A ring of mass m and radius r has three particles attached to the ring as shown in the figure. The centre of the ring has a speed Vo .The kinetic energy of system is
Answers
Answered by
0
Answer:
The answer will be 6mvo^2
Explanation:
According to the problem
the right most particle has the speed of 2^1/2vo
the lft most particle has the speed of 2^1/2vo
the middle particle has the speed of 2vo
It is mentioned that the speed of the ring in the center is v0
Therefore the kinetic energy of the ring= 1/2( Iw^2+ mvo^2) = 1/2(2mvo^2)
Therefore the kinetic energy of the three particles = 1/2(2m^2vo^2+m(2vo)^2+m^2vo^2) = 1/2(10mvo^2)
Therefore the total kinetic energy of the system = 1/2(2mvo^2+ 10mvo^2) = vo^2(m+5m)= 6mvo^2
Similar questions