Ian: New page: If $F=\mu R$ then the coin can remain on the surface. This is the maximum value that $F$ can take. That is $mr{{\omega }^{2}}=\mu R$ . This gives $\...$

Ian — Tue, 12 Oct 2010 14:14:03 GMT

New page: If <math>F=\mu R</math> then the coin can remain on the surface. This is the maximum value that <math>F</math> can take. That is <math>mr{{\omega }^{2}}=\mu R</math>. This gives <math>\...

New page

If <math>F=\mu R</math> then the coin can remain on the surface. This is the maximum value that <math>F</math> can take.

That is <math>mr{{\omega }^{2}}=\mu R</math>.

This gives

<math>\begin{align}
& 0\textrm{.}02\times r\times {{\pi }^{2}}=0\textrm{.}1176 \\
& r=\frac{0\textrm{.}1176}{0\textrm{.}02\times {{\pi }^{2}}}=0\textrm{.}596\text{ m or 59}\text{.6 cm} \\
\end{align}</math>

Note that for values of <math>r</math> greater than this the frictional force cannot keep the coin from sliding.

Solution 20.4b - Revision history

Ian: New page: If F=\mu R then the coin can remain on the surface. This is the maximum value that F can take. That is mr{{\omega }^{2}}=\mu R. This gives \...

Ian: New page: If $F=\mu R$ then the coin can remain on the surface. This is the maximum value that $F$ can take. That is $mr{{\omega }^{2}}=\mu R$ . This gives $\...$