A wheel at rest gains an angular momentum of 10.7kgm²/s in 8s. if the moment if inertia of the wheel is 2.2kgm², how many revolutions will be made before the wheel attains an angular velocity of 30rad/s

Revolutions made before attaining angular velocity of 30 rad/s:

θ = 3.92 revolutions

Explanation:

Given that:

L(final) = 10.7 kgm²/s

L(initial) = 0

time = 8s

Torque is the rate of change of angular momentum:

We know that

T = Iα

α = T/I

where I = moment of inertia = 2.2kgm²

α = 1.34/2.2

α = 0.61 rad/s²

We know that angular equation of motion is:

ω²(final) = ω²(initial) +2αθ

(30 rad/s)² = 0 + 2(0.61 rad/s²)θ

θ = (30 rad/s)²/ 2(0.61 rad/s²)

θ = 24.6 radians

Convert it into revolutions:

θ = 24.6/ 2π

θ = 3.92 revolutions

answer: the correct answer id option(d).

explanation:

snell's law refraction:

the angle of refraction =

the angle of incidence =

refractive index of air = = 1

refractive index of glycerol = 1.47

from the given options, the most closest answer to 63.72° is 65° of option (d).

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