Mechanics> Rotation...

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)

Ajay Kumar , 14 Years ago

Grade 11

4 Answers

vikas askiitian expert

Last Activity: 14 Years ago

velocity of left side particle is 2^1/2V_o

velocity of right side particle is 2^1/2V_o

velocity os particle at top is 2V_o

KE of ring = 1/2( Iw²+ mV_o²) = 1/2(2mV_o²)

KE of particles = 1/2(2M2V_o²+M(2V_o)²+M2V_o²) = 1/2(10MV_o²)

total KE = 1/2(2mV_o²+ 10MV_o²) = V_o²(m+5M)

Rakesh kumar

Last Activity: 8 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

Last Activity: 6 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+mv²

=6mv²

Rishi Sharma

Last Activity: 4 Years ago

Dear Student,
Please find below the solution to your problem.

Since slipping is absent vo = Rωo
Speed 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 = 2vo
KE(total) = sum of individual KE
KE = 0.5*mvo^2 + 0.5*(mR^2)(Rvo)^2 + 0.5*(3m)(2vo)^2 + 0.5*(m)(2vo)^2
= 6mvo^2
∴x=6

Thanks and Regards