# 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 2-kg 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}