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3-2.Motion in Plane
medium
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
A$\frac {a}{R}$
B$a^2R$
C$R(a^2 + b)$
D$R\sqrt {a^4 + b^2}$
Solution
$\alpha=\frac{\mathrm{d} \omega}{\mathrm{dt}}=-\mathrm{b}$
$a_{t}=R \alpha=-R b$
At $t=\frac{2 a}{b}, \omega=-a$
$\Rightarrow a_{r}=R \omega^{2}=R a^{2}$
$\mathrm{Now}$ $a=\sqrt{a_{t}^{2}+a_{r}^{2}}=R \sqrt{a^{4}+b^{2}}$
$a_{t}=R \alpha=-R b$
At $t=\frac{2 a}{b}, \omega=-a$
$\Rightarrow a_{r}=R \omega^{2}=R a^{2}$
$\mathrm{Now}$ $a=\sqrt{a_{t}^{2}+a_{r}^{2}}=R \sqrt{a^{4}+b^{2}}$
Standard 11
Physics
