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 10.0 seconds from the start? A) 27.6 rad B) 37.5 rad C) 40.7 rad D) 48.2 rad E) 51.2 rad

Question

Quizplus · Accepted Answer

To find the angular distance traveled by a point on the CD, we first need to calculate the time it takes for the CD to accelerate to 5.00 rad/sec. Using the formula for angular acceleration, $\alpha = \frac{\Delta \omega}{\Delta t}$, where $\Delta \omega = 5.00 \, \text{rad/sec}$ and $\alpha = 1.00 \, \text{rad/sec}^2$, we find that $\Delta t = \frac{\Delta \omega}{\alpha} = \frac{5.00}{1.00} = 5.00 \, \text{seconds}$.Since the question asks for the angular distance at 10.0 seconds from the start, this falls within the period when the CD is rotating at a constant angular velocity of 5.00 rad/sec (after the initial acceleration and before it begins to decelerate). The CD accelerates for the first 5.0 seconds and then rotates at a constant speed for the next 5.0 seconds (up to the 10-second mark).The angular distance traveled during the acceleration phase can be found using the formula $\theta = \frac{1}{2} \alpha t^2$, which gives $\theta = \frac{1}{2} \times 1.00 \times (5.00)^2 = 12.5 \, \text{rad}$.During the next 5.0 seconds of constant velocity, the angular distance is $\theta = \omega t = 5.00 \times 5.00 = 25.0 \, \text{rad}$.Adding these two distances gives the total angular distance traveled by a point on the CD as $12.5 \, \text{rad} + 25.0 \, \text{rad} = 37.5 \, \text{rad}$.

A CD with a Diameter of 12