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of Inertia and Torque
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, m2…mn are the
masses of the particles of the body situated at distances r1, r2 …rn from the
axis of rotation F1, F2 …Fn are the
forces acting in m1, m2…mn,
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
Further F = ma, \ t
= r1m1a1 + r2m2a2 + ….. + rnmnan
= r1m1 (r1a) + r2m2 (r2a) + ….. + rnmn (rna)
[... a = ra]
\ t = m1r12 a + m2r22 a + …. mnrn2 a
i = 1
= Ia [ ...
= I ]
\ t = Ia
Remember: Torque is a vector quantity.
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.
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.
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).
of Inertia of a thin Uniform Rod
About an axis passing through its centre and perpendicular to its
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,
\ M.I. of the element about C =
M.I. of the rod = I =
i.e. I =
B] About an axis through one end perpendicular to its
By the principle of parallel axes, Io = IC + Mh2
Iz1 = Iz + M
\ Iz1 =
4 Ml 2