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a uniform thin rod of length l and mass m is allowed to rotate on a frictionless pin passing through one end. The rod is released from rest in the horizontal position. a.) What is the speed of the center of gravity when the rod reaches its lowest position? b.) What is the tangential speed of the lowest point of the rod when the rod reaches its lowest position?

a uniform thin rod of length l and mass m is allowed to rotate on a frictionless pin passing through one end The rod is released from rest in the horizontal pos class=

Respuesta :

Answer:

Explanation:

Potential energy gets converted to rotational kinetic energy

a)         ½Iω² = mgh

½(mL²/3)ω² = mgL/2

       (L/3)ω² = g

              ω = [tex]\sqrt{3g/L}[/tex]

       v(CG) = (L/2) [tex]\sqrt{3g/L}[/tex]

Not sure if you wanted angular speed or tangential speed of the CG so I gave both.

b) v = Rω =  L [tex]\sqrt{3g/L}[/tex]

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