Solution 20.4a
From Mechanics
(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{a...) |
Current revision
Converting the angular speed to standard units.
\displaystyle \omega =30\text{ rpm }=\frac{30\times 2\pi }{60}=\pi \text{ rad}{{\text{s}}^{\text{-1}}}
For circular motion:
\displaystyle \begin{align} & F=mr{{\omega }^{2}} \\ & =0\textrm{.}02\times 0\textrm{.}25\times {{\pi }^{2}} \\ & =0\textrm{.}0493\text{ N} \end{align}
The force \displaystyle R on the coin normal to the surface is
\displaystyle R=9\textrm{.}8\times 0\textrm{.}02=0\textrm{.}196\text{ N}
Using the friction inequality for the coin to remain at rest on the surface:
\displaystyle \begin{align} & F\le \mu R \\ & F\le 0\textrm{.}6\times 0\textrm{.}196 \\ & F\le 0\textrm{.}1176\text{ N} \\ \end{align}
As 0.0493 is less than 0.1176 the coin remains in the same position on the turntable.
