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Gnoma [55]
3 years ago
7

A rotating wheel requires 3.05-s to rotate through 37.0 revolutions. Its angular speed at the end of the 3.05-s interval is 97.9

rad/s. What is the constant angular acceleration of the wheel?
Physics
1 answer:
Setler79 [48]3 years ago
5 0

Answer:

\alpha=14.2rad/s^2

Explanation:

The formula that relates angular displacement with angular acceleration is:

\Delta \theta=\omega_i t+\frac{\alpha t^2}{2}

We can obtain \omega_i from the definition of angular acceleration:

\alpha=\frac{\Delta \omega}{\Delta t}=\frac{\omega_f-\omega_i}{t}

\omega_i=\omega_f-\alpha t

Putting all together:

\Delta \theta=(\omega_f-\alpha t) t+\frac{\alpha t^2}{2}=\omega_f t-\frac{\alpha t^2}{2}

Which, since we want the angular acceleration, is:

\alpha=\frac{2(\omega_f t-\Delta \theta)}{t^2}

And for our values is:

\alpha=\frac{2((97.9rad/s)(3.05s)-(37(2\pi rad)))}{(3.05s)^2}=14.2rad/s^2

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<em>Translated from Spanish to English for other English speaking students to benefit.</em>

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