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# a ring of mass m and radius r has three particles attached to the ring as shown. the center of the ring has a speed V0. the kinetic energy of the system is : (slipping is absent)

509 Points
10 years ago

velocity of left  side particle is 21/2Vo

velocity of right side particle is 21/2Vo

velocity os particle at top is 2Vo

KE of ring = 1/2( Iw2+ mVo2) = 1/2(2mVo2)

KE of particles = 1/2(2M2Vo2+M(2Vo)2+M2Vo2) = 1/2(10MVo2)

total KE = 1/2(2mVo2+ 10MVo2) = Vo2(m+5M)

Rakesh kumar
11 Points
4 years ago
velocity of left side particle is 21/2Vovelocity of right side particle is 21/2Vovelocity os particle at top is 2VoKE of ring = 1/2( Iw2+ mVo2) = 1/2(2mVo2)KE of particles = 1/2(2M2Vo2+M(2Vo)2+M2Vo2) = 1/2(10MVo2)total KE = 1/2(2mVo2+ 10MVo2) = Vo2(m+5M)
Manish
13 Points
2 years ago
Kinetic energy of 3 particles +kinetic energy of ring
(1/2mv2+1/2mr2w2)+(1/2*2m*v2+1/2mr2w2)+1/2*m(v+v)2+1/2mr2w2)+(1/2mv2+1/2mr2w2)

=5mv2+mv
=6mv2
Rishi Sharma
askIITians Faculty 646 Points
one year ago
Dear Student,
Please find below the solution to your problem.

Since slipping is absent vo​ = Rωo​
Speed of the 2m particle : square root (vo​2+Rωo​2)​
= vo*square root (2​)​
Speed of the m(right) particle : square root(vo​2+Rωo​2) ​
= ​vo*square root(2)​
Speed of the m(top) particle : vo​+Rωo​ = 2vo​
KE(total) = sum of individual KE
KE = 0.5*​mvo^2​ + 0.5*​(mR^2)(Rvo​​)^2 + 0.5*​(3m)(2​vo​)^2 + 0.5*​(m)(2vo​)^2
= 6mvo^2​
∴x=6

Thanks and Regards