RBD



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  • @Ilham-Syed

    applying linear and angular impulse momentum theorem. linear -: -mVncm - (-mVcm) =I Vncm= new velocity of COM sign convention-: up=+ down=- hence-: -mLw2/2 + mLw1/2 =I angular impulse theorem about the first end of rod-: -mVncmL/2 + mL^2w2/12 -
    (-mL^2w1/3)=IL hence-: -mL^2w2/4 + mL^2w2/12 + mL^2w1/3=LI I= -mLw2/6 + mLw1/3 equating with previous equation-: -mLw2/2 + mLw1/2 = -mLw2/6 + mLw1/3 hence mLw1/6 = mLw2/3 hence -: w2=w1/2


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