In this experiment, you will investigate the balance point of a non-uniform rod. The rod is 80.0 cm long and has a heavy metal tip at one end. You balances ...
Question 1 Report
In this experiment, you will investigate the balance point of a non-uniform rod. The rod is 80.0 cm long and has a heavy metal tip at one end. You balances the rod on a knife-edge pivot and records the position of the balance point. You then hangs a 200 g mass at different positions along the rod and finds the new balance point each time. Fig. 31.1 shows the arrangement. Your results are in Table 31.1.
| Position of 200 g / cm | Balance point / cm |
|---|---|
| No mass | 55.0 |
| 10.0 | 43.5 |
| 20.0 | 44.8 |
| 30.0 | 46.2 |
| 40.0 | 47.8 |
| 60.0 | 51.8 |
(a) Record the balance point for each setup in Table 31.1. [1]
(b) Measure the shift in balance point when the 200 g mass is at 10.0 cm compared to no mass. [1]
(c) Use the balance point with no mass (55.0 cm) and a 200 g mass at 10.0 cm (balance at 43.5 cm) to calculate the mass of the rod. Show your working. [3]
(d) State why the balance point with no mass is not at 40.0 cm (the centre of the rod). [1]
(e) Plot a graph of balance point (y-axis) against position of 200 g mass (x-axis). [3]
(f) Describe one difficulty in finding the exact balance point. [1]
Answer Details
(a) All six values are recorded correctly in Table 31.1: the balance point with no mass is 55.0 cm, and the five balance points with the 200 g mass at positions 10.0, 20.0, 30.0, 40.0 and 60.0 cm are 43.5, 44.8, 46.2, 47.8 and 51.8 cm respectively. [1]
(b) The shift in balance point when the 200 g mass is placed at 10.0 cm, compared to no mass:
\[ \text{Shift} = 55.0 - 43.5 = 11.5 \text{ cm} \]
The balance point moves 11.5 cm towards the 0 cm end because the added mass creates an anticlockwise moment that pulls the equilibrium position towards itself. [1]
(c) Taking moments about the new balance point at 43.5 cm. At equilibrium, the sum of clockwise moments equals the sum of anticlockwise moments.
The 200 g mass hangs at 10.0 cm, which is to the left of the pivot at 43.5 cm. Its anticlockwise moment about the pivot:
\[ \text{Moment of 200 g mass} = 200 \times (43.5 - 10.0) = 200 \times 33.5 = 6700 \text{ g}\cdot\text{cm} \]
[1]
The weight of the rod acts at its centre of gravity (55.0 cm with no added mass), which is to the right of the pivot. Its clockwise moment:
\[ \text{Moment of rod} = M \times (55.0 - 43.5) = M \times 11.5 \]
[1]
Setting the moments equal:
\[ M \times 11.5 = 6700 \]
\[ M = \frac{6700}{11.5} = 583 \text{ g} \]
The mass of the rod is approximately 583 g (accept 580 to 585 g). [1]
(d) The rod is non-uniform because it has a heavy metal tip at one end. This extra mass shifts the centre of gravity towards that end, away from the geometric centre at 40.0 cm. For a uniform rod, the centre of gravity would be at the midpoint, but the metal tip moves it to 55.0 cm. [1]
(e) The graph below shows balance point (y-axis) plotted against the position of the 200 g mass (x-axis). The axes are labelled with units, all five data points are plotted, and a smooth line is drawn through the points. [3]
(f) The rod may rock or oscillate slightly on the knife-edge pivot, making it difficult to judge the exact position where it balances perfectly horizontal. Small air currents or vibrations from the bench can also disturb the balance. [1]
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