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 θ cos2t where θ is in radians and t is in seconds. Determine the magnitude of the acceleration of the particle when θ = 30°.

Question

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 heta =\cos { 2t } where heta is in radians and t is in seconds. Determine the magnitude of the acceleration of the particle when heta = 30°.

Accepted Answer

When [latex] heta=\frac{\pi}{6} \mathrm{rad}[/latex],

[latex]\frac{\pi}{6}=\cos 2 t \quad t=0.5099 \mathrm{~s}[/latex]

[latex]\begin{aligned}& \dot{ heta}=\frac{d heta}{d t}=-\left.2 \sin 2 t ight|_{t=0.5099 \mathrm{~s}}=-1.7039 \mathrm{~rad} / \mathrm{s} \& \ddot{ heta}=\frac{d^2 heta}{d t^2}=-\left.4 \cos 2 t ight|_{t=0.5099 \mathrm{~s}}=-2.0944 \mathrm{~rad} / \mathrm{s}^2 \& r=4 \quad \dot{r}=0 \quad \ddot{r}=0 \& a_r=\ddot{r}-r \dot{ heta}^2=0-4(-1.7039)^2=-11.6135 \mathrm{~in} . / \mathrm{s}^2 \& a_ heta=\ddot{r}+2 \dot{r} \dot{ heta}=4(-2.0944)+0=-8.3776 \mathrm{~in} . / \mathrm{s}^2 \& a=\sqrt{a_r^2+a_ heta^2}=\sqrt{(-11.6135)^2+(-8.3776)^2}=14.3 \mathrm{~in} . / \mathrm{s}^2\end{aligned}[/latex]

Question 12.155: A particle is moving along a circular path having a radius o...

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