MHT-CET : Physics Entrance Exam

### MHT - CET : Physics - Rotational motion Page 2

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Moment of Inertia and Torque

Consider a rigid body rotating about an axis passing through O and perpendicular to its plane. A torque t acting on the body produces an angular acceleration a. m1, m2mn are the masses of the particles of the body situated at distances r1, r2rn from the axis of rotation F1, F2 …Fn are the forces acting in m1, m2mn, respectively.
The torque (
t) acting on the body = Sum of the torques acting on each of the particles.
\ t = t1 + t2 + ….tn
But t = rF sinq, since and q = 90°
\ t = rF
\ t = r1F1 + r2F2 + ….rnFn

 Further F = ma, \ t = r1m1a1 + r2m2a2 + ….. + rnmnan = r1m1 (r1a) + r2m2 (r2a) + ….. + rnmn (rna) [... a = ra]

\ t = m1r12 a + m2r22 a + …. mnrn2 a

t =

 N miri2a å i = 1

= Ia  [ ...

 n miri2 å i = 1

= I ]

\ t = Ia

Remember: Torque is a vector quantity.

 Principle of Parallel Axes Statement: The moment of inertia of a body about any axis is equal to the sum of its moment of inertia about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes. Mathematically, Io = ICM + Mh2 where Io is the moment of inertia about an axis passing through O, ICM is the moment of inertia about a parallel axis through the centre of mass (C), M is the mass of the body and h is the distance between the two parallel axes.

 Principle of Perpendicular axes Statement: The moment of inertia of a plane lamina about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two mutually perpendicular axes in the plane of lamina and passing through the point where perpendicular axis cuts the lamina.   Mathematically, [Iz = Ix + Iy where Ix and Iy are the moments of inertia of the plane lamina about the X and Y axes (in its plane) and I2 is its moment of inertia about the z axis, (perpendicular to its plane).

Moment of Inertia of a thin Uniform Rod

A] About an axis passing through its centre and perpendicular to its length

Let M be the mass of a thin uniform rod of length l rotating about an axis perpendicular to its length and passing through its centre. The mass per unit length of the rod = M/l. The mass dm of an element of length dx at a distance x from the centre of mass,

dm =

 M l

dx.

\ M.I. of the element about C =

 M l

dx. x2

M.I. of the rod = I =

dx. x2

=

 M l

i.e. I =

 M l

 l 3 24

+

 l 3 24

i.e. I =

 Ml 2 l 2

B] About an axis through one end perpendicular to its length

By the principle of parallel axes, Io = IC + Mh2

Iz1 = Iz + M

 l 2

2

\ Iz1 =

 Ml 2 12

+

 Ml 2 4

=

 4 Ml 2 12

=

 Ml 2 3

\ Iz1 =

 Ml 2 3

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