A wheel accelerates from rest to $59 \mathrm { rad } / \mathrm { s }$ at a uniform rate of 58 rad/s²_. Through what angle (in radians)did the wheel turn while accelerating? A)30 rad B)24 rad C)60 rad D)38 rad

Question

Quizplus · Accepted Answer

Explanation:The angle turned by the wheel is given by:$$\theta = \frac{1}{2} \alpha t^2$$where $\alpha$ is the angular acceleration and $t$ is the time.We are given that $\alpha = 58 \mathrm { rad } / \mathrm { s }^2$ and the wheel accelerates from rest, so $v_i = 0$. We are also given that the final angular velocity is $\omega_f = 59 \mathrm { rad } / \mathrm { s }$. We can use the following equation to find the time:$$\omega_f = \omega_i + \alpha t$$Substituting in the given values, we get:$$59 \mathrm { rad } / \mathrm { s } = 0 + 58 \mathrm { rad } / \mathrm { s }^2 t$$Solving for $t$, we get:$$t = 1 \mathrm { s }$$Now we can substitute this value of $t$ into the equation for $\theta$:$$\theta = \frac{1}{2} (58 \mathrm { rad } / \mathrm { s }^2) (1 \mathrm { s })^2 = 29 \mathrm { rad }$$Therefore, the wheel turned through an angle of 29 radians while accelerating.

A Wheel Accelerates from Rest To 59rad/s59 \mathrm { rad } / \mathrm { s }59rad/s

A Wheel Accelerates from Rest To $59 \mathrm { rad } / \mathrm { s }$