Physics

Rotational Mechanics - 2 Session Rotational Mechanics - 2

Session Objectives

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

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

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

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

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

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

Moment of Inertia of a thin Rod dr x Y r L/2 M L

Moment of Inertia of a Disc Limits  0 to R

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

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

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

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

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.

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

Questions

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

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

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

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

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

Solution IZ = IX + IY

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

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

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

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

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

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

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

Solution Hence, answer is (c).

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

Solution Hence, answer is (b).

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

Solution Hence, answer is (c).

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

Solution Hence, answer is (d).

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

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

Solution

Thank you