A person taking in his hands spheres each of mass 2kg is standing on a table revolving with an angular velocity of 10 rad/s.his arms are extended and each sphere is at a distance of 1m from the axis of rotation .if the moment of inertia of the person and the table about the axis of rotation be 1 kg m^2 then what will be the kinetic energy of rotation of the whole system.
Answers
Let m= mass of a sphere = 2 kg
r= distance between sphere & axis of rotation = 1m
Is= moment of inertia of a sphere = mr²
It= moment of inertia of the person and table = 1 kgm²
ω=angular velocity of the system = 10 rad/s
Kinetic energy of the system = Kinetic energy of 2 spheres +
Kinetic energy of the table & person
=2(Kinetic energy of a sphere) +
Kinetic energy of table & person
=2[(1/2)Isω²] + (1/2)Itω²
=2[ (1/2)X (mr²)Xω²] + (1/2)Itω²
=2[ (1/2)X (2x1²)X10²] + [(1/2)x1x10²]
=2[ (1/2)X 2X100] + [(1/2)x1x100]
=2[100 + 50]
Kinetic energy of the system = 75 J