MOMENT OF INERTIA

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• Q1: What do you know about inertia?
  Ans: The property of a body by which it resists its state of rest or uniform motion is called inertia.
• Q2: Define Moment of Inertia.
  Ans: The product of the first moment of area (ax or ay) with centroidal distance (x or y) is called Second Moment of Area.
  Example: “ax²” or “ay²”
• Q3: Define First Moment of Inertia.
  Ans: The product of the area of an object and its centroidal distance is called the First Moment of Area.
  Example: “ax” or “ay”
• Q4: Define Second Moment of Inertia.
  Ans: The product of the first moment of area (ax or ay) with centroidal distance (x or y) is called Second Moment of Area.
  Example: “ax²” or “ay²”
• Q5: Define Polar Moment of Inertia.
  Ans: If you twist a shaft or rod, the polar moment of inertia tells us how difficult it is to twist it.
  Simple:
  

  • ➔ Larger polar moment of inertia → harder to twist
  • ➔ Smaller polar moment of inertia → easier to twist
• Q6: Write down the units of following in MKS and FPS systems.
  Ans:

  Terms	MKS	FPS
  First Moment of area	m3	ft3
  Second Moment of area	m4	ft4
  Moment of inertia	m4	ft4
  Radius of gyration	m	ft

• Q7: Define Radius of gyration.
  Ans: Radius of gyration is the imaginary distance from the centroid of the given axis to a point where the whole mass of a body is supposed to be concentrated. It tells about the stiffness of the columns.
• Q8: Express moment of inertia of a rectangular section. Also draw diagrams.
  Ans:
• Q9: State parallel axis theorem.
  Ans: Parallel Axis Theorem states:
  “The moment of inertia of a body about an axis parallel to the body passing through its centre is equal to the sum of the moment of inertia of the body about the axis passing through the centre and the product of the mass of the body times the square of the distance of between the two axes.”
• Q10: What is the perpendicular axis theorem?
  Ans: Perpendicular Axis Theorem states:
  “For any plane body, the moment of inertia about any of its axes which are perpendicular to the plane is equal to the sum of the moment of inertia about any two perpendicular axes in the plane of the body which intersect the first axis in the plane.”
• Q11: Define composite section. Give examples.
  Ans: Two or more basic shapes combine to form a composite section or a composite figure. Examples:
• Q12: Write moment of inertia of a rectangular hollow section.
  Ans:
• Q13: Write the moment of inertia of a right-angle triangle.
  Ans:
• Q14: Why does the moment of the area always have a positive value?
  Ans: It is because the second moment uses the square of the distance, and squares are positive.
• Q15: What is the alternate name of moment of inertia?
  Ans: The other names of moment of inertia are:

  • ➔ Second Moment of Area
  • ➔ Angular Mass
  • ➔ Rotational Inertia
• Q16: Express moment of inertia of I-section.
  Ans:
• Q17: What is a Fly Wheel?
  Ans: A fly wheel is a heavy rotating wheel used in machines to store rotational energy and maintain uniform motion.
• Q18: State the function of Fly Wheel.
  Ans: The functions of a fly wheel are:

  • ➔ It stores rotational energy.
  • ➔ It reduces fluctuations in speed.
  • ➔ It maintains uniform motion of machines.
  • ➔ It supplies energy when input energy decreases.
• Q19: State the energy stored in a Fly Wheel.
  Ans: A fly wheel stores rotational kinetic energy.
• Q20: State the formula to calculate the moment of inertia of the Fly Wheel.
  Ans: The formula is:
  
  I = Mk²
  
  where:
  I = moment of inertia
  M = mass
  k = radius of gyration
• Q21: State the formula to calculate the moment of inertia of a spherical body.
  Ans: For a solid sphere, the formula is:
  
  I = (2/5) MR²
  
  where:
  M = mass of sphere
  R = radius of sphere
• Q22: Find Moment of Inertia of a fly wheel of mass 10kg and radius of gyration 4m.
  Ans: Given:
  M = 10 kg
  k = 4 m
  
  Formula:
  I = Mk²
  
  Solution:
  I = 10 × (4)²
  I = 10 × 16
  I = 160 kg m²
• Q23: A 30kg fly wheel has a radius of gyration of 3m. Find the MOI.
  Ans: Given:
  M = 30 kg
  k = 3 m
  
  Formula:
  I = Mk²
  
  Solution:
  I = 30 × (3)²
  I = 30 × 9
  I = 270 kg m²
• Q24: Find MOI of a 100kg sphere having 50cm diameter.
  Ans: Given:
  M = 100 kg
  Diameter = 50 cm = 0.5 m
  Radius R = 0.25 m
  
  Formula:
  I = (2/5) MR²
  
  Solution:
  I = (2/5) × 100 × (0.25)²
  I = 40 × 0.0625
  I = 2.5 kg m²
• Q25: A solid sphere of radius 200cm has a mass of 200kg. Find MOI.
  Ans: Given:
  M = 200 kg
  R = 200 cm = 2 m
  
  Formula:
  I = (2/5) MR²
  
  Solution:
  I = (2/5) × 200 × (2)²
  I = 80 × 4
  I = 320 kg m²