a particle of mass m is rigidly attached to a uniform disc of mass m and radius R at its periphery ,: disc at this moment is rolling without slipping on a fixed horizontal surface. if the speed of the centre of the disc is v, then the total kinetic energy of the system at this instant will be what?
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The total kinetic energy of the system at this instant will be Vo2(m+5M)
Explanation:
Give data:
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 K.E = 1/2(2mVo2+ 10MVo2) = Vo2(m+5M)
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