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• Notes

$\begin{array}{l}{\text { A wheel of diameter } 40.0 \mathrm{cm} \text { starts from rest and rotates }} \\ {\text { with a constant angular acceleration of } 3.00 \mathrm{rad} / \mathrm{s}^{2} \text { . At the }} \\ {\text { instant the wheel has computed its second revolution, compute }} \\ {\text { the radial acceleration of a point on the rim in two ways: (a) }} \\ {\text { using the relationship } a_{\mathrm{rad}}=\omega^{2} r \text { and }(\mathrm{b}) \text { from the relationship }} \\ {a_{\mathrm{rad}}=v^{2} / r .}\end{array}$

$D=40 \mathrm{cm}$

$r=20 \mathrm{cm} \longrightarrow r=0.2 \mathrm{m}$

$\alpha_{z}=3 \text { rad/s }^2$

$w_{0 z}=0$

$2 rev=4 \pi \ r a d$

$w \longrightarrow a_{rad}=r \omega^2$

${\omega_{z}}^{2}={\omega_{0 z}}^{2}+2 \alpha_{z}\left(\theta-\theta_{0}\right) \rightarrow w_{z}=\sqrt{2 \alpha_{z}\left(\theta-\theta_{0}\right)}$

$=\sqrt{2 * 3(4 \pi)}= 8.68 rad/s$

$∴ a_{r a d}=r \omega^{2}=0.2* 8.68^{2}=15.1 \mathrm{m} / \mathrm{s}^{2}$

$V=r \omega=0.2 * 8.68=1.74 \mathrm{m} / \mathrm{s}$

$∴ a_{rad}=\frac{v^{2}}{r}=\frac{(1.74)^{2}}{0.2}=15.1 \mathrm{m} / \mathrm{s}^{2}$

$\begin{array}{l}{\text { An advertisement claims that a centrifuge }} \\ {\text { takes up only } 0.127 \mathrm{m} \text { of bench space but can produce a radial accel- }} \\ {\text { eration of } 3000 \mathrm{g} \text { at } 5000 \text { rev/min. Calculate the required radius of }} \\ {\text { the centrifuge. Is the claim realistic? }}\end{array}$

$a_{rad}=3000 g=3000(9.8)=29400 m / s^{2}$

$r=\frac{a_{rad}}{\omega^{2}} \rightarrow a_{rad}=r \omega^{2}$

$\omega=5000 \mathrm{rev} / \mathrm{min}\left(\frac{1 \mathrm{min}}{60 \mathrm{s}}\right)\left(\frac{2 \pi \mathrm{rad}}{1 \mathrm{r} \mathrm{ev}}\right)$

$\omega=523.6 \mathrm{rad} / \mathrm{s}$

$r=\frac{29400}{(523.6)^{2}}=0.107 \mathrm{m}$

$D=0.214 \mathrm{m}>0.127 \mathrm{m}$

$∴ \text { The claim is not realistic }$

$\begin{array}{l}{\text { According to the shop manual, when }} \\ {\text { drilling a } 12.7-\mathrm{mm} \text { -diameter hole in wood, plastic, or aluminum, }} \\ {\text { a drill should have a speed of } 1250 \text { rev/min. For a } 12.7-\mathrm{mm}-} \\ {\text { diameter drill bit turning at a constant } 1250 \mathrm{rev} / \min \text { , find (a) the }} \\ {\text { maximum linear speed of any part of the bit and (b) the maximum }} \\ {\text { radial acceleration of any part of the bit. }}\end{array}$

$\omega=1250 \mathrm{rev} / \mathrm{min} \longrightarrow 1250\left(\frac{\pi \ \mathrm{rad} / \mathrm{s}}{30 \ \mathrm{rev} / \mathrm{min}} \right)$

$r=\frac{12.7}{2} \mathrm{m}=\frac{12.7 * 10^{-3}}{2} \mathrm{m}$

$V=r \omega \longrightarrow \quad\left(\frac{12.7 * 10^{-3}}{2}\right)\left(1250 * \frac{\pi}{30}\right)=0.831 \mathrm{m/s}$

$a_\text { rad }=\frac{V^{2}}{r}=r \omega^{2}$

$a_{rad}=\frac{(0.831)^{2}}{\left(\frac{12.7 * 10^{-3}}{2}\right)}= 109 \mathrm{m} / \mathrm{s}^{2}$

$=r \omega ^2$