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

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

Step-by-step 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!