Question

Three point particles P, Q, R move in a 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 at which they all meet is:

• A. 2/3 sec
• B. 1/6 sec
• C. 1/2 sec
• D. 3/2 sec

Answer 1

Answer: (C)

1/2 sec

Answer 2

The problem asks for the time when three point particles P, Q, and R, moving in a circle with constant angular velocities, all meet. The angular velocities are given as $\omega_P = 5\pi$ rad/sec, $\omega_Q = 2\pi$ rad/sec, and $\omega_R = 3\pi$ rad/sec. The initial positions are "as shown in the figure". Since the figure is not provided, we must infer a common or standard configuration for the initial positions.

Let the initial angular positions of P, Q, and R be $\phi_P, \phi_Q, \phi_R$ respectively, measured from a common reference line (e.g., the positive x-axis). At time 't', their angular positions will be:

$\theta_P(t) = \phi_P + \omega_P t = \phi_P + 5\pi t$

$\theta_Q(t) = \phi_Q + \omega_Q t = \phi_Q + 2\pi t$

$\theta_R(t) = \phi_R + \omega_R t = \phi_R + 3\pi t$

For the three particles to meet at time 't', their angular positions must be the same, modulo $2\pi$ (i.e., they must be at the same physical point on the circle). This means:

$\theta_P(t) \equiv \theta_Q(t) \pmod{2\pi}$

$\theta_P(t) \equiv \theta_R(t) \pmod{2\pi}$

So, we can write:

$\phi_P + 5\pi t = \phi_Q + 2\pi t + 2n_1\pi$ (where $n_1$ is an integer)

$\phi_P + 5\pi t = \phi_R + 3\pi t + 2n_2\pi$ (where $n_2$ is an integer)

Rearranging these equations:

$(5\pi - 2\pi)t = (\phi_Q - \phi_P) + 2n_1\pi \implies 3\pi t = (\phi_Q - \phi_P) + 2n_1\pi$ (Equation 1)

$(5\pi - 3\pi)t = (\phi_R - \phi_P) + 2n_2\pi \implies 2\pi t = (\phi_R - \phi_P) + 2n_2\pi$ (Equation 2)

Dividing by $\pi$:

$3t = \frac{\phi_Q - \phi_P}{\pi} + 2n_1$

$2t = \frac{\phi_R - \phi_P}{\pi} + 2n_2$

A common and plausible initial configuration for such problems, when not explicitly shown, is that the particles are positioned at cardinal points, e.g., P at $0$ radians, Q at $3\pi/2$ radians (270 degrees), and R at $\pi$ radians (180 degrees). Let's assume this configuration:

$\phi_P = 0$

$\phi_Q = 3\pi/2$

$\phi_R = \pi$

Substitute these initial positions into the equations:

$3t = \frac{3\pi/2 - 0}{\pi} + 2n_1 \implies 3t = 3/2 + 2n_1$ (Equation 1')

$2t = \frac{\pi - 0}{\pi} + 2n_2 \implies 2t = 1 + 2n_2$ (Equation 2')

From Equation 2': $2t = 1 + 2n_2$. This implies $1 + 2n_2$ must be an even integer. No, $1+2n_2$ is always an odd integer. So $2t$ must be an odd integer. This means $t$ must be of the form (odd integer)/2. The smallest positive value for $t$ occurs when $n_2 = 0$, which gives $2t = 1 \implies t = 1/2$ sec.

Now, substitute $t = 1/2$ sec into Equation 1':

$3(1/2) = 3/2 + 2n_1$

$3/2 = 3/2 + 2n_1$

$0 = 2n_1 \implies n_1 = 0$

Since we found integer values for both $n_1$ and $n_2$ ($n_1=0, n_2=0$) for $t=1/2$ sec, this means that with the assumed initial configuration, the particles will meet at $t=1/2$ sec.

Let's verify the positions at $t=1/2$ sec:

$\theta_P(1/2) = 0 + 5\pi (1/2) = 5\pi/2 = 2\pi + \pi/2 \equiv \pi/2 \pmod{2\pi}$

$\theta_Q(1/2) = 3\pi/2 + 2\pi (1/2) = 3\pi/2 + \pi = 5\pi/2 = 2\pi + \pi/2 \equiv \pi/2 \pmod{2\pi}$

$\theta_R(1/2) = \pi + 3\pi (1/2) = \pi + 3\pi/2 = 5\pi/2 = 2\pi + \pi/2 \equiv \pi/2 \pmod{2\pi}$

All three particles are at the angular position $\pi/2$ radians (or 90 degrees) at $t=1/2$ sec.

This result matches one of the given options.