2* 1.5 * (.850/2)^2a small ball with mass 1.50 kg is mounted on one end of a rod 0.850 m long and of negligible mass. the system rotates in a horizontal circle about the other end of the rod at 5060 rev/min.

Answer

given,

mass of the rod = 1.50 Kg

length of rod = 0.85 m

rotational velocity = 5060 rev/min

now calculating the rotational inertia of the system.

where L is the length of road, we will take whole length of rod because mass is at  the end of it.      

I = 1.084 kg.m²                        

hence, the rotational inertia the system is equal to I = 1.084 kg.m²

well clustered together.

points close on a scatter diagram