Physics, asked by Sayantana, 2 months ago


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 5pi, 2pi & 3pi respectively, in the same sense. The time interval after which they all meet is:-
(1) 2/3 sec
(2) 1/6 sec
(3) 1/2 sec
(4) 3/2 sec

pls explain with correct explanation and answer...
(irrelevants and incorrect will get report)​

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Answered by Anonymous
3

Angular velocity of P, \sf W_P = 5π rad/s

Angular velocity of Q, \sf W_Q = 2π rad/s

Angular velocity of R, \sf W_R = 3π rad/s

Now, let's assume P as the origin. So,

• Angle travelled by P, θ = \sf W_P × t = 5πt - (i)

• Angle travelled by Q, θ = \sf W_Q × t = 2πt - (ii)

• Angle travelled by R, θ = \sf W_R × t = 3πt - (iii)

Let's use the rule of elimination, in equation (i), (ii) and (iii). So,

• 5πt = 5π × \dfrac{2}{3} = \dfrac{10π}{3}

• 2πt = 2π × \dfrac{2}{3} + \dfrac{π}{2} = \dfrac{4π}{3}

• 3πt = 3π × \dfrac{2}{3} + π = 2π

Similarly, when t was either \dfrac{1}{6} or \dfrac{1}{2}, the value of θ was different. The value of θ should be between 0 and 2π. Now, we put the value of t = \dfrac{3}{2} . Now,

• 5πt = 5π × \dfrac{3}{2} = \dfrac{15π}{2} - (a)

• 2πt = 2π × \dfrac{3}{2} + \dfrac{π}{2}= \dfrac{7π}{2} - (b)

• 3πt = 3π × \dfrac{3}{2} + π = \dfrac{11π}{2} - (c)

But the value of θ is different in the above equations. As we know,

\dfrac{15π}{2} = 6π + \dfrac{3π}{2}

\dfrac{7π}{2} = 2π + \dfrac{3π}{2}

\dfrac{11π}{2} = 4π + \dfrac{3π}{2}

 \bold{Hope\;it \; helps\;!}

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