A CD with a diameter of $12.0 \mathrm{~cm}$ starts from rest and with a constant angular acceleration of $1.0 \mathrm{rad} / \mathrm{sec}^{2}$ acquires an angular velocity of $5.0 \mathrm{rad} / \mathrm{sec}$. The CD continues rotating at $5.0 \mathrm{rad} / \mathrm{sec}$ for 15.0 seconds and then slows to a stop in 12.0 second with a constant angular acceleration. What is the linear speed of a point $4.0 \mathrm{~cm}$ from the center at the time 2.0 seconds from the start?
A) $0.18 \mathrm{~m} / \mathrm{s}$
B) $0.060 \mathrm{~m} / \mathrm{s}$
C) $0.080 \mathrm{~m} / \mathrm{s}$
D) $0.10 \mathrm{~m} / \mathrm{s}$
E) $0.14 \mathrm{~m} / \mathrm{s}$

Question

Quizplus · Accepted Answer

The linear speed $v$ of a point on a rotating object is given by $v = r\omega$, where $r$ is the radius from the center to the point, and $\omega$ is the angular velocity. At 2.0 seconds from the start, the angular velocity can be found using $\omega = \omega_0 + \alpha t$, where $\omega_0$ is the initial angular velocity (0 rad/s, since it starts from rest), $\alpha$ is the angular acceleration (1.0 rad/s$^2$), and $t$ is the time (2.0 s). Thus, $\omega = 0 + 1.0 \times 2.0 = 2.0$ rad/s. The radius $r$ is 4.0 cm or 0.04 m. Therefore, $v = 0.04 m \times 2.0 rad/s = 0.080 m/s$.

A CD with a Diameter Of 12.0 cm12.0 \mathrm{~cm}12.0 cm Starts from Rest and with a Constant Angular Acceleration Of Rest

A CD with a Diameter Of $12.0 \mathrm{~cm}$ Starts from Rest and with a Constant Angular Acceleration Of Rest