Engineering Mechanics

Chapter 26 : Helical Springs and Pendulums „„„„„ 543

Now let I 0 = Mass moment of inertia of the body about O

α = Angular acceleration of the body.

∴ Disturbing moment due to angular displacement

= I 0 α ...(ii)

Equating two moments i.e. equations (i) and (ii),

mgh θ = I 0 α

∴

0

mgh

I

θ

α=

We know that periodic time,

Displacement

22

Acceleration

t

θ

=π =π

α

0

0

22

I

mgh mgh

I

θ

=π =π

θ

...(iii)

and frequency of motion,
0

11

2

mgh

n

tI

==

π

...(iv)

Note. The above formula for periodic time (or frequency of motion may also be expressed in
terms of radius of gyration (k) as discussed below :

We know from the theorem of parallel axis (Art. 7·12) that the mass moment of inertia of the
pendulum about O,
I 0 = Ig + mh^2 = mk^2 + mh^2 = m (k^2 + h^2 )

Now substituting the value of I 0 in equation (iii),

()^2222

22

mk h k h

t

mgh gh

++

=π =π

Similarly 22

11

2

gh

n

t kh

==
π +
Example 26.13. A uniform straight rod of length 600 mm and mass 250 g is smoothly
pivoted about a point, which is 40 mm from one end. Find the period of small oscillation about the
pivot, if the rod can turn freely in the vertical plane.

Solution. Given : Length of rod (l ) = 600 mm = 0.6 m ; Mass of rod (m)
=250 g = 0·25 kg. and distance between the point of suspension and the centre of
gravity of the body (h) = 300 – 40 = 260 mm = 0·26 m

We know that mass moment of inertia of the rod about the pivot G,

2

2

0

0·25 (0·3)

0·25 (0·26)

3

I =+= 0·0244 kg-m^2

and period of small oscillation, 2 0
I
t
mgh

=π

0·0244

2

0·25 9·8 0·26

=π

××

s

= 1·23 s Ans.

Fig. 26.8.