A1 kg ball is hung at the end of a rod 1 m long. if the system balances at a point on the rod one third of the distance from the end holding the mass, what is the mass of the rod?

The mass of the rod on which balances the ball hung at one end is 2 kg.

The given parameters;

The one-third distance from the point holding the mass is calculated as follows;

p = ¹/₃ x 1 = 0.33 m

0                                 0.33 m                                                        1 m

------------------------------Δ----------------------------------------------------------

↓                                                                                                      ↓

W                                                                                                 1 kg

Take moment about the pivot;

W(0.33) = 1(1 - 0.33)

W(0.33) = 1(0.67)

Thus, the mass of the rod on which balances the ball hung at one end is 2 kg.

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2 kg

Explanation:

Assuming the rod's mass is uniformly distributed, the center of mass is at half the length.

Sum of the moments at the balance point:

-(Mg)(L/3) + (mg)(L/2 − L/3) = 0

(Mg)(L/3) = (mg)(L/2 − L/3)

(Mg)(L/3) = (mg)(L/6)

2M = m

M = 1 kg, so m = 2 kg.

The mass of the rod is 2 kg.

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.