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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)
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)
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)
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)
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+mv2 =6mv2
Dear Student,Please find below the solution to your problem.Since slipping is absent vo = RωoSpeed of the 2m particle : square root (vo2+Rωo2)= vo*square root (2)Speed of the m(right) particle : square root(vo2+Rωo2) = vo*square root(2)Speed of the m(top) particle : vo+Rωo = 2voKE(total) = sum of individual KEKE = 0.5*mvo^2 + 0.5*(mR^2)(Rvo)^2 + 0.5*(3m)(2vo)^2 + 0.5*(m)(2vo)^2= 6mvo^2∴x=6Thanks and Regards
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