Question

Interactive LearningWare 8.1 reviews the approach that is necessary for solving problems such as this one. A motorcyclist is traveling along a road and accelerates for 4.36 s to pass another cyclist. The angular acceleration of each wheel is 6.10 rad/s2, and, just after passing, the angular velocity of each is 75.2 rad/s, where the plus signs indicate counterclockwise directions. What is the angular displacement of each wheel during this time?

Answer 1

Answer:

The angular displacement of each wheel is 269.92 rad

Explanation:

Given:

Angular acceleration $\alpha = 6.10 \frac{rad}{s^{2} }$

Time to pass cyclist $t = 4.36$ s

Angular velocity $\omega _{f} = 75.2 \frac{rad}{s}$

According to the equation of kinematics,

  $\omega _{f} = \omega _{i} + \alpha t$

   $\omega _{i} = \omega _{f} - \alpha t$

   $\omega _{i} = 75.2 - 6.10 \times 4.36$

  $\omega _{f} = 48.60$ $\frac{rad}{s}$

For finding angular displacement,

    $\omega _{f} ^{2} - \omega _{i} ^{2} = 2 \alpha \theta$

Where $\theta =$ angular displacement,

  $\theta = \frac{\omega _{f}^{2} - \omega _{i} ^{2} }{2\alpha }$

  $\theta = \frac{5655.04 - 2361.96 }{2\times 6.10 }$

  $\theta = 269.92$ rad

Therefore, the angular displacement of each wheel is 269.92 rad