Question

A wheel rotates about a fixed axis with an initial angular velocity of 13 rad/s. During a 8-s interval the angular velocity increases to 57 rad/s. Assume that the angular acceleration was constant during this time interval. How many revolutions does the wheel turn through during this time interval

Answer 1

The number of revolutions is 44.6.

Step-by-step explanation:

We can find the revolutions of the wheel with the following equation:

`\theta = \omega_(0)t + (1)/(2)\alpha t^(2)`

Where:

`\omega_(0)`: is the initial angular velocity = 13 rad/s

t: is the time = 8 s

α: is the angular acceleration

We can find the angular acceleration with the initial and final angular velocities:

`\omega_(f) = \omega_(0) + \alpha t`

Where:

`\omega_(f)`: is the final angular velocity = 57 rad/s

`\alpha = (\omega_(f) - \omega_(0))/(t) = (57 rad/s - 13 rad/s)/(8 s) = 5.5 rad/s^(2)`

Hence, the number of revolutions is:

`\theta = \omega_(0)t + (1)/(2)\alpha t^(2) = 13 rad/s*8 s + (1)/(2)*5.5 rad/s^(2)*(8 s)^(2) = 280 rad*(1 rev)/(2\pi rad) = 44.6 rev`

Therefore, the number of revolutions is 44.6.

I hope it helps you!