The "Laboratory" Frame Part III

Once again we shall set up our inertial frame ABC, and this time we'll examine the behaviour of an unequal pair of masses in it (Fig 4).

Fig 4.  An unequal pair of masses

Centrifugal Force and Kinetic Energy data

The unequal mass-pair of Fig 4 is joined together by a rod (black) of negligible mass which pivots and rotates about the center of rotation as shown. For the values given we have:—

10kg mass:   Centrifugal Force = mv²/r  = 10 × 1² / 0.1 = 100 N
                          Kinetic Energy   = ½mv²  = ½ × 10 × 1² = 5 J

1kg mass:    Centrifugal Force = mv²/r   = 1 × 10² / 1    = 100 N
                         Kinetic Energy   = ½mv²  = ½ × 1 × 10² = 50 J

So we can have equal, balanced centrifugal forces, even though the masses have very different energies. By causing the mass-pair to travel along the inertial line ABC, we can achieve different start and finish velocities in the laboratory frame for both masses. Will that deliver any net energy?

Fig 5.  Mass-pair travelling along the (straight) inertial line ABC

Analysis

We shall assume that the mass-pair is "weightless", e.g. it has a constant-force spring to Earth (not shown) acting at the center of rotation, and neutralising its weight in the inertial frame. Since the spring is always active from A to C, and undergoes no net change in length, there can be no net energy change in it.

We start the mass-pair at point A, rotating as before, such that its center of gravity (at the center of rotation) is following the straight-line (purple) trajectory ABC in the inertial frame. For simplicity, we shall start it travelling downwards at Vy = -1m/s in the laboratory frame. Then, in that frame, we have:—

10kg mass:   Kinetic Energy = ½mv² = ½ × 10 × (1 + 1)² = 20 J

1kg mass:      Kinetic Energy = ½mv² = ½ × 1 × (10 - 1)²  = 40.5 J

When the mass-pair reaches point C, where its center of gravity is travelling upwards at Vy = 1m/s in the laboratory frame, we have:—

10kg mass:   Kinetic Energy = ½mv² = ½ × 10 × (1 - 1)²  = 0 J

1kg mass:      Kinetic Energy = ½mv² = ½ × 1 × (10 + 1)² = 60.5 J

So, although the masses each have quite different start and finish energies, there is no net energy loss or gain. More generally, in the inertial frame ABC, all we did was to take the rotating mass pair from a given energy environment at A, into another environment at C that had the same energy as A.