Solution 20.1
From Mechanics
(Difference between revisions)
(New page: Firstly, the angular speed must be converted to standard units. <math>\omega =400\text{ rpm }=\text{ }\frac{400\times 2\pi }{\text{60}}=\frac{40\pi }{3}\text{ rad}{{\text{s}}^{\text{-1}}}...) |
(New page: Firstly, the angular speed must be converted to standard units. <math>\omega =400\text{ rpm }=\text{ }\frac{400\times 2\pi }{\text{60}}=\frac{40\pi }{3}\text{ rad}{{\text{s}}^{\text{-1}}}...) |
Current revision
Firstly, the angular speed must be converted to standard units.
\displaystyle \omega =400\text{ rpm }=\text{ }\frac{400\times 2\pi }{\text{60}}=\frac{40\pi }{3}\text{ rad}{{\text{s}}^{\text{-1}}}
\displaystyle \begin{align} & T=mr{{\omega }^{2}} \\ & =2\times 0\textrm{.}5\times {{\left( \frac{40\pi }{3} \right)}^{2}} \\ & =1755\text{ N} \end{align}