Question

When a slice of buttered toast is accidentally pushed over the edge of a counter, it rotates as it falls. If the distance to the floor is 62 cm and for rotation less than 1 rev, what are the (a) smallest and (b) largest angular speeds that cause the toast to hit and then topple to be butter-side down? Assume free-fall acceleration to be equal to 9.81 m/s2.

Answer 1

Answer:

The smallest angular velocity is 4.42 rad/s and the largest angular velocity is 13.15 rad/s for the toast to hit and then topple to be butter-side down

Explanation:

62 cm = 0.62 m

Knowing that the vertical distance traveled is s = 0.62 m and gravitational acceleration is g = 9.81 m/s2, we can calculate the time of falling

$s = gt^2/2$

$0.62 = 9.81t^2/2$

$t^2 = 2*0.62/9.81 = 0.13$

$t = \sqrt{0.13} = 0.36 s$

If there's only 0.36s to rotate, and for the toast to hit and then topple to be butter-side down, then the angle of rotation must be between 90 and 270 degrees (Suppose that it starts with butter up and it rotates less than 1 revolution, or 360 degrees):

In other words (and units):

$\frac{\pi}{2} \leq \theta \leq \frac{3\pi}{2}$ rad

As $\omega = \theta / t$:

$\frac{\pi}{2} / t \leq \theta / t \leq \frac{3\pi}{2} / t$ rad/s

$\frac{\pi}{2*0.36} \leq \omega \leq \frac{3\pi}{2*0.36}$ rad/s

$4.42 \leq \omega \leq 13.25$ rad/s

So the smallest angular velocity is 4.42 rad/s and the largest angular velocity is 13.15 rad/s for the toast to hit and then topple to be butter-side down