Question

Four particles each of mass m are placed at four corner of square of side l. The radius of gyration

Answer 1

Given :

Mass of the particles = m

Length of the side of square = l

To Find :

Radius of gyration

Solution :

• The distance of particles placed at corners from the center of the mass of the square is l/√2

• The total moment of inertia of the system is given by

         $I=m\times (\frac{1}{\sqrt{2} })^{2}+ m\times (\frac{1}{\sqrt{2} })^{2}+ m\times (\frac{1}{\sqrt{2} })^{2}+ m\times (\frac{1}{\sqrt{2} })^{2}$

        $I=4 m\times (\frac{1}{\sqrt{2} })^{2}$

        $I=2ml^{2}$

• By comparing this with the equation I=mK² we get k=l/√2

The radius of gyration of the system is l/√2

 

Answer 2

Explanation:

please repeat we compare mk2 =2ml2