Answer
The piece travels a horizontal distance of 7.68 meters.
Work Step by Step
We can find the angular speed $\omega$ when the piece breaks loose.
$\omega^2 = 2\theta \alpha$
$\omega = \sqrt{2\theta \alpha}$
$\omega = \sqrt{(2)(155~rev)(2.00~rev/s^2)}$
$\omega = 24.9~rev/s$
We can find the horizontal speed when the piece breaks off.
$v = (24.9~rev/s)(2\pi)(0.120~m)$
$v = 18.77~m/s$
We can find the time it takes the piece to drop a vertical distance of 0.820 meters.
$y = \frac{1}{2}at^2$
$t = \sqrt{\frac{2y}{g}} = \sqrt{\frac{(2)(0.820~m)}{9.80~m/s^2}}$
$t = 0.409~s$
We can find the horizontal distance $x$.
$x = v~t = (18.77~m/s)(0.409~s)$
$x = 7.68~m$
The piece travels a horizontal distance of 7.68 meters.
