So one rod has no mass except for the mass sitting on the free end. The other ro

Question

So one rod has no mass except for the mass sitting on the free end. The other rod has a mass that is uniform throughout. How do I find the rotational KE for rod A and rod B?

Two thin rods of length L are rotating with the same angular speed ? (in rad/s) about axes that pass perpendicularly through one end. Rod A is massless but has a particle of mass 0.67 kg attached to its free end. Rod B has a mass of 0.67 kg, which is distributed uniformly along its length. The length of each rod is 0.84 m, and the angular speed is 5.0 rad/s. Find the kinetic energies of rod A with its attached particle and of rod B.

KE a= ______ KE b= _____

Explanation / Answer

Data: Mass of the particle, M = 0.67 kg Mass of rod B, m = 0.67 kg Length of each rod, L = 0.84 m Angular speed, = 5.0 rad / s Solution: The moment of inertia for rod A is just that of the attached particle, since the rod itself is massless.
For rod A with its attached particle, then, the moment of inertia is                              IA = M L 2   Moment of inertia for rod B is .                              IB = ( 1/3) m L 2 Rotational kinetic energy is   KE = ( 1/2) I 2 (a) Kinetic energy of rod A:                KE A = ( 1/2) I A 2                              = (0.5)( M L 2 ) 2                            = ( 0.5 ) ( 0.67 kg) [ (0.84 m)2 ] (5.0 rad)2                             = 5.91 J   (b) Kinetic energy of rod B:               KE B = ( 1/2) I B 2                              = (0.5) ( m L 2 /3 ) 2                            = ( 0.5 ) (0.67 kg) [ (0.84 m)2 / 3 ] ( 5.0 rad/s)2                             = 1.97  J                              = ( 0.5 ) ( 0.67 kg) [ (0.84 m)2 ] (5.0 rad)2                             = 5.91 J   (b) Kinetic energy of rod B:               KE B = ( 1/2) I B 2                              = (0.5) ( m L 2 /3 ) 2                            = ( 0.5 ) (0.67 kg) [ (0.84 m)2 / 3 ] ( 5.0 rad/s)2                             = 1.97  J                               = 5.91 J   (b) Kinetic energy of rod B:               KE B = ( 1/2) I B 2                              = (0.5) ( m L 2 /3 ) 2                            = ( 0.5 ) (0.67 kg) [ (0.84 m)2 / 3 ] ( 5.0 rad/s)2                             = 1.97  J                              = (0.5) ( m L 2 /3 ) 2                            = ( 0.5 ) (0.67 kg) [ (0.84 m)2 / 3 ] ( 5.0 rad/s)2                             = 1.97  J                              = ( 0.5 ) (0.67 kg) [ (0.84 m)2 / 3 ] ( 5.0 rad/s)2                             = 1.97  J                               = 1.97  J  

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