Question

A wheel rotates with a constant angular acceleration of 3.50 rad/s2. If the angular speed of the wheel is 2.00 rad/s at t = 0,

(a) through what angle does the wheel rotate between t = 0 and t = 2.00 s? Give your answer in radians and in revolutions.
(b) What is the angular speed of the wheel at t = 2.00 s?
(c) What angular displacement (in revolutions) results while the angular speed found in part (b) doubles?

Answer 1

a) θ = 11 rad, θ = 1.75 rev., b) w = 9 rad / s, c) θ = 7.17 rev

Step-by-step explanation:

This is a rotation kinematics exercise

θ = θ₀ + w₀ t + ½ α t²

They indicate the initial angular velocity w₀ = 2.00 rad / s, the angular acceleration α = 3.50 rad / s² and that at the initial instant θ₀ = 0

a) let's find the rotated angle

θ = 0 + 2.00 2.00 +1/2 3.5 2²

θ = 11 rad

let's reduce 2π rad = 1 rev

θ = 11 rad (1 rev / 2π rad)

θ = 1.75 rev.

b) angular velocity

w = w₀ + α t

w = 2.00 + 3.50 2

w = 9 rad / s

c) the angular displacement to reach this speed

w² = w₀² + 2 α θ

in this case they indicate that w = 2 9 = 18 rad / s

θ =
`(w^2 - w_o^2)/(2 \alpha )`

θ =
`(18^2 - 2^2 )/(2 \ 3.5 )`

θ = 45.7 rad

let's reduce to rev

θ = 45.7 rad (1rev / 2π rad)

θ = 7.17 rev