A CD with a diameter of 12.0 cm starts from rest and with a constant angular acceleration of 1.00 rad/sec² acquires an angular velocity of 5.00 rad/sec. The CD continues rotating at 5.00 rad/sec for 15.0 seconds and then slows to a stop in 12.0 second with a constant angular acceleration. What is the angular distance traveled by a point 4.00 cm from the center, at the time 25.0 seconds from the start? A) 107 rad B) 193 rad C) 205 rad D) 237 rad E) 274 rad

Question

Quizplus · Accepted Answer

Explanation:The angular distance traveled by a point 4.00 cm from the center, at the time 25.0 seconds from the start is:$$\theta = \omega_i t + \frac{1}{2} \alpha t^2$$where $\omega_i$ is the initial angular velocity, $\alpha$ is the angular acceleration, and $t$ is the time.In the first 15.0 seconds, the CD rotates with a constant angular velocity of 5.00 rad/sec. So, the angular distance traveled during this time is:$$\theta_1 = \omega_i t = (5.00 \text{ rad/sec})(15.0 \text{ s}) = 75.0 \text{ rad}$$In the next 12.0 seconds, the CD slows to a stop with a constant angular acceleration. So, the angular distance traveled during this time is:$$\theta_2 = \frac{1}{2} \alpha t^2 = \frac{1}{2}(-1.00 \text{ rad/sec}^2)(12.0 \text{ s})^2 = -72.0 \text{ rad}$$The total angular distance traveled by a point 4.00 cm from the center, at the time 25.0 seconds from the start is:$$\theta = \theta_1 + \theta_2 = 75.0 \text{ rad} - 72.0 \text{ rad} = 3.00 \text{ rad}$$So, the correct answer is A.

A CD with a Diameter of 12