Question

A particle of mass 'm' is projected on horizontal ground with an initial velocity of 'u' making an angle θ with horizontal. Find out the angular momentum of particle about the point of projection when, it just strikes the ground.

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Solution

As shown in figure , Just strikes the ground ux = ucos(θ) , uy=usin(θ), Time of flight = 2usin(θ)g, ∴ Range R=ucos(θ)× 2usin(θ)g,⇒R=u2sin(2θ)g,Angular Momentum ¯L=r×mu,Here r=R=u2sin(2θ)g^i, mu=m(ucos(θ)^i−usin(θ)^j)∴¯L=mu3sin(2θ)g×(cos(θ)^i×^i−sin(θ)^i×^j)⇒¯L=mu3sin(θ)sin(2θ)g(−^k)

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