The position vector of a particle of mass m= 6kg is given as `vec(r)=[(3t^(2)-6t) hat(i)+(-4t^(3)) hat(j)] m`. Find:
(i) The force `(vec(F)=mvec(a))` acting on the particle.
(ii) The torque `(vec(tau)=vec(r)xxvec(F))` with respect to the origin, acting on the particle.
(iii) The momentum `(vec(p)=mvec(v))` of the particle.
(iv) The angular momentum `(vec(L)=vec(r)xxvec(p))` of the particle with respect to the origin.

Answer

(i) `36hat(i)-144 t hat(j)` (ii) `(-288 t^(3)+864 t^(2)) hat(k)` (iii) `(36t-36)hat(i)-72 t^(2) hat(j)` (iv) `-72 t^(4)+288 t^(3)hat(k)`

Explanation

`vecv=(dvecr)/(dt)=(6t-6)hati+(-12t^2)hatj m//s`
`a=(dvecr)/(dt)=(6t-24tjatj)m//s^2`
(i)`vecF=mveca=6(6hati-24hatj)=(36hati-144thatj)N`
(ii)`vectau=vecrxxvecF=[(3t^2-6t)hati+(-4t^3)hatj]xx[36hati-144thatj]`
`=[(-144xx3t^2)+(144xx6t^2)+144t^3]hatk`
`=(-288t^3+864t^2)hatk`
(iii)`vecp=mvecv=6[(6t-6)hati+(-12t^2)hatj]`
`[36(t-1)hati-72t^2hatj]`
(iv)`vecL=vecrxxvecp=[(3t^2-6t)hati+(-4t^3)hatj]xx[36(t-1)hati-72t^2hatj]`
`=[-72t^4+288t^3]hatk`