A slender uniform rod of mass M and length
$fraction numerator 3 straight g over denominator 2 calligraphic l end fraction space cos space straight theta$
$fraction numerator 2 space straight g over denominator 3 space calligraphic l end fraction space cos space straight theta$
$fraction numerator 3 straight g over denominator 2 space calligraphic l end fraction space sin space straight theta$
$fraction numerator 2 space straight g over denominator 3 space calligraphic l end fraction space sin space straight theta$

Solution

fraction numerator 3 straight g over denominator 2 space calligraphic l end fraction space sin space straight theta

Taking torque about pivot τ = Iα

mg space sin space straight theta space calligraphic l over 2 space equals space fraction numerator straight m calligraphic l squared over denominator 3 end fraction space straight alpha straight alpha space equals space fraction numerator 3 straight g over denominator 2 space calligraphic l end fraction space sinθ

Some More Questions From Units and Measurement Chapter

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