Agrinding wheel is a uniform cylinder with a radius of 7.80 cm and a mass of 0.550 kg.

part a
calculate its moment of inertia about its center. express your answer to three significant figures and include the appropriate units.
part b
calculate the applied torque needed to accelerate it from rest to 1750 rpm in 7.40 s .

take into account a frictional torque that has been measured to slow down the wheel from 1500 rpm to rest in 58.0 s .

a. I = 167.31 x 10 ⁻³ kg*m²

b. T = 4.59 kg * m² / s²

Explanation:

The moment of inertia of a uniform cylinder:

a.

r = 7.8 cm * 1 m / 100 cm = 0.078 m

I = ½ * m * r²  

I = ½ * 0.55 kg * (0.078²m)

I = 167.31 x 10 ⁻³ kg*m²

b.

T = Iα’ + Iα,    

α’ = ω’/t = 1750 rpm * (2π/60) / 7.40s  =  24.76 rad/s²

α = ω/t = 1500 rpm * (2π/60) / 58  = 2.71 rad/s²

T = (167.31 x 10⁻³ kg*m²)* (24.76 + 2.71 ) rad / s²  

T = 4.59 kg * m² / s²

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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.