher than Towards the end the drops will get more spaced out because of acceleration due to gravity, so there will be less drops towards the bottom compared to the top. Since each drop is equal mass, there is more mass towards the top and this means that the center of mass will be at least higher than half way.

For the system of three particles shown, which have masses M, 2M, and 3M as indicated, where is the center of mass located?

between the particle of mass 2M and the particle of mass

Two particles of masses m1 and m2 (m1<m2) are located 10 meters apart. Where is the center of mass of the system located?

more than 5 meters but less than 10 meters from the particle of mass m1 This is a torque problem, similar to two children on a teeter-totter. If the teeter-totter is adjusted so both children can have their feet off the ground and board does not move, the fulcrum is at the center of mass. Mass 1 is on the left end of the 10 m mass 1 * distance from fulcrum = mass 2 * its distance from the fulcrum Eq. #1.. m1 * d1 = m2 * d2 Since they are sitting 10 m apart d1 + d2 = 10 Eq. #2.. d2 = 10 - d1 Substitute into Eq. #1 m1 * d1 = m2 * (10 - d1) divide both side by m1 d1 = m2/m1 * (10 - d1) distribute right side d1 = 10 * m2/m1 - (m2/m1* d1) add (m2/m1* d1) to both sides d1 + (m2/m1* d1) = 10 * m2/m1 factor d1 out of left side d1 (1 + m2/m1) = 10 * m2/m1 divide both sides by (1 + m2) d1 = 10 * m2/m1 ÷ (1 + m2/m1)