A light rod 1.00 m in length rotates about an axisperpendicular to its length an

Question

A light rod 1.00 m in length rotates about an axisperpendicular to its length and passing through its center. twoparticles of masses 4.00 kg and 3.00 kg are connected to the endsof the rod
neglecting the mass of the rod, what is the sytems's kineticenergy when its angular speed is 2.50 rad/s? _____________J? if the mass of the the rod is taken to be 2.20 kg, what is thekinetic energy? ___________________J A light rod 1.00 m in length rotates about an axisperpendicular to its length and passing through its center. twoparticles of masses 4.00 kg and 3.00 kg are connected to the endsof the rod
neglecting the mass of the rod, what is the sytems's kineticenergy when its angular speed is 2.50 rad/s? _____________J? if the mass of the the rod is taken to be 2.20 kg, what is thekinetic energy? ___________________J

Explanation / Answer

Ke= I To find moment of inertia I we have to use parallel axistheorem It=Ii   since we assume theu arepoint masses we have I i= miRi2 For case 1 It= m1R12 +m2R22 It = 4.0 (0.5)2 +3.0(0.5)2   = 1.75 kg m2 Ke= It= 2.5 x 1.75 Ke= 4.375 J - For case 2 I rod = 1/12 mL2 It= m1R12 +m2R22   + 1/12mL2 It= 1.75 + (1/12) 2.2 (1.0)2 It= 1.93 kg m2 K= 1.93 x 2.5 Ke= 4.83 J K= 1.93 x 2.5 Ke= 4.83 J

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