Three particles of mass m,2m,3m repectively are rigidly attached to a ring of mass m and radius r which rolls on a horizobtal surface without slippong velocity of center of mass of the system is
Answers
The question is incomplete.
However,
Given:
Particle a mass = m
Particle b mass = 2 m
Particle c mass = 3 m
Ring mass = m
Ring radius = r
Velocity of the centre of a ring = v0
To find:
The kinetic energy
Solution:
Velocity of a = √2 v
Velocity of b = 2 v
Velocity of c = √v
By formula,
Kinetic energy = 1/2 m v^2
Substituting the values,
We get,
Kinetic energy of the particles= (1/2 × 2 m × 4 v^2) + (1/2 × m × v^2) + (1/2 × 3 m × 2 v^2)
Hence, The total Kinetic energy of system = 9 m v^2
Answer:9mv^2
Explanation:
Given:
Particle a mass = m
Particle b mass = 2 m
Particle c mass = 3 m
Ring mass = m
Ring radius = r
Velocity of the centre of a ring = v0
To find:
The kinetic energy
Solution:
Velocity of a = √2 v
Velocity of b = 2 v
Velocity of c = √v
By formula,
Kinetic energy = 1/2 m v^2
Substituting the values,
We get,
Kinetic energy of the particles= (1/2 × 2 m × 4 v^2) + (1/2 × m × v^2) + (1/2 × 3 m × 2 v^2)
Hence, The total Kinetic energy of system = 9 m v^2