Moment of inertia for particle – problems and solutions
1. Two balls connected by a rod, as shown in the figure below. Ignore rod’s mass. Mass of ball P is 600 gram and mass of ball Q is 400 gram. What is the moment of inertia of the system about AB?
Known :
The axis of rotation is AB.
m_{p} = 600 gram = 0.6 kg, m_{q} = 400 gram = 0.4 kg
r_{p} = 20 cm = 0.2 m, r_{q} = 50 cm = 0.5 m
Wanted : The moment of inertia of the system
Solution :
I = m_{p} r_{p}^{2} + m_{q} r_{q}^{2}
I = (0.6 kg)(0.2 m)^{2} + (0.4 kg)(0.5 m)^{2}
I = (0.6 kg)(0.04 m^{2}) + (0.4 kg)(0.25 m^{2})
I = 0.024 kg m^{2} + 0.1 kg m^{2}
I = 0.124 kg m^{2}
2. The Rod of AB with mass of 2kg rotated about point A, the moment of inertia of the rod is 8 kg m^{2}. If rotated about point O (AO = OB),what is the moment of inertia of the rod.
Known :
Mass of rod AB (m) = 2 kg
If rotated about point A so that the radius of rotation (r) = length of AB = r then the moment of inertia (I) = 8 kg m^{2}
Wanted: If rotated about point O so that the radius of rotation (r) = length of AO = length of OB = 1/2 r then what is the moment of inertia of the rod.
Solution :
I = m r^{2}
8 kg m^{2} = (2 kg) r^{2}
8 m^{2} = (2) r^{2}
r^{2 }= 8 m^{2} / 2
r^{2 }= 4 m^{2}
r = 2 meters
If rotated about point O so ½ r = 1 meter, then the moment of inertia :
I = m r^{2 }= (2 kg)(1 m)^{2} = (2 kg)(1 m^{2}) = 2 kg m^{2}
3. Two balls connected by a rod as shown in figure below. Ignore rod’s mass. What is the moment of inertia of the system.
Known :
Mass of ball A (m_{A}) = 200 gram = 0.2 kg
Mass of ball B (m_{B}) = 400 gram = 0.4 kg
Distance between ball A and the axis of rotation (r_{A}) = 0
Distance between ball B and the axis of rotation (r_{B}) = 25 cm = 0.25 m
Wanted : Moment of inertia of the system
Solution :
The moment of inertia of ball A :
I_{A} = (m_{A})(r_{A}^{2}) = (0.2)(0)^{2} = 0
The moment of inertia of ball B :
I_{B} = (m_{B})(r_{B}^{2}) = (0.4)(0.25)^{2} = (0.4)(0.0625) = 0.025 kg m^{2}
The moment of inertia of system :
I = I_{A }+ I_{B} = 0 + 0.025 = 0.025 kg m^{2} = 25 x 10^{3} kg m^{2}
4. Four particles with different mass, shown in figure below. Determine the moment of inertia of the system about the horizontal line P.
Solution
The axis of rotation is the horizontal line P.
Known :
Mass of particle A (m_{A}) = m
Mass of particle B (m_{B}) = 2m
Mass of particle C (m_{C}) = 3m
Pass of particle D (m_{D}) = 4m
Distance between particle A and the axis of rotation (r_{A}) = b
Distance between particle B and the axis of rotation (r_{B}) = b
Distance between particle C and the axis of rotation (r_{C}) = 2b
Distance between particle D and the axis of rotation (r_{D}) = 2b
Wanted : The moment of inertia of the system about the horizontal line P
Solution :
I = m_{A} r_{A}^{2} + m_{B} r_{B}^{2 }+ m_{C} r_{C}^{2 }+ m_{D }r_{D}^{2 }
I = (m)(b)^{2 }+ (2m)(b)^{2} + (3m)(2b)^{2 }+ (4m)(2b)^{2 }
I = mb^{2 }+ 2 mb^{2} + (3m)(4b^{2}) + (4m)(4b^{2})
I = mb^{2 }+ 2 mb^{2} + 12 mb^{2 }+ 16 mb^{2 }
I = 31 mb^{2 }
5. Four particles connected by a rod. Ignore rod’s mass. Determine the moment of inertia about the axis of rotation through particle m_{1} and m_{2}, as shown in figure below.
Known
Mass of particle 1 (m_{1}) = 1/4 kg
Mass of particle 2 (m_{2}) = 1/2 kg
Mass of particle 3 (m_{3}) = 1/4 kg
Mass of particle 4 (m_{4}) = 1/4 kg
Distance between particle 1 and the axis of rotation (r_{1}) = 0
Distance between particle 2 and the axis of rotation (r_{2}) = 0
Distance between particle 3 and the axis of rotation (r_{3}) = 10 cm = 10/100 m = 1/10 m
Distance between particle 4 and the axis of rotation (r_{4}) = 10 cm = 10/100 m = 1/10 m
Wanted : The moment of inertia
Solution :
I = m_{1} r_{1}^{2} + m_{2} r_{2}^{2 }+ m_{3} r_{3}^{2 }+ m_{4 }r_{4}^{2 }
I = (1/4)(0)^{2 }+ (1/2)(0)^{2} + (1/4)(1/10)^{2 }+ (1/4)(1/10)^{2 }
I = 0 + 0 + (1/4)(1/100) + (1/4)(1/100)
I = 1/400 + 1/400
I = 2/400
I = 1/200 kg.m^{2}
 What is the moment of inertia for a particle?
 Answer: For a single particle of mass $m$ at a distance $r$ from an axis of rotation, its moment of inertia $I$ is given by $I=m×r$.
 Why is moment of inertia often referred to as the “rotational analogue” of mass?
 Answer: Just as mass is a measure of an object’s resistance to changes in its translational motion (due to Newton’s second law), the moment of inertia is a measure of an object’s resistance to changes in its rotational motion.
 How does changing the distance of a particle from its axis of rotation affect its moment of inertia?
 Answer: The moment of inertia is proportional to the square of the distance from the axis of rotation. If you double the distance, the moment of inertia will increase by a factor of four.
 Why is the square of the distance (r^{2}) used in the formula for the moment of inertia instead of just the distance?
 Answer: The square of the distance is used because of the way kinetic energy in rotation works. In rotational motion, every particle of an object contributes to the rotational kinetic energy based on both its mass and its distance from the axis squared.
 How does the moment of inertia of a particle change if its mass is tripled while keeping the distance from the axis constant?
 Answer: If the mass is tripled and the distance is kept constant, the moment of inertia will also triple because it is directly proportional to mass.
 Can a particle have a moment of inertia of zero? If so, under what condition?
 Answer: Yes, a particle will have a moment of inertia of zero if it is located directly on the axis of rotation, making its distance $r$ from the axis equal to zero.
 Why do different objects with the same mass and size have different moments of inertia when rotating about different axes?
 Answer: The distribution of mass about the axis of rotation determines the moment of inertia. Even if two objects have the same mass and size, their mass distributions relative to the axis of rotation may differ, leading to different moments of inertia.
 Is the moment of inertia scalar or vector quantity?
 Answer: Moment of inertia is a scalar quantity. However, for rigid bodies with complex shapes and multiple axes of rotation, it is represented by a tensor.
 If two particles, each of mass $m$, are located at distances $r$ and $r$ from the axis of rotation, what is the combined moment of inertia?
 Answer: The moment of inertia is additive for separate particles. Thus, the combined moment of inertia $I=m×r_{1}+m×r_{2}$.

How does the moment of inertia relate to the conservation of angular momentum?
 Answer: Angular momentum $L$ is the product of moment of inertia $I$ and angular velocity $ω$, represented by the equation $L=I×ω$. If no external torques act on a system, the angular momentum will remain constant. This means if the moment of inertia changes (as in a figure skater pulling in their arms), the angular velocity must adjust to keep the product constant.