Two beads P & Q move along two wires straight and semi circle. At some instant both are shown in figure having same horizontal component of velocity V. Find relation of time taken to reach A by beads :
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In comparative motion problems, remember that the straight-line path between two points is the shortest distance.
A particle moving along this straight path will have the highest possible average velocity component in that direction compared to any other particle moving along a longer, curved path between the same two points, assuming comparable speeds.
Step 1: Understanding the Question:
We are asked to compare the time taken by two beads, P and Q, to travel to a common final point A. Both beads start from a position where they have the same horizontal component of velocity, V. Bead P moves along a straight path (a chord), while bead Q moves along a curved, circular path (an arc). Step 2: Key Formula or Approach:
The time taken to travel a certain displacement is given by the relation:
\[ \text{time} = \frac{\text{displacement}}{\text{average velocity}} \]
We will analyze the motion of the beads in the horizontal (X) direction. Let the horizontal displacement from the starting point to point A be \(d_x\). This displacement is the same for both beads. Therefore, the time taken is inversely proportional to the average velocity in the horizontal direction, \(v_{avg,x}\). Step 3: Detailed Explanation:
Let's analyze the average horizontal velocity for each bead. For Bead P: Bead P travels along a straight path. The most direct path between its start and end point in the horizontal direction is this straight line. Its horizontal velocity component is given as V. We can consider its average horizontal velocity to be constant and equal to V.
\[ v_{P, avg, x} = V \]
For Bead Q: Bead Q is constrained to move along a circular arc. The length of this arc is greater than the length of the straight path taken by P to cover the same horizontal distance. Although its horizontal velocity component is initially V, its velocity vector must continuously change direction to follow the curve. A particle that travels along a curved path to get between two points will have a lower average velocity component along the straight line connecting those points compared to a particle that takes the straight path directly. The curved path is a less efficient way to achieve horizontal displacement.
Therefore, the average velocity of Q in the horizontal direction will be less than the average horizontal velocity of P.
\[ v_{Q, avg, x}<v_{P, avg, x} \]
Step 4: Final Answer:
The time taken is \(t = \frac{d_x}{v_{avg,x}}\). Since the horizontal displacement \(d_x\) is the same for both beads, the time taken is inversely proportional to the average horizontal velocity.
Because \(v_{P, avg, x}>v_{Q, avg, x}\), it follows that the time taken by P will be less than the time taken by Q.
\[ t_p<t_Q \]
