Question

A wheel rotating with a constant angular acceleration turns through 19 revolutions during a 3 s time interval. Its angular velocity at the end of this interval is 18 rad/s. What is the angular acceleration of the wheel? Note that the initial angular velocity is not zero. Answer in units of rad/s 2 .

Answer 1

Answer:

The magnitude of the angular acceleration of the wheel is 14.53 rad/s².

Explanation:

The angular acceleration can be found by using the following equation:

$\omega_{f}^{2} = \omega_{0}^{2} + 2\alpha \Delta \theta$  (1)

Where:

$\omega_{f}$: is the final angular velocity = 18 rad/s

$\omega_{0}$: is the initial angular velocity

α: is the angular acceleration =?

Δθ = 19 rev*(2π/1 rev) = 119.4 rad

The initial angular velocity can be found knowing that the wheel turns through 19 revolutions during a 3 s time interval:

$\omega_{f} = \omega_{0} + \alpha t$

Where:  

t: is the time = 3 s

By solving the above equation for ω₀ we have:

$\omega_{0} = \omega_{f} - \alpha t$   (2)

Now, by entering equation (2) into (1) we have:

$\omega_{f}^{2} = (\omega_{f} - \alpha t)^{2} + 2\alpha \Delta \theta$                        

$\omega_{f}^{2} = \omega_{f}^{2} - 2\omega_{f} \alpha t + (\alpha t)^{2} + 2\alpha \Delta \theta$                              

$(9\alpha)^{2} + 130.8 \alpha = 0$  

By solving the above equation for "α" we have:

α = -14.53    

The minus sign means that the wheel is decelerating.

Hence, the angular acceleration of the wheel is -14.53 rad/s².                                                                                            

I hope it helps you!