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A particle starting from rest, moves in a circle of radius ' $r$ '. It attains a velocity of $\mathrm{v}_0 \mathrm{~m} / \mathrm{s}$ in the $\mathrm{n}^{\text {th }}$ round. Its angular acceleration will be
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Verified Answer
The correct answer is:
$\frac{v_0^2}{4 \pi n r^2} \mathrm{rad} / \mathrm{s}^2$
From third equation of motion for circular motion
$$
\omega^2-\omega_0^2=2 \alpha \theta
$$
where, $\omega=$ final angular velocity of particle
$\omega_0=$ initial angular velocity
$\alpha=$ angular acceleration and
$\theta=$ angular displacement
Here, $\omega=\frac{\mathrm{v}_0}{\mathrm{r}} \mathrm{rad} / \mathrm{s}$ (where, $\mathrm{r}$ radius of the circle)
$\omega_0=0$ (initially particle is at rest)
$\theta=2 \pi n$ (for $n$ rounds)
Substituting these values in Eq. (i), we get
$$
\left(\frac{\mathrm{v}_0}{\mathrm{r}}\right)^2-0=2 \alpha(2 \pi \mathrm{n}) \Rightarrow \alpha=\frac{\mathrm{v}_0^2}{4 \pi \mathrm{nr}^2} \mathrm{rad} / \mathrm{s}^2
$$
$$
\omega^2-\omega_0^2=2 \alpha \theta
$$
where, $\omega=$ final angular velocity of particle
$\omega_0=$ initial angular velocity
$\alpha=$ angular acceleration and
$\theta=$ angular displacement
Here, $\omega=\frac{\mathrm{v}_0}{\mathrm{r}} \mathrm{rad} / \mathrm{s}$ (where, $\mathrm{r}$ radius of the circle)
$\omega_0=0$ (initially particle is at rest)
$\theta=2 \pi n$ (for $n$ rounds)
Substituting these values in Eq. (i), we get
$$
\left(\frac{\mathrm{v}_0}{\mathrm{r}}\right)^2-0=2 \alpha(2 \pi \mathrm{n}) \Rightarrow \alpha=\frac{\mathrm{v}_0^2}{4 \pi \mathrm{nr}^2} \mathrm{rad} / \mathrm{s}^2
$$
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