a light rod of length l is pivoted at upper end.two masses (each m) are attached to rod one at middle and other at free end .what horizontal velocity must be imparted to lower end mass .so tht the rod may just take up horizontal position underrot 6gl 5

Suppose the velocity should be v v . So the initial kinetic energy is

12mv2 frac{1}{2}mv^2

When it will be horizontal the potential energy is mgl2+mgl=mg3l2 mgfrac{l}{2}+mgl=mgfrac{3l}{2} .

Now this should be equal to

v2=g3lv=3gl v^2=g3l Rightarrow v=sqrt{3gl}

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