A grindstone of radius 4.0 m is initially spinning with an angular speed of 8.0 rad/s. The angular speed is then increased to 12 rad/s over the next 4.0 seconds. Assume that the angular acceleration is constant. Through how many revolutions does the grindstone turn during the 4.0-second interval?

6.36

Explanation:

Given

Radius of grindstone, r = 4 m

Initial angular speed of grindstone, w(i) = 8 rad/s

Final angular speed of the grindstone, w(f) = 12 rad/s

Time used, t = 4 s

Angular acceleration of the grinder,

α = Δw / t

α = w(f) - w(i) / t

α = (12 - 8) / 4

α = 4/4 = 1 rad/s²

Number of complete revolution in 4s =

Δθ = w(i).t + 1/2.α.t²

Δθ = 8 * 4 + 1/2 * 1 * 4²

Δθ = 32 + 1/2 * 16

Δθ = 32 + 8

Δθ = 40 rad/s

40 rad/s = 40/2π rpm = 6.36 rpm

Therefore, the grindstone does 6.36 revolutions during the 4 s interval

c. newton's third law of motion