HSC Board

A big dumb-bell is prepared by using a uniform rod of mass 60 g and length 20 cm. Two identical solid thermocol spheres of mass 25 g and radius 10 cm each are at the two ends of the rod. Calculate moment of inertia of the dumb bell when rotated about an axis passing through its centre and perpendicular to the length.

As per the given condition we can list out the data as

mass of rod = M= 60g

length of the rod = L = 20cm

mass of sphere , m = 25g

radius of sphere , r= 10cm

a dumbell is formed with rod length 20cm, at both the ends 10 cm radius sphere are attatched ,

The moment of inertia of the rod is given by

$I_{r}=\frac{ML^{2}}{12}$

$I_{r}=\frac{60\mathrm{x}20^{2}}{12}$

$I_{r}=2000 g cm^{^{2}}$

Moment of inertia of sphere through diameter ,

$I_{s}=\frac{2mr^{2}}{5}$

$I_{s}=\mathrm{\frac{2 x 25 x 10^{2}}{5}}$

$I_{s}=1000 gcm^{2$

Total moment of inertia of the dumbel is ,

I = Moment of inertia of rod + 2( Moment of Inertia of sphere + mh² )

h = 20cm = half distance of rod + radius of sphere

I = 2000 + 2( 1000 + 25 x 20² )

I = 2000 + 2( 1000 + 25 x 400)

I = 2000 + 2( 1000 + 10000 )

I = 2000 + 2(11000)

I = 2000 + 22000

I = 24000 gcm²

I = 0.0024 kg m²

HSC Board rotational dynamics