Question

A wheel that is rotating at 33.3 rad/s is given an angular acceleration of 2.15 rad/s 2. Through what angle has the wheel turned when its angular speed reaches 72.0 rad/s

Answer 1

Answer:

The angle through which the wheel turned is 947.7 rad.

Explanation:

initial angular velocity, $\omega _i$ = 33.3 rad/s

angular acceleration, α = 2.15 rad/s²

final angular velocity, $\omega_f$ = 72 rad/s

angle the wheel turned, θ = ?

The angle through which the wheel turned can be calculated by applying the following kinematic equation;

$\omega_f^2 = \omega_i^2 + 2\alpha \theta\\\\\theta = \frac{\omega_f^2\ -\ \omega_i^2}{2\alpha } \\\\\theta = \frac{(72)^2\ -\ (33.3)^2}{2(2.15)}\\\\\theta = 947.7 \ rad$

Therefore, the angle through which the wheel turned is 947.7 rad.