Q.Explain the method to find the center of gravity of an irregularly shaped lamina.

Step 1. Why a center of gravity exists at all.

An irregular lamina is really a collection of a great many individual point masses, each pulled straight down by its own tiny share of the Earth's gravity. Because the lamina is minuscule compared to the size of the Earth, all these individual weights act as essentially parallel downward forces. The resultant of a whole system of parallel forces always acts through one single fixed point (relative to the body) — this point is the center of gravity, and it is the point at which the lamina's entire weight can be taken to act, whatever orientation the lamina is held in. (For a body of ordinary size on Earth, the gravitational field is uniform enough that this point coincides with the center of mass.) Two independent experimental methods can locate it.

Step 2. Method (i) — balancing by pivoting.

Support the lamina on a fine, sharp pivot (such as a needle point) at a trial location, and release it. If the lamina tips over, the weight and the pivot's normal reaction are not acting along the same vertical line, so there is a net torque about the pivot. Move the pivot to a new trial point and try again, repeating by trial and error, until a location is found where the lamina, once released, stays perfectly horizontal without tipping.

Step 3. Why the balance point IS the center of gravity.

At that special balance point, the pivot's upward normal reaction force exactly cancels the downward weight, and — because the lamina stays horizontal with no net torque about the pivot — the two forces must also act along the very same vertical line. Since the weight always acts vertically through the center of gravity, this means the pivot itself must be located directly at (beneath) the center of gravity. The lamina is now in static equilibrium (zero net force, zero net torque), and the located point is the CG.

Step 4. Method (ii) — suspension.

Instead, suspend the lamina freely from a point $P$ near its edge using a string, and let it hang freely until it stops swinging. A freely-hanging body always settles into the orientation where its center of gravity lies directly below the suspension point — if it did not, the weight (acting through the CG) would have a nonzero torque about $P$, and the lamina would keep rotating until that torque vanished. So the vertical line drawn straight down from $P$, marked on the lamina and extended as $PP'$, must pass through the center of gravity.

Step 5. Repeating from further points.

A single suspension only constrains the CG to lie somewhere along the one line $PP'$ — it does not yet fix an exact point. Suspend the lamina again from a second point $Q$ near a different edge, and mark the new vertical line $QQ'$. The center of gravity must also lie on this second line, so it is located exactly at the intersection of $PP'$ and $QQ'$. A third suspension from a point $R$, giving a third line $RR'$, is then done purely as a check — since the CG is a single fixed point of the lamina, this third line must also pass through the very same intersection point already found.

Empirical torque-balance methods — free horizontal balance under a point pivot, and the plumb-line intersection from multiple suspension points

• Thinking a single suspension is enough to fix the center of gravity — one plumb line only constrains the CG to lie SOMEWHERE along that line; at least two lines from different suspension points are needed to pin down the actual point, with a third used as a cross-check.
• Forgetting that a fresh plumb line must be drawn relative to the NEW suspension point each time, rather than reusing the geometry of a previous trial — each line only passes through the CG because it was drawn straight down from that specific suspension point.
• Confusing the pivoting method's balance point with the lamina's plain geometric center — for an irregular shape these are generally different points, because the geometric center ignores how the mass is actually distributed across the lamina.