In the Cavendish balance apparatus shown in the figure below, suppose that m 1 =

Question

In the Cavendish balance apparatus shown in the figure below, suppose that m1 = 1.04 kg, m2 = 22.7 kg, and the rod connecting the pairs is 28.5 cm long.

(a) If, in each pair, m1 and m2 are 12.7 cm apart, find the net force.
N

(b) If, in each pair, m1 and m2 are 12.7 cm apart, find the net torque (about the rotation axis) on the rotating part of the apparatus.
N · m

(c) Does it seem that the torque in part (b) would be enough to easily rotate the rod?

Suggest some ways to improve the sensitivity of this experiment.

2 The deflection of the laser beam indicates how far the fiber has twisted. Once the instrument is 1 Gravitation pulls the small masses toward the large calibrated, this result gives a value for G masses, causing the vertical quartz fiber to twist. Mirror Laser beam The small balls reach a new equilibrium position Laser when the elastic force exerted by the twisted quartz fiber balances the gravitational force between the masses. Quartz fiber Scale Large mass (m2) Small mass (m

Explanation / Answer

a)

m1 = 1.04kg

m2 = 22.7kg

r = distance between the pair of masses = 12.7 cm = 0.127 m

gravitational force between the pair of masses is given as

Fg = Gm1 m2 /r2

Fg = (6.67 * 10^-11) (1.04) (22.7) / (0.127)^2

Fg = 9.76 * 10-8 N

b)

L = length of rod = 28.5 cm = 0.285 m

Net Torque = Fg x L = 9.76*10^-8*0.285 = 2.78x 10-8 Nm

c)

yes it would be enough

--------------

No torque is very less

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