A wheel starts from rest and has an angular acceleration that is given by $\alpha$ (t) = (6.0 rad/s⁴)t^2. After it has turned through 10 rev its angular velocity is: A) 63 rad/s B) 75 rad/s C) 89 rad/s D) 130 rad/s E) 210 rad/s

Question

Quizplus · Accepted Answer

To solve this problem, we need to use the equations of angular motion:

ω = ω0 + αt (1)

θ = θ0 + ω0t + 1/2 αt^2 (2)

ω^2 = ω0^2 + 2α(θ - θ0) (3)

where ω is the final angular velocity, ω0 is the initial angular velocity (which is zero in this case), α is the angular acceleration, θ is the final angle (which is 10 revolutions or 20π radians), and θ0 is the initial angle (which is also zero).

We are given α = 6.0 rad/s^2, and we can find t from the equation F1F1F10(t) = (6.0 rad/s^2)t, which simplifies to t = 2 seconds.

Using equation (1), we can find the final angular velocity:

ω = ω0 + αt = 0 + (6.0 rad/s^2)(2 s) = 12 rad/s

However, this is only the final angular velocity after 2 seconds. We need to use equation (3) to find the final angular velocity after 10 revolutions:

ω^2 = ω0^2 + 2α(θ - θ0)

ω^2 = 0 + 2(6.0 rad/s^2)(20π rad)

ω^2 = 240π rad/s^2

ω = √(240π) ≈ 77.25 rad/s

The closest answer choice to this value is B) 75 rad/s.

A Wheel Starts from Rest and Has an Angular Acceleration α\alphaα

A Wheel Starts from Rest and Has an Angular Acceleration $\alpha$