A wheel 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 seconds. At t=12, the kinetic energy of the wheel is 36.0 J. For an axis through its center, what is the moment of inertia for the wheel?

Answer:

Initial angular velocity of the wheel = ₁ = 0 rad/s (Starts from rest)

Angular velocity of the wheel after 12 sec = ₂

Number of revolutions made by the wheel in 12 sec = n = 8.2

Time period = T = 12 sec

Angle turned by the wheel in 12 sec = 

Angular acceleration of the wheel = 

 = 2n

 = 2(8.2)

 = 51.522 rad

 = ₁T + T²/2

51.522 = (0)(12) + (12)²/2

 = 0.716 rad/s²

₂ = ₁ + T

₂ = 0 + (0.716)(12)

₂ = 8.592 rad/s

Kinetic energy of the wheel at t=12 sec = E = 36 J

Moment of inertia of the wheel = I

E = I₂²/2

36 = I(8.592)²/2

I = 0.975 kg.m²

Moment of inertia for the wheel = 0.975 kg.m²