‹‹ Previous | Page 1 | Page 2 | Page 3 | Next ››

Moment of Inertia of a Thin Uniform Disc

A) About an axis perpendicular to its plane and passing through its centre Consider a thin disc of radius R and mass M. Mass per unit area of the disc, s = M/p R^{2} Consider a ring of radius x and thickness dx.

\ The mass of the ring of radius x and width dx is dm = s 2p x.dx. Moment of inertia of the ring about the given axis, dI = (s . 2px.dx) x^{2} = s 2p x^{3}dx

\ Moment of inertia of the disc = I =

s2px^{3}dx

i.e. I = 2p s

x^{3}dx = s p .

R^{4}

2

substituting for s, I =

M

pR^{2}

.

pR^{4}

=

1

MR^{2}

\ I_{z} =

Corresponding radius of gyration is given by the relation, I = MK^{2}

\

MR^{2}= MK^{2}

\ K =

R

B] About a diameter: By the principle of perpendicular axes, I_{z} = I_{x} + I_{y} = 21 (^{.}.^{.} I_{x} = I_{y} = I_{d})

But I_{z} =

MR^{2} = 2 × I_{d}

\ I_{d} =

4

C] About a Tangent Perpendicular to its Plane By principle of parallel axes,

I_{o} =

I_{c} + Mh^{2}

+ MR^{2}

(^{.}.^{.} I_{c} = I_{z} =

)

3

D] About any Tangent in the Plane by the disc By principle of parallel axes,

I_{d} + Mh^{2}

5

Angular Momentum (L)

Definition (1) Angular momentum (L) of a rotating body about its axis of rotation is defined as the product of the moment of inertia of the body about that axis and the angular velocity of the body. \L = Iw

(2) Angular momentum of a rotating body is given by

where is the position vector

and is the linear momentum of the particle. ^{.}.^{.} = , \ = ´ \ The direction of L is given by the right hand rule.

Unit of Angular Momentum: Kilogram-metre^{2}/second (kg -m^{2}/s) Dimensions of Angular Momentum: [M^{1}L^{2}T^{ -1}] Expression for Angular Momentum:

Consider a rigid body rotating with uniform angular velocity w, about an axis through O and perpendicular to its plane. Let m_{1}, m_{2}, …m_{n} be the masses of the constituent particles of the body and r_{1}, r_{2} ….r_{n} be their respective distances from the axis of rotation. Angular momentum of the body = Sum of the angular momenta of the constituent particles. \ L = L_{1} + L_{2} + …. + L_{n} = m_{1}v_{1}r_{1} + m_{2}v_{2}r_{2} + ….m_{n}v_{n}r_{n} \ L = r_{1} × m_{1} (wr_{1} ) + r_{2} × m_{2} (wr_{2}) + ….r_{n}m_{n}(wr_{n}) [^{.}.^{.} v = rw]

\ L = (m_{1} r^{2}_{2} + m_{2} r^{2}_{2} + ….. m_{n} r_{n}^{2} ) w =

n

m_{i}r_{i}^{2}. w

å

i = 1

\ L = Iw

^{.}.^{.}

m_{i}r_{i}^{2} = I

Remember: Angular momentum is a vector quantity.

Law of Conservation of Angular Momentum

Statement: If the resultant external torque acting on a body is zero, its angular momentum remains constant. Examples: 1. Ballet dancers: Can increase or decrease their speed of rotation by holding their arms close to their bodies or spreading them out. 2. Diver: While diving, the diver holds his arms and legs close to his body so that his M.I. decreases causing his speed of rotation to increase. 3. Skater: A skater who is gyrating about himself can change his speed of rotation by either spreading out his arms or holding them close to his body.