Ian: New page: Converting the angular speed to standard units. $\omega =30\text{ rpm }=\frac{30\times 2\pi }{60}=\pi \text{ rad}{{\text{s}}^{\text{-1}}}$ For circular motion: $\begin{a...$

Ian — Tue, 12 Oct 2010 14:02:34 GMT

New page: Converting the angular speed to standard units. <math>\omega =30\text{ rpm }=\frac{30\times 2\pi }{60}=\pi \text{ rad}{{\text{s}}^{\text{-1}}}</math> For circular motion: <math>\begin{a...

New page

Converting the angular speed to standard units.

<math>\omega =30\text{ rpm }=\frac{30\times 2\pi }{60}=\pi \text{ rad}{{\text{s}}^{\text{-1}}}</math>

For circular motion:

<math>\begin{align}
& F=mr{{\omega }^{2}} \\
& =0\textrm{.}02\times 0\textrm{.}25\times {{\pi }^{2}} \\
& =0\textrm{.}0493\text{ N}
\end{align}</math>

The force <math>R</math> on the coin normal to the surface is

<math>R=9\textrm{.}8\times 0\textrm{.}02=0\textrm{.}196\text{ N}</math>

Using the friction inequality for the coin to remain at rest on the surface:

<math>\begin{align}
& F\le \mu R \\
& F\le 0\textrm{.}6\times 0\textrm{.}196 \\
& F\le 0\textrm{.}1176\text{ N} \\
\end{align}</math>

As 0.0493 is less than 0.1176 the coin remains in the same position on the turntable.

Solution 20.4a - Revision history

Ian: New page: Converting the angular speed to standard units. \omega =30\text{ rpm }=\frac{30\times 2\pi }{60}=\pi \text{ rad}{{\text{s}}^{\text{-1}}} For circular motion: \begin{a...

Ian: New page: Converting the angular speed to standard units. $\omega =30\text{ rpm }=\frac{30\times 2\pi }{60}=\pi \text{ rad}{{\text{s}}^{\text{-1}}}$ For circular motion: $\begin{a...$