The problem of the frog and the rolling pipe is a beautiful symphony of kinematics, relative motion, and geometry. At first glance, it seems like a standard projectile motion question, but the constraint "touching it only at the top" elevates it to a true JEE Advanced masterpiece. Let's break down the physics step-by-step.
The Vertical Constraint
Clearing the Pipe
Imagine the scenario: a cylindrical pipe of radius r is rolling towards a frog. To save itself, the frog leaps into the air. The problem states that the frog passes over the pipe, touching it only at the very top.
What does this physical constraint tell us about the frog's trajectory?
Since the frog jumps from the ground and grazes the top of the pipe, the maximum height of its parabolic path must be exactly equal to the diameter of the pipe. Therefore, the maximum height H is 2r.
We know from standard projectile kinematics that the maximum height is governed entirely by the initial vertical velocity
uy. The formula is:
H=2guy2
Setting this equal to our required height
2r, we can easily solve for the vertical velocity:
With the vertical velocity locked in, calculating the total air-time
T of the frog is straightforward. The time of flight is simply twice the time taken to reach the maximum height:
This elegant result immediately confirms that option (a) is correct. But the real challenge lies ahead.
The Horizontal Challenge
Avoiding a Crash
We know the frog reaches the correct height, but how do we ensure it doesn't crash into the sides of the rolling pipe before or after reaching the top?
Analyzing this from the ground frame is messy because both the frog and the pipe are moving. The secret weapon here is to shift our perspective. Let's jump onto the pipe and observe the world from its frame of reference.
In the pipe's frame, the pipe is completely stationary. The frog, however, is now flying towards us. If the frog jumps with a horizontal velocity
ux towards the pipe, and the pipe is moving towards the frog with velocity
v, the relative horizontal velocity of the frog is:
vrel=ux+v
The Elegance of Curvature
Now, visualize the frog's trajectory in this relative frame. It is still a parabola, and it must graze the stationary circular pipe exactly at its apex.
For the parabola to touch the circle only at the top and not intersect it anywhere else, the parabola must be "flatter" than the circle at that point. In rigorous mathematical terms, the radius of curvature of the parabola at its highest point must be greater than or equal to the radius of the pipe r.
The formula for the radius of curvature
Rc at the apex of a projectile's path is the square of the horizontal velocity divided by the perpendicular acceleration (which is just
g):
Rc=gvrel2
Applying our geometric constraint
Rc≥r, we get:
g(ux+v)2≥r
Taking the square root of both sides, we find the critical condition for the frog's horizontal velocity:
This tells us the minimum horizontal speed the frog must muster to safely clear the rolling pipe without a collision.
The Final Leap
With both the time of flight and the condition for horizontal velocity in hand, we can finally determine the horizontal range R of the frog's jump in the ground frame.
The range is simply the horizontal velocity multiplied by the time of flight:
R=uxT
Substituting our minimum condition for
ux and our calculated
T:
Rearranging this slightly gives us the final beautiful expression:
This perfectly matches option (d). By seamlessly blending vertical kinematics with the geometric power of the radius of curvature in a relative frame, we have conquered this problem. It is a brilliant reminder of why shifting reference frames is one of the most powerful tools in a physicist's arsenal.
