1. Physics

2. Rotational Mechanics - 2
Session
Rotational Mechanics - 2

3. Session Objectives

4. Session Objective Rotational Kinetic energy Moment of Inertia
Standard Results of moment of Inertia
Radius of gyration
Parallel-axis Theorem
Perpendicular axis theorem

5. Rotational Kinetic Energy
Energy possessed due to rotation of a body or a system of particles about an axis of rotation.

6. Rotational Kinetic Energy –Discrete System
For an ith particle of a rigid body, m2 m3 m1 r2 r3 r1
For a discrete body

7. Rotational Kinetic Energy –Coetaneous System
For a continuous body dm r

8. Moment of Inertia
This property opposes a change in the state of uniform rotation
Particle
I=mr2
Continuous body m2 m3 m1 r2 r3 r1
Discrete System
I=mr2

9. Moment of Inertia It has dimensions [ML-2] Its SI unit is kg-m2
It is a tensor quantity
Depends on mass,shape of the body
Depends on the distribution of mass for same size and shape

10. Moment of Inertia of a thin Roddr x Y r
L/2 M L

11. Moment of Inertia of a Disc
Limits  0 to R

12. Standard Result Y x Z R b a Y x Z R Y x Z R Z x Y b a

13. Standard Results Y x Z R X Y Z O R Z Y x R

14. Radius of Gyration
It is the distance whose square when multiplied by mass gives the moment of inertia of a rigid body.
It is denoted by k

15. Radius of Gyration
A real rigid body can be replaced by a point mass for rotational motion k
For ring

16. Theorem of Parallel Axis
To calculate moment of inertia
about an axis
Icm I a
The two axes should be parallel to each other.
One of them should pass through CM.

17. Theorem of Perpendicular Axis
Applicable for plane lamina
IZ=IX + IY X Y Z
X-Y have to be in a plane perpendicular to Z-axis.
All the three axis should be perpendicular to each other
Ix  Iy + Iz
Iy  Ix + Iz

18. Questions

19. Class Exercise - 1
Two circular discs A and B of equal masses and thickness are made of metals with densities dA and dB (dA > dB). If their moments of inertia about an axis passing through the centre and normal to the circular faces be IA and IB, then
IA = IB (b) IA > IB
(c) IA < IB (d) IA > IB or IA < IB

20. Solution
[As MA = MB]
IB > IA as dA > dB
Hence, answer is (b).

21. Class Exercise - 2
With usual notation, radius of
gyration is given by

22. Solution
I = Mk2 [k radius of gyration]
Hence, answer is (b).

23. Class Exercise - 3
The radius of gyration of a disc of mass M and radius R about one of its diameters is

24. Solution
IZ = IX + IY

25. Solution
[k Radius of gyration]
Hence, answer is (a).

26. Class Exercise - 4
The moment of inertia of a circular disc about one of its diameters is I. Its moment of inertia about an axis perpendicular to the plane of the disc and passing through its centre is

27. Solution
IX = IY = I
IZ = IX + IY = 2I
Hence, answer is (b)

28. Class Exercise - 5
Four spheres of diameter 2a and mass M are placed with their centres on the four corners of a square of side b. The moment of inertia of the system about an axis along one of the sides of the square is

29. Solution
[Parallel axis theorem]
Hence, answer is (b).

30. Class Exercise - 6
The moment of inertia of a circular ring of radius R and mass M about a tangent in its plane is

31. Solution
IZ = MR2
IZ = IX + IY [Perpendicular axis theorem]

32. Solution
Hence, answer is (c).

33. Class Exercise - 7
The adjoining figure shows a disc of mass M and radius R lying in the X-Y plane with its centre on X-axis at a distance ‘a’ from the origin. Then the moment of inertia of the disc about the X-axis is

34. Solution
Hence, answer is (b).

35. Class Exercise - 8
In question 7, the moment of inertia of the disc about the Y-axis is

36. Solution
Hence, answer is (c).

37. Class Exercise - 9
In question 7, the moment of inertia of the disc about the Z-axis is

38. Solution
Hence, answer is (d).

39. Class Exercise - 10
The moment of inertia of a cylinder of radius R, length and mass M about an axis passing through its centre of mass and normal to its length is

40. Solution Mass of disc of thickness
Choose an elemental disc at a distance x from the CM of cylinder.
Mass of disc of thickness

41. Solution

42. Thank you