Question

Do not hurt yourself; don't strike your hand against anything. Within these limitations, describe what you do to give your hand a large acceleration. Compute an order-of-magnitude estimate of this acceleration, stating the quantities you measure or estimate and their values.

Short answer

The order of magnitude${10}^2{\text{m/s}}^{\text{2}}$ .

Step-by-step solution

Definition of Acceleration

The rate at which an object's velocity changes with respect to time is called acceleration. Accelerations are measured in terms of vectors. The orientation of the net force applied to an item determines the orientation of its acceleration.

Step 2: Compute an order-of-magnitude estimate of this acceleration, stating the quantities you measure or estimate and their values.

By rotating our hand in a circular manner, it may be accelerated. Now start by calculating the radius of the circular motion, which will equal the length of the hand. The circumference of the circle and the number of rotations per second are then calculated.

Using the centripetal acceleration formula,

$$\begin{gathered} a_c=\frac{v^2}{r} \\ =\frac{{\left(2\pi \text{r/t}\right)}^2}{r} \\ =\frac{4{\pi }^{2}r}{t^2} \end{gathered}$$

The order of magnitude of acceleration can be estimated.

Ifthelength of the hand is$\text{r}=0.6\text{m}$ ,

centripetal acceleration,

$$\begin{gathered} {\text{a}}_{\text{e}}={\left(\frac{\left(2\right)(3.14)(0.6\text{m})}{(1\text{s})}\right)}^2\left(\frac{1}{0.6}\right) \\ \approx 0.39\times {10}^2{\text{m/s}}^{\text{2}} \end{gathered}$$

.

The order of magnitude is ${10}^2{\text{m/s}}^{\text{2}}$.