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A particle moves along a circle $\left(\frac{20}{\pi}\right) \mathrm{m}$ with constant tangential acceleration. If the velocity of the particle is $80 \mathrm{~m} / \mathrm{s}$ at the end of the second revolution after after motion has begun, the tangential acceleration is:
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Verified Answer
The correct answer is:
$40 \mathrm{~m} / \mathrm{s}^2$
From question
$$
\begin{aligned}
& r=\frac{20}{\pi} m, v=80 \mathrm{~m} / \mathrm{s} \\
& \therefore \theta=2 \text { revolutions }=2 \times 2 \pi=4 \pi \mathrm{rad} \\
& \therefore \omega^2=\omega_0{ }^2+2 \alpha \theta \\
& \therefore \omega^2=2 \alpha \theta \ldots(\mathrm{i}) \text { since } \omega_0=0 \\
& \text { Since } \alpha=\frac{a}{r} \text { and } \omega=\frac{\mathrm{V}}{r}
\end{aligned}
$$
Putting these values in eq. (i)
$$
\begin{aligned}
& \frac{V^2}{r^2}= \frac{2 a \theta}{r} \\
& \Rightarrow a= \frac{V^2}{2 r \theta}=\frac{80 \times 80}{2 \times\left(\frac{20}{\pi}\right) \times 4 \pi} \\
&=40 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
$$
$$
\begin{aligned}
& r=\frac{20}{\pi} m, v=80 \mathrm{~m} / \mathrm{s} \\
& \therefore \theta=2 \text { revolutions }=2 \times 2 \pi=4 \pi \mathrm{rad} \\
& \therefore \omega^2=\omega_0{ }^2+2 \alpha \theta \\
& \therefore \omega^2=2 \alpha \theta \ldots(\mathrm{i}) \text { since } \omega_0=0 \\
& \text { Since } \alpha=\frac{a}{r} \text { and } \omega=\frac{\mathrm{V}}{r}
\end{aligned}
$$
Putting these values in eq. (i)
$$
\begin{aligned}
& \frac{V^2}{r^2}= \frac{2 a \theta}{r} \\
& \Rightarrow a= \frac{V^2}{2 r \theta}=\frac{80 \times 80}{2 \times\left(\frac{20}{\pi}\right) \times 4 \pi} \\
&=40 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
$$
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