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Juli2301 [7.4K]
3 years ago
7

A wheel rotates with a constant angular acceleration of  rad/s2. During a certain time interval its angular displacement is  r

ad. At the end of the interval its angular velocity is 2 rad/s. Its angular velocity at the beginning of the interval is:
Physics
1 answer:
Hatshy [7]3 years ago
3 0

Answer:

The angular velocity at the beginning of the interval is \pi\sqrt{2}\ rad/s.

Explanation:

Given that,

Angular acceleration \alpha=\pi\ rad/s^2

Angular displacement \theta=\pi\ rad

Angular velocity \omega =2\pi\ rad/s

We need to calculate the angular velocity at the beginning

Using formula of angular velocity

\alpha =\dfrac{\omega^2-\omega_{0}^2}{2\theta}

\omega_{0}^2=\omega^2-2\alpha\theta

Where, \alpha = angular acceleration

\omega = angular velocity

Put the value into the formula

\omega_{0}^2=(2\pi)^2-2\times\pi\times\pi

\omega=\sqrt{2\pi^2}

\omega_{0}=\pi\sqrt{2}\ rad/s

Hence, The angular velocity at the beginning of the interval is \pi\sqrt{2}\ rad/s.

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