Describe an experiment to determine the position of the centre of gravity of an irregularly shaped plane lamina

Objective 📐

Describe an experiment that will let you find the position of the centre of gravity of an irregularly shaped plane lamina.

Analogy & Example 🤹‍♂️

Think of a seesaw. If you place a toy at the exact middle, the seesaw stays level. The middle point is like the centre of gravity. For a flat object that isn’t a perfect rectangle, we need a trick to find that “balance point.”

Experiment Setup 🔬

1. Materials:

   • Irregular flat lamina (e.g., a piece of cardboard cut into a shape)
   • Thin wooden stick or ruler (≈ 30 cm)
   • String or thread (≈ 50 cm)
   • Clothespin or small clamp
   • Scale or ruler for measuring distances
   • Notebook and pen

2. Procedure:

   1. Place the stick on a flat table and mark its centre (point C).
   2. Attach the string to the stick at point C and tie the other end to the lamina so that the string hangs vertically.
   3. Hold the lamina with the string so that it is free to rotate about point C.
   4. Move the lamina until it comes to rest (it will balance at a point where the torque is zero).
   5. Mark the point on the lamina where the string touches the table; call this point P.
   6. Measure the horizontal distance d from the lamina’s edge to point P.
   7. Repeat the experiment with the string attached at a different point on the stick (e.g., 5 cm to the left of C) and record the new distance d'.

3. Data Recording:

   Attachment Point on Stick	Distance d (cm)
   Centre (C)	12.3
   5 cm left of C	8.7

Theory & Calculations 🧮

The centre of gravity (CG) is the point where the total torque due to gravity is zero. When the lamina balances on the string, the line of action of its weight passes through the attachment point on the stick.

Let r be the distance from the attachment point to the CG along the lamina, and L the distance between the two attachment points on the stick. Using the principle of moments:

\$\frac{r}{L} = \frac{d'}{d}\$

Rearranging gives the CG position relative to the first attachment point:

\$r = L \times \frac{d'}{d}\$

Insert the measured values to find r, then locate the CG on the lamina.

Exam Tips 🏆

• Show a clear diagram of the experiment, labeling the stick, string, lamina, and measured distances.
• Explain why the torque must be zero at equilibrium.
• Use the ratio of distances (as in the formula above) to avoid needing the mass of the lamina.
• Remember to state any assumptions (e.g., lamina is thin, weight acts at CG).
• Check units: all distances should be in the same units (cm or m).