There is an analogy between quantities in rotational and translational dynamics - AQA - A-Level Physics - Question 1 - 2017 - Paper 6

To find the total moment of inertia, we need to sum the moment of inertia of the jib and the contributions from the trolley and load:

1. The moment of inertia of the jib is given as: $I_{jib} = 2.6 \times 10^{2} \, \text{kg m}^2$

2. The moment of inertia for the trolley and load can be calculated using the formula: $I_{trolley/load} = m \times r^2$ where:

   • mass ($m$) = $2.2 \times 10^{3} \, \text{kg}$
   • distance ($r$) = $35 \, \text{m}$

   Therefore, $I_{trolley/load} = (2.2 \times 10^{3} \, \text{kg}) \times (35 \, \text{m})^2 = 2.7 \times 10^{6} \, \text{kg m}^2$

3. Adding the individual moments of inertia: $I_{total} = I_{jib} + I_{trolley/load}$ $I_{total} = 2.6 \times 10^{2} + 2.7 \times 10^{6} \approx 2.9 \times 10^{6} \, \text{kg m}^2$