Question

A wheel rotates through an angle of 15.6 rad as it slows down uniformly from 22.0 rad/s to 13.5 rad/s. What is the magnitude of the angular acceleration of the wheel

Answer 1

Answer:

α = -9.67 rad/s²

Explanation:

the magnitude of the angular acceleration of the wheel can be calculated using the expression below;

ω² = ω₀² + 2α(θ- θ₀)..............eqn(1)

ω² = ω₀² + 2αΔθ...............eqn(2)

Δθ= θ- θ₀= 15.6 rad

ω = 13.5 rad/s

ω₀ = 22.0 rad/s

where ω₀= initial angular velocity

ω= final angular velocity

We can now input all the parameters into the equation(2) above

(13.5)² = (22.0)² + 2αΔθ

2αΔθ= (13.5)² - (22.0)²

2αΔθ= 182.25 - 484

2αΔθ=301.75

But our Δθ= 15.6 rad

2α *15.6 rad = 301.75

2α=301.75/15.6

2α=19.343

α=9.67

the magnitude of the angular acceleration of the wheel is 9.67 rad/s²