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A particle of mass `m` is released from rest at point `A` in Fig., falling parallel to the (vertical) `y`-axis. Find the angular momentum of the particle at any time `t` with respect to the same origin `O`. |
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Answer» `vecL=vecrxxvecp` The magnitude of `vecL` is given by `L=rp sintheta(hatk)` in this example velocity of the particle at any time `v="gt",rsintheta=x_(0)` and `p=mv=m"gt"` `:.vecL=mgx_(0)t(hatk)`……..i The right hand rule shows that `vecL` is parallel to torque, i.e. direcrected perpedicular to paper in wards. |
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