 Michael Maggard

2022-01-02

A particle is moving along a circular path having a radius of 4 in. such that its position as a function of time is given by is in radians and t is in seconds. Determine the magnitude of the acceleration of the particle when $u={30}^{\circ }$. Annie Levasseur

Given:
r=4 in, $\theta =\mathrm{cos}\left(2t\right)$,
Acceleration in this case is equal to
$\stackrel{\to }{a}={a}_{r}{\stackrel{\to }{u}}_{r}+{a}_{\theta }{\stackrel{\to }{u}}_{\theta }$
${a}_{r}=\stackrel{¨}{r}-r\stackrel{˙}{{\theta }^{2}}$
$={a}_{\theta }=r\stackrel{¨}{\theta }+2\stackrel{˙}{r}\stackrel{˙}{\theta }$
Since the radius in circular motion is constant: $\stackrel{˙}{r}=\stackrel{¨}{r}=0$, and time-derivatives of position are equal to: $\stackrel{˙}{\theta }=-2\mathrm{sin}\left(2t\right),\stackrel{¨}{\theta }=-4\mathrm{cos}\left(2t\right)$
If we want to find out in what time the particle will have $\theta =30=\frac{\pi }{6}$, we use following formula:
$\theta =\mathrm{cos}\left(2t\right)$
$\frac{\pi }{6}=\mathrm{cos}\left(2t\right)$
$2t=1.0197$
$t=0.51s$
Acceleration is finally equal to:
$\stackrel{\to }{a}={a}_{r}{\stackrel{\to }{u}}_{r}+{a}_{\theta }{\stackrel{\to }{u}}_{\theta }$
$=-r{\stackrel{˙}{\theta }}^{2}{\stackrel{\to }{u}}_{r}+r\stackrel{¨}{\theta }{\stackrel{\to }{u}}_{\theta }$
$=-4\cdot {\left(-2\mathrm{sin}\left(2\cdot 0.51\right)\right)}^{2}{\stackrel{\to }{u}}_{r}+4\cdot -4\mathrm{cos}\left(2\cdot 0.51\right){\stackrel{\to }{u}}_{\theta }$
$=-11.62{\stackrel{\to }{u}}_{r}-8.37{\stackrel{\to }{u}}_{\theta }$
Magnitude of the acceleration is:
$a=\sqrt{{a}_{r}^{2}+{a}_{\theta }^{2}}$
$=\sqrt{{\left(11.62\right)}^{2}+{\left(8.37\right)}^{2}}$
$a=14.32\frac{in}{{s}^{2}}$ Jordan Mitchell

Step 1
Given data:
The radius of the circular path, $r=4$ in.
Position of a particle, $\theta =\mathrm{cos}2t$
Let,
${a}_{r}=$ radial acceleration
${u}_{r}=$ unit vector along radial direction
${a}_{\theta }=$ acceleration along $\theta$ direction
${u}_{\theta }=$ unit vector along $\theta$ direction
Step 2
Now calculating the time (t) when $\theta =30°$
$\theta =\mathrm{cos}2t$
$\frac{\pi }{6}=\mathrm{cos}2t$
$2t=1.0197$
$t=0.51s$
Waiting cclerton in veto form
$\stackrel{\to }{a}={a}_{r}{\stackrel{―}{u}}_{r}+{a}_{\theta }{\stackrel{―}{u}}_{\theta }$
$=-r{\stackrel{˙}{\theta }}^{2}{\stackrel{―}{u}}_{r}+r\stackrel{¨}{\theta }{\stackrel{―}{u}}_{\theta }$
$=-4\cdot {\left[-2\mathrm{sin}\left(2\cdot 0.51\right)\right]}^{2}{\stackrel{―}{u}}_{r}+4\left[-4\mathrm{cos}\left(2\cdot 0.51\right)\right]{\stackrel{―}{u}}_{\theta }$
$=-11.62{\stackrel{―}{u}}_{r}-8.37{\stackrel{―}{u}}_{\theta }$
Step 3
Finally the magnitude of the acceleration of the particle (a) when $\theta =30°$ is calculated as:

$a=\sqrt{{a}_{r}^{2}+{a}_{\theta }^{2}}$
$=\sqrt{{11.62}^{2}+{8.37}^{2}}$
$=14.32\frac{in}{{s}^{2}}$
Hence the required time (t) when $\theta ={30}^{\circ }$ is 0.51 seconds and the magnitude of the acceleration of the particle when $\theta =30°$ is $14.32\frac{in}{{s}^{2}}$. Vasquez

When

$\begin{array}{}\theta =\frac{\pi }{6}rad,\frac{\pi }{6}=2\mathrm{cos}2tt=0.5099s\\ \stackrel{˙}{\theta }\frac{d\theta }{dt}=-2\mathrm{sin}2t\mid t=0.5099s=-1.7039\frac{rad}{s}\\ \stackrel{¨}{\theta }\frac{{d}^{2}\theta }{d{t}^{2}}=-4\mathrm{cos}2t\mid t=0.5099s=-2.0944\frac{rad}{{s}^{2}}\\ r=4,\stackrel{˙}{r}=0,\stackrel{¨}{r}=0\\ {a}_{r}=\stackrel{¨}{r}-r{\stackrel{˙}{\theta }}^{2}=0-4\left(-1.7039{\right)}^{2}=-11.6135\frac{in.}{{s}^{2}}\\ {a}_{\theta }=r\stackrel{¨}{\theta }+2\stackrel{˙}{r}\stackrel{˙}{\theta }=4\left(-2.0944\right)+0=-8.3776\frac{in.}{{s}^{2}}\\ a=\sqrt{{a}_{r}^{2}+{a}_{\theta }^{2}}=\sqrt{\left(-11.6135{\right)}^{2}+\left(-8.3776{\right)}^{2}}=14.3\frac{in.}{{s}^{2}}\end{array}$

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