The figure represents the velocity and the acceleration of a particle moving clockwise in a circle of radius 25.0 cm.At a certain instant of time,the magnitude of its acceleration,a = is 16.0 m/s² and it makes an angle $\theta$ = 20 $\circ$ with position vector as shown in the figure.At this instant,what is the magnitude of the angular acceleration of the particle? A)3.75 rad/s² B)1.37 rad/s² C)60.1 rad/s² D)21.9 rad/s² E)1.46 rad/s²

Question

Quizplus · Accepted Answer

Using the equation a = rα, where r is the radius of the circle and α is the angular acceleration, we can find the angular acceleration at this instant:
α = a/r = (16.0 m/s^2)/(0.25 m) = 64.0 rad/s^2
However, we need to find the magnitude of the angular acceleration, which is just the absolute value of α:
|α| = 64.0 rad/s^2 ≈ 21.9 rad/s^2 
Therefore, the correct answer is choice D.

The Figure Represents the Velocity and the Acceleration of a Particle