let’s find out, the time period of a torsional
pendulum. a torsional pendulum is made by
suspending, any rigid body from a stiff wire.
like say we’re having a stiff wire, with
its coefficient, of stiffness, equal to c.
we suspend, say a disc from the stiff wire,
such that the disc remains horizontal, in
iquilibrium. say the disc is of radius r and
its mass is m. from iquilibrium if the disc
is say, rotated anti-clock-wise by an angle,
theta. now, if it is twisted by an angle theta,
the wire will also be twisted. and due to
the stiffness, the wire will exert clock-wise
restoring torque on the disc, which will be
given as c theta. we can write, if, disc is
twisted, by an angle theta, from iquilibrium
position, wire exerts, a restoring torque
on disc, which is given as, minus c-theta,
where c is the stiffness coefficient, and
stiffness coefficient is just giving us an
idea, that how much torque will act, if wire
is twisted by an angle-theta from one end
related to the other end. obviously at suspension
point wire is rigidly fixed, so it’ll be
relative angle-theta by which the overall
wire is twisted. so restoring torque will
be c-theta, and angular acceleration we can
write as, torque upon moment of inertia. so
it’ll be minus c-theta upon, i. and here,
for s echch m we know, that angular acceleration
of body can be written as minus omega square
theta. so by comparing, we get, the angular
frequency of s echch m will be equal to, root
of c by i. so time period of angular s echch
m, of disc can be given as, 2 pi by omega,
which is written as, 2 pi root of, i by c.
this is the result we use for time period
of, angular s echch m, of a rigid body, which
is suspended by a stiff wire. in case of a
disc, we take moment of inertia to be half
m r square, and for any other rigid body respective
moment of inertia should be taken here.
