Question

A 4.50-kg wheel that is 34.5 cm in diameter rotates through an angle of 13.8 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

Use the formula

`{\omega_f}^2-{\omega_i}^2=2\alpha\Delta\theta`

`\implies\left(13.5(\rm rad)/(\rm s)\right)^2-\left(22.0(\rm rad)/(\rm s)\right)^2=2\alpha(13.8\,\mathrm{rad})`

`\implies\alpha=-10.9(\rm rad)/(\mathrm s^2)`

so the magnitude is 10.9 rad/s^2.

Answer 2

-10.9 rad/s²

Step-by-step explanation:

ω² = ω₀² + 2α(θ - θ₀)

Given:

ω = 13.5 rad/s

ω₀ = 22.0 rad/s

θ - θ₀ = 13.8 rad

(13.5)² = (22.0)² + 2α (13.8)

α = -10.9 rad/s²