Two particles of masses 'm' and '2m' are
connected by massless string of length 3 meter
and whole system is rotating about their centre of
mass with angular speed 'w'. The kinetic energy of
system is
(1) m omega ^2
(3) 6 m omega ^2
(4) 3 m omega^2
Answers
answer : option (4) 3 m omagea²
explanation : first of all, we should find out centre of mass .
If we assume system of m and 2m is along horizontal line. where m is placed at origin and then 2m is placed (3, 0).
C.M = (3 × 2m + m × 0)/(m + 2m)
= 2 metres from mass m or (3 - 2) = 1m from mass 2m
now, moment of inertia of system of masses about centre of mass, I = I1 + I2
= m(2)² + 2m(1)²
= 4m + 2m = 6m
now, kinetic energy = 1/2 I w²
= 1/2 (6m) w²
= 3m w²
hence, option (4) is correct choice .
Answer:
Explanation:
Kinetic energy of the system = 1/2 (Iω²)
Where I is the moment of inertia of the system
To find the moment of inertia of the system we need to locate its centre of mass
Let centre of mass be located at distance x from the particle of mass m
then
(m+2m)x = m×0 + 2m × 3
or x = 6/3 = 2m (from the mass m)
The moment of inertia will be calculated around the axis passing through the centre of mass
Moment of Inertia I = m × 2² + 2m × 1² = 4m + 2m = 6m
Kinetic Energy = (1/2)Iω² = 1/2 × 6m × ω² = 3mω²