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Derive the equation of angular momentum in the case of rotational motion about a fixed axis. |
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Answer» Solution :ANGULAR MOMENTUM of a particle of rigid body is in the direction of a fixed Z-axis `therefore vec(l_(1))=vec(r_(1))xxvec(p_(1))` If the rigid body consisting of n particle, then the vector sum of angular momentum of all particle will be obtain in the direction of Z-axis. `therefore vec(L_(Z))=vec(l_(1))+vec(l_(2))+vec(l_(3))+....vec(l_(n))` `=(vec(r_(1))xxvec(p_(1)))+(vec(r_(2))xxvec(p_(2)))+(vec(r_(3))xxvec(p_(3)))+....(vec(r_(n))xxvec(p_(n)))` `=underset(i=1)overset(n)sumvec(r_(i))xxvec(p_(i))"where "i=1,2,3,...,n` `=underset(i=1)overset(n)sumvec(r_(i))xxm_(i)vec(v)_(i)` `=underset(i=1)overset(n)summ_(i)r_(i)^(2)omegahatk"" [because v_(i)=r_(i)omega]` `=underset(i=1)overset(n)sumI_(i)omegahatk"where "underset(i=1)overset(n)summ_(i)r_(i)^(2)=I` but for all particle `SigmaI_(i)=I`, `vec(L_(Z))=Iomegahatk` where `hatk` is a UNIT vector in Z-axis. but angular momentum `vec(L_(1))` perpendicular to the axis in rigid body, then total angular momentum `vecL=vec(L_(Z))+vec(L_(_|_))` but `vec(L_(_|_))=0` `therefore vecL=vec(L_(Z))` `vecL=Iomegahatk` `therefore L=Iomega` SIMILAR to `p=mv` in linear motion. |
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