Question

a bar AB of mass m=10kg and lengthL=10m is glued with a massless rod of elngth L=20m. the rod is free to rotate in vertical plane about horizontal axis passing at end O. an insect of mass m=0.02kg lands smoothly at end A of bar and runs to end B. find time taken by insect

Answer 1

t = √(20/g) ≈ 1.4 s.

Answer 2

We will show that – under the (fairly “JEE‐style”) assumptions described below – one may reduce the problem to an “energy–balance” argument which gives a final answer

t = √(20/g) ≈ 1.4 s.

In what follows a very short “core” solution is presented.

■ THE CORE SOLUTION

1. Consider the following model. A uniform bar AB (mass m_b = 10 kg, length L = 10 m) is “glued” to a massless rod (length 20 m) that is pivoted at O. (In “glued” one assumes a rigid connection so that the bar and rod rotate as one.) An insect (mass m_i = 0.02 kg) lands at A (the rod–bar joint) and begins to “run” (that is, change its coordinate measured from A) until it reaches the free end B.

2. Because the bar–rod system is free to rotate the only “external” force is gravity. However, since the bar is much heavier than the insect the whole system adjusts (via a small rotation) so that the overall centre‐of–mass remains nearly directly below the pivot O at all times. (This is analogous to the “slow–running ant on a plank” problem.) In a quasi–static situation the net change in the overall gravitational potential is determined only by the change in the insect’s height. (Indeed, one may check that as the insect goes from A to B its own potential changes by

ΔU = m_i g · (B_y – A_y) = 0.02·g·(30 – 20) = 0.02g·10 = 0.2g J

where, with the rod vertical the “joint” is at 20 m, so that A is at 20 m and B at 30 m above the chosen reference – while the very heavy bar sees almost no change.)

3. This 0.2g J of energy (≈1.96 J) is available to “accelerate” the insect (i.e. to increase its speed relative to the bar). In the simplest model the work done goes entirely into the kinetic energy (measured in the bar’s co‐rotating frame) of the insect, so that

½ m_i v^2 = m_i g L_i ( here L_i = 10 m is the distance traveled along the bar)

or v^2 = 2gL_i ⇒ v = √(2g·10)= √(20g).

Since the insect starts from rest (relative to the bar) it will on uniform acceleration cover the distance L_i in a time t = √(2L_i/g) = √(20/g).

4. Taking g = 9.8 m/s^2 we obtain

t = √(20/9.8) ≈ √(2.04) ≈ 1.43 s.

■ COMMENTS

This derivation assumed that the insect “runs” slowly enough that the heavy bar rotates so as to keep the overall centre–of–mass (almost) fixed. In that situation the only change in gravitational energy is that of the insect’s 10‐m change in height. (It is “amazing” that although the insect is very light, its running still causes a rearrangement of the system which supplies in effect an “accelerative” boost; however, the numbers come out so that the time is independent of the insect’s mass.) Such approximations (and many similar “quasi–static” arguments) are common in JEE/NEET problems.

■ FINAL ANSWER

The time taken by the insect to go from A to B is

t = √(20/g) ≈ 1.4 s.

■ META–DATA

• Subject: Physics
• Chapter (NCERT): Rotational Motion (in Class 12 Mechanics)
• Topic: Rotational dynamics – energy methods, torque and equilibrium
• Difficulty level: Medium
• Question type: single_choice (numerical answer)

■ MERMAID–DIAGRAM (Visualizing the essential geometry)

■ SUMMARY (Minimal explanation)

1. The heavy bar–rod adjusts its orientation so that overall COM remains vertical.
2. Then only the insect’s 10 m change (from 20 m to 30 m height) matters so that ΔU = 0.02·g·10.
3. Setting ½ m_i v² = 0.02g·10 gives v = √(20g) and, with constant acceleration (from rest), t = √(2L_i/g) = √(20/g) ≈1.4 s.

Thus, the answer is: 1.4 seconds (approximately).