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3-2.Motion in Plane
hard
Three point particles $P, Q, R$ move in circle of radius $‘r’$ with different but constant speeds. They start moving at $t = 0$ from their initial positions as shown in the figure. The angular velocities (in rad/ sec) of $P, Q$ and $R$ are $5\pi , 2\pi$ & $3\pi$ respectively, in the same sense. The time interval after which they are at same angular position.
A$2/3\, sec$
B$1/6\, sec$
C$1/2\, sec$
D$3/2\, sec$
Solution
relative angular speed of $\mathrm{P}$ and $\mathrm{Q}$ particles, $5 \pi-2 \pi=3 \pi$
They will meet at
$t_{1}=\frac{2 \pi+\pi / 2}{3 \pi}=\frac{2.5}{3} s$
$t_{2}=\frac{4 \pi+\pi / 2}{3 \pi}=\frac{4.5}{3}=\frac{3}{2} s$
$t_{3}=\frac{6 \pi+\pi / 2}{3 \pi}=\frac{6.5}{3} s$
Relative angular speed of $\mathrm{P}$ and $\mathrm{R}$ particles, $5 \pi-3 \pi=2 \pi$
They will meet at
$t_{4}=\frac{2 \pi+\pi}{2 \pi}=\frac{3}{2} s$
$t_{5}=\frac{4 \pi+\pi}{2 \pi}=\frac{5}{2} s$
$t_{6}=\frac{6 \pi+\pi}{2 \pi}=\frac{7}{2} s$
All particle will meet at time, $T=\frac{3}{2} s$
They will meet at
$t_{1}=\frac{2 \pi+\pi / 2}{3 \pi}=\frac{2.5}{3} s$
$t_{2}=\frac{4 \pi+\pi / 2}{3 \pi}=\frac{4.5}{3}=\frac{3}{2} s$
$t_{3}=\frac{6 \pi+\pi / 2}{3 \pi}=\frac{6.5}{3} s$
Relative angular speed of $\mathrm{P}$ and $\mathrm{R}$ particles, $5 \pi-3 \pi=2 \pi$
They will meet at
$t_{4}=\frac{2 \pi+\pi}{2 \pi}=\frac{3}{2} s$
$t_{5}=\frac{4 \pi+\pi}{2 \pi}=\frac{5}{2} s$
$t_{6}=\frac{6 \pi+\pi}{2 \pi}=\frac{7}{2} s$
All particle will meet at time, $T=\frac{3}{2} s$
Standard 11
Physics
