Question

6. (BONUS) Two particles each with mass m = 0.4 kg, are fastened to each other, and to a rotation axis at 0, by the two thin rods, each of length d and mass M = 1.5 kg as shown below. The combination rotates around the rotation axis with angular speed w = 0.2 rad/s. The total moment of inertia of the system measured about O is 2.3 x 10-4 kg m?. (Hint: The moment of inertia of a thin rod with length L about axis through center perpendicular to length is I = ML. The moment of inertia of a particle of mass m and radius r is I = mr?). What is the value of length d ? (10 pt)

Answer 1

Moment of inertia of the combined rods = (M/3) (2d)² ......................(1)

( Both rods each having length d are combined by fastening the particle .

Hence combination of two rods is considered as single rod of length 2d .

Mass of this combination of tow rods will be M = 2 $\times$ 1.5 = 3 kg )

Moment of inertia of particle near to axis = m $\times$ d² ..........................(2)

Moment of inertia of particle at extreme end of combined rod = m $\times$ 4d² ................(3)

Moment of inertia of the system = (M/3) (2d)² + m $\times$ d² + m $\times$ 4d²

By substituting M = 3 kg and m = 0.4 kg in the above equation, we get

6 d² = 2.3 $\times$ 10^-3 ;

we get d $\approx$ 2 cm