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Natalka [10]
3 years ago
12

Interactive LearningWare 8.1 reviews the approach that is necessary for solving problems such as this one. A motorcyclist is tra

veling 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?
Physics
1 answer:
STALIN [3.7K]3 years ago
8 0

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

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