The centripetal acceleration of a particle in uniform circular motion is $18 \mathrm{~ms}^{-2}$. If the…

The centripetal acceleration of a particle in uniform circular motion is $18 \mathrm{~ms}^{-2}$. If the radius of the circular path is 50 cm , the change in velocity of the particle in a time of $\frac{\pi}{18} \mathrm{~s}$ is
  1. $9 \mathrm{~ms}^{-1}$
  2. $2 \mathrm{~ms}^{-1}$
  3. $3 \mathrm{~ms}^{-1}$
  4. $6 \mathrm{~ms}^{-1}$

Solution

In circular motion, $\begin{aligned} & a_c=\frac{v^2}{r} \\ & \therefore \quad v=\sqrt{a_c \cdot r}=\sqrt{18 \times 50 \times 10^{-2}}=3 \mathrm{~m} / \mathrm{s} \end{aligned}$
Also, $w=\frac{\mathrm{v}}{\mathrm{r}}=\frac{3}{50 \times 10^{-2}}=6 \mathrm{rad} / \mathrm{s}$ $\begin{aligned} & \therefore \theta=\omega t=6 \times \frac{\pi}{18}=\frac{\pi}{3} \\ & \therefore \quad \Delta v=2 v \sin \frac{\theta}{2}=2 \times 3 \sin \frac{\pi}{6}=2 \times 3 \times \frac{1}{2}=3 \mathrm{~m} / \mathrm{s} \end{aligned}$

Asked in: AP EAMCET 2024 (21 May Shift 1)

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