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A particle is moving in a circle of radius $5 \mathrm{~cm}$ with uniform speed and completes the circle in $5 \mathrm{~s}$. What is the magnitude of linear acceleration?
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Verified Answer
The correct answer is:
$0.8 \pi^2 \mathrm{~cm} / \mathrm{s}^2$
Radius of the circle, $r=5 \mathrm{~cm}$
Time period, $T=5 \mathrm{~s}$
$\therefore \quad T=\frac{2 \pi}{\omega}$ (where, $\omega$ is angular velocity.)
$$
\Rightarrow \quad \omega=\frac{2 \pi}{5} \mathrm{rad} / \mathrm{s}
$$
$\therefore$ Linear speed, $v=\omega r=\frac{2 \pi}{5} \times 5=2 \pi \mathrm{cm} / \mathrm{s}$
As, the particle is in uniform circular motion, the centripetal force is acting on it.
$\therefore$ Centripetal acceleration is $v^2 / r$
$\therefore$ Linear acceleration is also given as
$$
a=\frac{v^2}{r}=\frac{4 \pi^2}{5}=0.8 \pi^2 \mathrm{~cm} / \mathrm{s}^2
$$
Time period, $T=5 \mathrm{~s}$
$\therefore \quad T=\frac{2 \pi}{\omega}$ (where, $\omega$ is angular velocity.)
$$
\Rightarrow \quad \omega=\frac{2 \pi}{5} \mathrm{rad} / \mathrm{s}
$$
$\therefore$ Linear speed, $v=\omega r=\frac{2 \pi}{5} \times 5=2 \pi \mathrm{cm} / \mathrm{s}$
As, the particle is in uniform circular motion, the centripetal force is acting on it.
$\therefore$ Centripetal acceleration is $v^2 / r$
$\therefore$ Linear acceleration is also given as
$$
a=\frac{v^2}{r}=\frac{4 \pi^2}{5}=0.8 \pi^2 \mathrm{~cm} / \mathrm{s}^2
$$
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