Answer:
The magnitude of the radial acceleration is 0.754 rad/s²
Explanation:
Given;
radius of the flywheel, r = 0.2 m
initial angular velocity of the flywheel, [tex]\omega_i = 0[/tex]
angular acceleration of the flywheel, a = 0.900 rad/s².
angular distance, θ = 120⁰
the angular distance in radian = [tex]\frac{120}{180} \pi = \frac{2 \pi}{3} \ rad[/tex]
Apply the following kinematic equation to determine the final angular velocity;
[tex]\omega _f^2 = \omega _i ^2 + 2\alpha \theta\\\\\omega _f^2 = 0 + 2(0.9)(\frac{2\pi}{3} )\\\\\omega _f^2 = 3.7699\\\\\omega _f = \sqrt{ 3.7699} \\\\ \omega _f = 1.942 \ rad/s[/tex]
The magnitude of the radial acceleration is calculated as;
[tex]\alpha _r = \omega ^2r\\\\\alpha _r = (1.942)^2 (0.2)\\\\\alpha _r =0.754 \ rad/s^2[/tex]
Therefore, the magnitude of the radial acceleration is 0.754 rad/s²
