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)

Question

vikas askiitian expert · Answer

velocity of left side particle is 2 1/2 V o velocity of right side particle is 2 1/2 V o velocity os particle at top is 2V o KE of ring = 1/2( Iw 2 + mV o 2 ) = 1/2(2mV o 2 ) KE of particles = 1/2(2M2V o 2 +M(2V o ) 2 +M2V o 2 ) = 1/2(10MV o 2 ) total KE = 1/2(2mV o 2 + 10MV o 2 ) = V o 2 (m+5M)

Rakesh kumar · Answer

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 · Answer

Kinetic energy of 3 particles +kinetic energy of ring (1/2mv 2 +1/2mr 2 w 2 )+(1/2*2m*v 2 +1/2mr 2 w 2 )+1/2*m(v+v) 2 +1/2mr 2 w 2 )+(1/2mv 2 +1/2mr 2 w 2 ) =5mv 2 +mv 2 =6mv 2

Rishi Sharma · Answer

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

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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)

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)

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