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The slendar rod of masss m and length L is released from rest on a smoooth (frictionless) surface when 0= 60°. Determine the

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Answer #1

Okay, so we can solve it using work every theorem,

The final velocity of Centre of Mass which will move in a circular path will give us its rotational velocity (omega)

Solution Initial Ve=0 pusifon at Rest 0-60 Location of centre of Mass from Groound. = CA = I singo CA = singo Ос CA = (OC)OPE= tengah.com APE = 48x (10) x (-0:18304) --14:64010 WIDE - APE= - (-14.6404).- 146401. WID = OKE= kefinal - Kenntial = meOkay, I am assuming that this will suffice but for all queries please feel free to use the comment section. Happy learning.

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