A rolling wheel of diameter of 68 cm slows down uniformly from 8.4 m/s to rest over a distance of 115 m.What is the magnitude of its angular acceleration if there was no slipping? A)1.8 rad/s² B)0.90 rad/s² C)5.7 rad/s² D)11 rad/s²

Question

Quizplus · Accepted Answer

The angular acceleration can be found using the relation between linear acceleration and angular acceleration, $a = \alpha r$, where $a$ is the linear acceleration, $\alpha$ is the angular acceleration, and $r$ is the radius of the wheel. First, we find the linear acceleration using the equation $v^2 = u^2 + 2as$, where $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration, and $s$ is the distance. Given that the wheel comes to rest, $v = 0$, $u = 8.4$ m/s (initial velocity), and $s = 115$ m, we can solve for $a$. Rearranging the equation gives $a = -u^2 / (2s)$. Substituting the given values, $a = -(8.4)^2 / (2 \times 115)$ m/s$^2$. Then, to find the angular acceleration, $\alpha = a / r$, where $r = 68/2 = 34$ cm = 0.34 m (radius of the wheel). Substituting the values, we find $\alpha$, which matches the value given in option B, indicating that the angular acceleration is 0.90 rad/s$^2$.

A Rolling Wheel of Diameter of 68 Cm Slows Down