A wheel accelerates from rest to $59 \mathrm { rad } / \mathrm { s }$ at a uniform rate of $58 \mathrm { rad } / \mathrm { s } ^ { 2 }$ 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

The wheel's angular acceleration is 58 rad/s^2 and its final angular velocity is 59 rad/s. We can use the following equation to find the angle through which the wheel turned while accelerating:$$	heta = \frac{1}{2} \alpha t^2$$where:* θ is the angle turned (in radians)* α is the angular acceleration (in rad/s^2)* t is the time (in seconds)We don't know the time, but we can use the fact that the wheel starts from rest to find it. The initial angular velocity is 0 rad/s, so we have:$$59 \mathrm { rad } / \mathrm { s } = 0 \mathrm { rad } / \mathrm { s } + 58 \mathrm { rad } / \mathrm { s } ^ { 2 } t$$Solving for t, we get:$$t = 1 \mathrm { s }$$Now we can substitute this value of t into the equation for θ:$$	heta = \frac{1}{2} (58 \mathrm { rad } / \mathrm { s } ^ { 2 }) (1 \mathrm { s } )^2 = 29 \mathrm { rad }$$Therefore, the wheel turned through 29 radians while accelerating.

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

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