Question

In the Cavendish balance apparatus shown in Figure 13.4 in the textbook, suppose that m1 = 1.00kg , m2 = 28.0kg , and the rod connecting each of the pairs of masses is 32.0cm long. Once the system reaches equilibrium, each pair of masses, m1 and m2, are separated by a distance of 11.0cm center-to-center.

Part A

Find the magnitude of the net force on one of the small masses, m1 .

Part B

Find the gravitational torque (about the rotation axis) on the rotating part of the apparatus. Ignore any forces on the connecting rods in your calculation.

Part C

Suggest some ways to improve the sensitivity of this experiment.

Answer 1

(a) F = G M m / r² = .

.

         =   6.673 x 10^-11 * 28* 1 /0.11²   =   1.544 x 10^-7N

.

(b) total torque = rF + rF     (one from each side)

.

            = 2 r F   = 2 * (L/2) F  =   L F = 0.32 * 1.544 x10^-7 = 4.94 x10^-8 N-m