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Setler [38]
3 years ago
6

A bicycle wheel with a radius of 0.42 m accelerates uniformly for 6.8 s from an initial angular velocity of 5.5 rad/s to a final

angular velocity of 6.7 rad/s. What was its angular acceleration
Physics
1 answer:
Art [367]3 years ago
8 0

Answer:

The angular acceleration required  is 0.1765 rad/ s^2

Explanation:

The radius of the bicycle wheel has a radius of 0.42 m.

The acceleration is for time, t =  6.8 seconds.

Initial angular velocity is given as  \omega_{0}  = 5.5 rad/s

Final angular velocity is given as \omega_{f} = 6.7 rad/s

Therefore from the formula for angular speed we get

\omega_{f} = \omega_{0} + (\frac{d\omega}{dt} \times t),   where t is the time in seconds.

Therefore we get

6.7 =  5.5 + (6.8 × \frac{d\omega}{dt} )

Therefore we get the angular acceleration, \frac{d\omega}{dt} = \frac{(6.7 - 5.5 }{6.8}  = 0.1765 rad/ s^2

The angular acceleration required  is 0.1765 rad/ s^2

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