91Ó°ÊÓ

The speck of clay is rotating on the potter’s wheel. The speed of clay is dependent on the radius of the wheel and the time taken by the wheel.The acceleration of the clay varies with the velocity of the clay and the radius of the wheel.

The given data can be listed below as,

• The rotation of the wheel is, $N=45\mathrm{rpm}$.
• The diameter of the wheel is, $D=35\mathrm{cm}\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)=0.35\mathrm{m}$.

The speed of the clay can be expressed as,

$\begin{array}{c}v=\frac{2\mathrm{Ï€}r}{T}\\ v=\left(\frac{2\mathrm{Ï€}\left(\frac{D}{2}\right)}{\frac{1}{N}}\right)\end{array}$

Here, r is the radius of the wheel, T is the time taken by the wheel.

Substitute the values in the above equation.

$\begin{array}{c}v=\frac{2\mathrm{π}×\left(\frac{0.35\mathrm{m}}{2}\right)}{\frac{1}{45\mathrm{rpm}}\left(\frac{60\mathrm{s}}{1\min }\right)}\\ v=0.825\mathrm{m}/\mathrm{s}\end{array}$

The centripetal acceleration can be expressed as,

$a_{\mathrm{c}}=\frac{v^2}{r}$

Here, v is the speed of the clay, r is the radius of the potter’s wheel.

Substitute the values in the above equation.

$\begin{array}{c}{a}_{\mathrm{c}}=\frac{{\left(0.825\mathrm{m}/\mathrm{s}\right)}^2}{\left(\frac{0.35\mathrm{m}}{2}\right)}\\ ≈3.89\mathrm{m}/{\mathrm{s}}^2\end{array}$

Thus, the magnitude of the acceleration of a speck of clayis $3.89\mathrm{m}/{\mathrm{s}}^2$.