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Obtain tau=Ialpha from angular momentum of rigid body. |
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Answer» Solution :The total angular momentum of a motion of rigid body about a fixed axis `vecL=VEC(L_(z))+vec(L_(_|_))` Differentiating w.r.t. time `(dvecL)/(DT)=(DVEC(L_(z)))/(dt)+(dvec(L_(_|_)))/(dt)` but `(dvec(L_(z)))/(dt)=tauhatk and (dvec(L_(_|_)))/(dt)=0` `therefore (dvec(L))/(dt)=tauhatk` but `vec(L_(Z))=Iomegahatk` `therefore (d(Iomega))/(dt)=tauhatk` `I(domega)/(dt)=tau` `therefore Ivecalpha=vectauimpliesvectau=Ivecalpha` Scalar form `tau=Ialpha` is similar to `F=ma` in linear motion. |
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