Question
Particle A is moving with constant angular velocity of $3 \pi \mathrm{rad} \mathrm{s}^{-1}$ along the horizontal circle in counter clockwise direction. Particle B starts from rest at $t=0$ and moves along the same circle but clockwise and with angular acceleration of $\pi / 4 \mathrm{rad} \mathrm{s}^{-2}$. At $\mathrm{t}=0$, they are at the ends of a diameter as shown. Find the time at which they meet. (Take $\sqrt{38}=6 \frac{1}{6}$ ).
Step 1
For any time $t$, the angular displacement of particle A, $\theta_A$, is given by $\theta_A = \omega t$, where $\omega = 3\pi \, \mathrm{rad/s}$ is the constant angular velocity of particle A. So, we have $\theta_A = 3\pi t$. Show more…
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