Question

In the Cavendish balance apparatus shown in Figure 13.4 in the textbook, suppose that m1 = 1.10 kg , m2 = 25.0 kg , and the rod connecting each of the pairs of masses is 31.0 cm long. Once the system reaches equilibrium, each pair of masses, m1 and m2, are separated by a distance of 12.0 cm 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

Top view of Cavendish appratus is shown below. Rod connecting masses m1 is free to rotate about vertical axis passing through O
     

A) Net force on one of the small mass m1, is gravitational force from nearby mass m2.
This force Fg = Gm1 m2 / r² Here G is gravitational constant 6.67x10^-11 , r is distance between masses.
hence Fg = 6.67x10^-11 x 1.1 x25 / 0.12²  = 1.27x10^-7 N
B) As same force acts on both m1 masses in oppsite direction, net force on rotating system is zero, but there is net troque on this system given by
Fg xL where L is length of rod connecting two m1 masses.
Hence Net Torque on rotating system is 1.27x10^-7 x 0.31 = 3.9x10^-8 Nm

C) In equilibrium Torque due to trosional spring = torque due to gravitational force
K = Gm1 m2 L / r²
increasing sensitivity means increasing value of . To do so we can decrease value of k or r, or increase m1 , m2 or L, that is length of rod connecting masses