Solution 20.4b
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(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>\...)
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If \displaystyle F=\mu R then the coin can remain on the surface. This is the maximum value that \displaystyle F can take.
That is \displaystyle mr{{\omega }^{2}}=\mu R.
This gives
\displaystyle \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}
Note that for values of \displaystyle r greater than this the frictional force cannot keep the coin from sliding.
