Notebook

Determination of unknown mass by proinciple of moments¶

Theory:¶

Apply the principle of moments to a metre rule to first determine its mass and then determine the mass of an unknown object.

Apparatus:¶

Loop a 200 g (1.96N) mass over the metre rule and adjust it until the ruler is horizontal.

Note down the distance, Ɩ, of the mass from the pivot. The mass (or weight) of the metre rule can now be calculated using the principle of moments:

0.20 × metre rule weight = Ɩ × 1.96

Now remove the 200 g mass and replace it with the unknown weight, W, and again adjust the position of the weight until the ruler balances. Measure the distance, d, of the unknown weight from the pivot. The unknown weight can again be calculated by applying the principle of moments:

0.20 × metre rule weight = d × unknown weight

The unknown weight can be converted into a mass (in kilograms) by dividing by 9.81. This can then be checked using a top pan balance.

In [5]:

# Importing the necessary libraries
from matplotlib import pyplot as plt
import numpy as np
#from prettytable import PrettyTable

# Preparing the data to be computed and plotted
dt = np.array([
          
    [100, 2.7],
    [250, 5.5],
    [400, 8.7],
    [550, 11.4],
    [700, 14.6],
    [850, 17.4]
])

# Preparing X and y data from the given data
x = dt[:, 0].reshape(dt.shape[0], 1)
X = np.append(x, np.ones((dt.shape[0], 1)), axis=1)
y = dt[:, 1].reshape(dt.shape[0], 1)

# Calculating the parameters using the least square method
theta = np.linalg.inv(X.T.dot(X)).dot(X.T).dot(y)

print(f'The parameters of the line: {theta}')

# Now, calculating the y-axis values against x-values according to
# the parameters theta0 and theta1
y_line = X.dot(theta)

# Plotting the data points and the best fit line
plt.scatter(x, y)
plt.plot(x, y_line, 'r')
plt.title('Best fit line using regression method')
plt.xlabel('Distance $ d$/mm')
plt.ylabel('Force $F$ /N')

fig, ax = plt.subplots()
table = ax.table(cellText=dt, loc='center', colLabels=('$ Distance $ d$/mm$','Force $F$ /N'))
table.set_fontsize(14)
table.scale(1,4)
ax.axis('off')


plt.show()

The parameters of the line: [[0.01971429]
 [0.68571429]]

Conclusion¶

In comparison, our calculated resistivity is 0.13 Ohm/m vs a databook value of 0.105 Ohm/m This is well within the expected variance of an A-level practical.

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