State what is meant by the moment of inertia of an object about an axis - AQA - A-Level Physics - Question 1 - 2020 - Paper 5

A solid disc pulley of mass $M$ and radius $R$ is supported in bearings which have negligible friction. A string of negligible mass is wrapped around the circumference of the pulley. A load of mass $0.5M$ is fixed to the free end of the string. The string does not slip on the pulley.

The moment of inertia of the pulley about the axis of rotation is $0.5MR^2$.

When the student releases the pulley, the load accelerates downwards uniformly and is at a velocity $v$ after moving a distance $h$.

To show that the acceleration of the load is $0.5g$, we use the principles of energy conservation:

1. The energy lost due to the falling mass is equal to the energy gained by the pulley plus the energy gained by the load: $E_{lost} = E_{pulley} + E_{kinetic}$

2. For the load, we have: $E_{lost} = 0.5Mgh$

3. The rotational kinetic energy of the pulley is given by: E_{pulley} = rac{1}{2} I rac{v^2}{R^2} = rac{1}{2} (0.5MR^2) rac{v^2}{R^2} = rac{1}{4}Mv^2

4. The kinetic energy of the load is: E_{kinetic} = rac{1}{2} 0.5Mv^2 = rac{1}{4}Mv^2

5. Setting these equal gives: 0.5Mgh = rac{1}{4}Mv^2 + rac{1}{4}Mv^2

6. Simplifying results in: 0.5Mgh = rac{1}{2}Mv^2

7. Dividing through by $M$ yields: gh = rac{1}{2}v^2

8. Solving for acceleration $a$, knowing that $v^2 = 2ah$, gives: $g = 0.5a$ Therefore, the acceleration of the load is: $a = 0.5g$