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${}^{\mathbf{76}\mathbf{}\mathbf{cm}}$and for rotation less than${}^{\mathbf{1}\mathbf{}\mathbf{rev}\mathbf{}}$, what are the (a) smallest and (b) largest angular speeds that cause the toast to hit and then topple to be butter-side down?

Short answer

1. The smallest angular speed of the toast to land butter side down is${}^{4.0\mathrm{rad}/\mathrm{s}}$.

2. The largest angular speed of the toast to land butter side down${}^{11.9\mathrm{rad}/\mathrm{s}}$.

Step-by-step solution

Given data

The vertical distance to the floor, $\mathrm{y}=76\mathrm{cm}$

The rotation of the toast,${}^{\mathrm{\theta }<1\mathrm{rev}}$

Understanding the angular velocity

Angular velocity represents the rate at which an object rotates about an axis. Angular velocity is a vector quantity, with direction given by right hand rule.

The expression for angular velocity is given as:

$\mathrm{\omega }=\frac{∆\mathrm{\theta }}{∆\mathrm{t}}$ … (i)

Here, $∆\theta$is the angular displacement and is the time interval.

The expression for the second equation of motion is given as:

$\mathrm{y}={\mathrm{v}}_0∆\mathrm{t}+\frac{1}{2}\mathrm{g}∆{\mathrm{t}}^2$

Here, $v_0$is the initial linear velocity and $g$is the acceleration due to gravity.

Determination of the time taken by a toast to fall to the floor

The initial velocity of the toast at maximum height is zero.${\mathrm{v}}_0=0\mathrm{m}/\mathrm{s}$

Using equation (ii), the time taken to fall to floor is calculated as:

$$\begin{gathered} \mathrm{y}=\left(0\right)∆\mathrm{t}+\frac{1}{2}\mathrm{g}∆{\mathrm{t}}^2 \\ ∆\mathrm{t}=\sqrt{\frac{2\mathrm{y}}{\mathrm{g}}} \\ =\sqrt{\frac{2\times 0.76\mathrm{m}}{9.8\mathrm{m}/{\mathrm{s}}^2}} \\ =0.40\mathrm{s} \\ \mathrm{Thus},\mathrm{the}\mathrm{time}\mathrm{it}\mathrm{takes}\mathrm{to}\mathrm{fall}\mathrm{on}\mathrm{the}\mathrm{florr}\mathrm{is}0.40\mathrm{s}. \end{gathered}$$

(a) Determination of smallest angular speed

The minimum angle through which the toast can be turned to land butter side down is,

$$\begin{gathered} ∆{\mathrm{\theta }}_{\min }=\frac{\mathrm{\pi }}{2}\mathrm{rad} \\ \mathrm{Using}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{the}\mathrm{minimum}\mathrm{angular}\mathrm{speed}\mathrm{is}\mathrm{calculated}\mathrm{as}: \\ \mathrm{\omega }=\frac{∆{\mathrm{\theta }}_{\min }}{∆\mathrm{t}} \\ =\frac{{\displaystyle \frac{\mathrm{\pi }}{2}}\mathrm{rad}}{0.40\mathrm{s}} \\ \approx 4.0\mathrm{rad}/\mathrm{s} \end{gathered}$$

Thus, the smallest angular speed of the toast to land butter side down is$4.0\mathrm{rad}/\mathrm{s}$

(b) Determination of the largest angular speed

The maximum angle through which the toast can be turned to be butter side down is,

$∆{\mathrm{\theta }}_{\max }=\frac{3\mathrm{\pi }}{2}\mathrm{rad}$

Using equation (i), the maximum angular speed is calculated as:

$$\begin{gathered} {\mathrm{\omega }}_{\max }=\frac{∆{\mathrm{\theta }}_{\max }}{∆\mathrm{t}} \\ =\frac{{\displaystyle \frac{3\mathrm{\pi }}{2}}\mathrm{rad}}{0.40\mathrm{s}} \\ \approx 11.9\mathrm{rad}/\mathrm{s} \end{gathered}$$

Thus, the largest angular speed of the toast to land butter side down is $11.9\mathrm{rad}/\mathrm{s}$