Awheel is turning about an axis through its center with constant angular acceleration. starting from rest, at t = 0, the wheel turns through 8.20 revolutions in 12.0 s. at t = 12.0 s the kinetic energy of the wheel is 36.0 j. for an axis through its center, what is the moment of inertia of the wheel?

I=0.489 kg.m²

Explanation:

Given that

Initial speed of wheel (ω₁)= 0 rad/s

θ = 8.2 rev in t= 12 s

We know that

1 rev = 2π rad

8.2 rev = 16.4 π rad

θ =16.4 π rad

Lets take angular acceleration of wheel is α rad/s²

We know that if angular acceleration is constant

Now by putting the values

α =0.715 rad/s²

The final speed of the wheel at 12 s

ω₂=ω₁ + α  t

ω₂ = 0 + 0.715 x 12

ω₂ =8.58 rad/s

The final kinetic energy of the wheel

I=0.489 kg.m²

This is moment of inertia of the wheel .

 L = 6944.44 m

Explanation:

For this exercise, let's look for the current consumed by each train drop

         P = I V

         I = P / V

         I = 6000/25000

         I = 0.24 A

This is the current consumed by each train down.

Let's calculate the current that carries the cables

         V = I R

         I₂ = V / R

         I₂ = 25000/15

         I₂ = 1666.67 A

As the resistance value increases linearly with the length of the cable, let's use a rule of proportions, to find the total length

         L = 1666.67 (1 / 0.24)

         L = 6944.44 m

This is the length that a downturn can travel before the current is insufficient to move it