Question

A uniform bar AB of length `6a` has been placed on a horizontal smooth table of width 5 a as shown in the figure. Length 2a of the bar is overhanging. Mass of the bar is 4m. An insect of mass m is sitting at the end A of the bar. The insect walks along the length of the bar to reach its other end B.

(a) Will the bar topple when the insect reaches end B of the bar ?
(b) After the insect reaches at B, another insect of mass M lands on the end A of the bar. Find the largest value of M which will not topple the bar.

Answer 1

Correct Answer - (a) No
(b) 85 m

Answer 2

(a) Since there is no external force, in horizontal direction on the system, its COM will not move. Let the bar move a distance x towards left by the time 1st insect reaches B.

Then 4mx = m(6a - x) ⇒ \(\frac {6}{5}a = x\)

\(\therefore\) Length hanging out of the table (on right) is \(2a-\frac {6a}{5}= \frac {4a}{5}\)

Length hanging out on left = a/5

The bar will topple if the vertical line through centre of mass of the system passes outside the table. The COM of the system lies at a distance

\(\frac {m\times 0 +4 m \times 3a}{5m}\) from end B

\(=\frac {12}{5}a > \frac {4a}{5}\)

The bar will not topple.

(b) When the second insect with large mass sits at end A, the bar has a tendency to topple about C (see figure). If M increases COM of the system shifts to left. M is maximum (for not toppling) when COM is at C

Distance of COM from \(A = \frac {M\times 0 +4 m \times 3a + m \times 6a}{M+5m}\)

\(\Rightarrow \frac {a}{5} = \frac{18 ma}{M+5m}\)

⇒ M + 5m = 90m

⇒ m = 85 m