Question

A wheel rotates with a constant angular acceleration of  rad/s2. During a certain time interval its angular displacement is  rad. At the end of the interval its angular velocity is 2 rad/s. Its angular velocity at the beginning of the interval is:

Answer 1

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 $\pi\sqrt{2}\ rad/s$.

Answer 2

The angular velocity at the beginning of the interval is  $w=\pi\sqrt{2} \frac{rad}{sec}$

What will be the angular velocity of the wheel?

It is given that

Angular acceleration $\alpha =\pi \ \frac{rad}{sec^2}$

Angular displacement  $\theta=\pi \ rad$

Angular velocity $w=2\pi \frac{rad}{sec}$

Now for finding angular velocity at the beginning

$\alpha =\dfrac{w^2-w_0^2}{2\theta}$

$w_0^2=w^2-2\alpha \theta$

$w_0^2=(2\pi^2)-2\times\pi \times\pi$

$w_0^2=\sqrt{2\pi^2}$

$w_0=\pi\sqrt{2}\frac{rad}{sec}$

Thus the angular velocity at the beginning of the interval is  $w=\pi\sqrt{2} \frac{rad}{sec}$

To know more about angular velocity follow

https://brainly.com/question/1452612