KINEMATICS ChYU 3

Solution to Problem 3

Problem Analysis: Particle P (speed $u$) chases Particle Q (speed $v$, circular path radius $R$). P always stays on the line connecting the center $O$ and Q.

Kinematics of P:
Since P, O, and Q are collinear, P must rotate with the same angular velocity as Q. $$ \omega = \frac{v}{R} $$ Let $r$ be the distance of P from the center $O$. The velocity of P has two orthogonal components:

1. Radial velocity ($\frac{dr}{dt}$): Moving outward to catch Q.

2. Tangential velocity ($v_{\tau} = r\omega$): Due to rotation.

The total speed of P is given as constant $u$. $$ u^2 = \left( \frac{dr}{dt} \right)^2 + (r\omega)^2 $$ Substituting $\omega = v/R$: $$ u^2 = \left( \frac{dr}{dt} \right)^2 + \left( \frac{rv}{R} \right)^2 $$

Solving for Time:
Isolate $\frac{dr}{dt}$: $$ \frac{dr}{dt} = \sqrt{u^2 – \frac{v^2 r^2}{R^2}} $$ $$ dt = \frac{dr}{\sqrt{u^2 – \left(\frac{vr}{R}\right)^2}} $$ To find the catch time, we integrate from $r=0$ (center) to $r=R$ (rim).

Let $k = v/R$. The integral becomes: $$ T = \int_0^R \frac{dr}{\sqrt{u^2 – k^2 r^2}} $$ Using the standard integral $\int \frac{dx}{\sqrt{a^2 – x^2}} = \sin^{-1}(x/a)$: To match the form, factor out $k$: $$ \sqrt{u^2 – k^2 r^2} = k \sqrt{\frac{u^2}{k^2} – r^2} $$ $$ T = \frac{1}{k} \left[ \sin^{-1}\left( \frac{r}{u/k} \right) \right]_0^R $$ Substitute $k = v/R$: $$ T = \frac{R}{v} \sin^{-1}\left( \frac{r v}{u R} \right) \Bigg|_0^R $$ $$ T = \frac{R}{v} \sin^{-1}\left( \frac{R v}{u R} \right) = \frac{R}{v} \sin^{-1}\left( \frac{v}{u} \right) $$

Part (b) Calculation:
Given: $v = 4$ m/s, $u = 8$ m/s, $R = 84$ m. $$ T = \frac{84}{4} \sin^{-1}\left( \frac{4}{8} \right) $$ $$ T = 21 \sin^{-1}(0.5) $$ We know $\sin^{-1}(0.5) = 30^\circ = \frac{\pi}{6}$ radians. $$ T = 21 \times \frac{\pi}{6} = \frac{7\pi}{2} $$ Using $\pi = \frac{22}{7}$: $$ T = \frac{7}{2} \times \frac{22}{7} = \frac{22}{2} = 11 \text{ s} $$

Answer: 11 s