A particle moves in a circular path of radius | vedclass

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Question

$\alpha=\frac{\mathrm{d} \omega}{\mathrm{dt}}=-\mathrm{b}$

$a_{t}=R \alpha=-R b$

At $t=\frac{2 a}{b}, \omega=-a$

$\Rightarrow a_{r}=R \omega^

Vedclass · Accepted Answer

$R\sqrt {a^4 + b^2}$

Vedclass · Answer

$\alpha=\frac{\mathrm{d} \omega}{\mathrm{dt}}=-\mathrm{b}$

$a_{t}=R \alpha=-R b$

At $t=\frac{2 a}{b}, \omega=-a$

$\Rightarrow a_{r}=R \omega^{2}=R a^{2}$

$\mathrm{Now}$       $a=\sqrt{a_{t}^{2}+a_{r}^{2}}=R \sqrt{a^{4}+b^{2}}$

A particle moves in a circular path of radius $R$ with an angular velocity $\omega = a -bt$ where $a$ and $b$ are positive constants and $t$ is time. The magnitude of the acceleration of the particle after time $\frac {2a}{b}$ is

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