3. • In order to determine the motion of a rigid body, under the action of external forces,
it is usually convenient to replace the rigid body by two masses placed at a fixed
distance apart, in such a way that,
1. the sum of their masses is equal to the total mass of the body ;
2. the centre of gravity of the two masses coincides with that of the body
3. the sum of mass moment of inertia of the masses about their centre of gravity is
equal to the mass moment of inertia of the body.
• When these three conditions are satisfied, then it is said to be an equivalent
dynamical system. Consider a rigid body, having its centre of gravity at G
4. Let m = Mass of the body,
KG = Radius of gyration about its centre of gravity G,
m1 and m2 = Two masses which form a dynamical equivalent system,
l1 = Distance of mass m1 from G,
l2 = Distance of mass m2 from G,
5.
6.
7. A connecting rod is suspended from a point 25 mm above the centre
of small end, and 650 mm above its centre of gravity, its mass being
37.5 kg. When permitted to oscil- late, the time period is found to be
1.87 seconds. Find the dynamical equivalent system constituted of two
masses, one of which is located at the small end centre.
Solution.
Given : h = 650 mm = 0.65 m ; l1 = 650–25 = 625 mm = 0.625 m ; m =
37.5 kg ; tp = 1.87 s
First of all, let us find the radius of gyration (kG) of the connecting rod
(considering it is a compound pendulum), about an axis passing
through its centre of gravity, G. We know that for a compound
pendulum, time period of oscillation (tp),
8. It is given that one of the masses is located at the small end centre. Let the
other mass is located at a distance l2 from the centre of gravity G, as shown
in Fig. 15.19. We know that, for a dynamicallyequivalentsystem,
l1.l2 = (kG)2
