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A particle at rest starts moving with constant angular acceleration $4 \mathrm{rad} / \mathrm{s}^2$ in circular path. At what time the magnitudes of its tangential acceleration and centrifugal acceleration will be equal?
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Verified Answer
The correct answer is:
$0.5 \mathrm{~s}$
In rotational motion,
$\begin{aligned}
& \omega=\omega_0+\alpha \mathrm{t} \\
& \omega=\alpha \mathrm{t}
\end{aligned}$
$\text { ( } \because \omega_0=0 \text {; particle at rest.) }$
$\therefore \quad$ Centrifugal acceleration $\mathrm{a}=\omega^2 \mathrm{r}$
$\therefore \quad \mathrm{a}=\alpha^2 \mathrm{t}^2 \mathrm{r}$
Tangential acceleration $\mathrm{a}_{\mathrm{t}}=\alpha \times \mathrm{r}$
Given: $\mathrm{a}=\mathrm{a}_{\mathrm{t}}$
$\Rightarrow \alpha^2 \mathrm{t}^2 \mathrm{r}=\alpha \mathrm{r}$
$\mathrm{t}^2=\frac{1}{\alpha}=\frac{1}{4}$
$\therefore \quad \mathrm{t}=\frac{1}{2}=0.5 \mathrm{~s}$
$\begin{aligned}
& \omega=\omega_0+\alpha \mathrm{t} \\
& \omega=\alpha \mathrm{t}
\end{aligned}$
$\text { ( } \because \omega_0=0 \text {; particle at rest.) }$
$\therefore \quad$ Centrifugal acceleration $\mathrm{a}=\omega^2 \mathrm{r}$
$\therefore \quad \mathrm{a}=\alpha^2 \mathrm{t}^2 \mathrm{r}$
Tangential acceleration $\mathrm{a}_{\mathrm{t}}=\alpha \times \mathrm{r}$
Given: $\mathrm{a}=\mathrm{a}_{\mathrm{t}}$
$\Rightarrow \alpha^2 \mathrm{t}^2 \mathrm{r}=\alpha \mathrm{r}$
$\mathrm{t}^2=\frac{1}{\alpha}=\frac{1}{4}$
$\therefore \quad \mathrm{t}=\frac{1}{2}=0.5 \mathrm{~s}$
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