The position vector of a particle of mass m=6kg is given as →r=[(3t2−6t)^i+(−4t3)^j]m. Find
(i) the force (→F=m→a) acting on the particle.
(ii) the torque (→τ=→r×→F) with respect to the origin, acting on the particle.
(iii) the momentum (→p=m→v) of the particle.
(iv) the angular momentum (→L=→r×→p) of the particle with respect to the origin.
→v=d→rdt=(6t−6)^i+(−12t2)^jm/s
→a=d→vdt=(6^i−24t^j)m/s2
(i)→F=m→a=6(6^i−24t^j)=(36^i−144t^j)N
(ii)→τ=→r×→F=[(3t2−6t)^i+(−4t3)^j]×[36^i−144t^j]
=[(−144×3t2)+(144×6T2)+144t3]^k
=(−288t3+864t2)^k
(iii)→p=m→v=6[(6t−6)^i+(−12t2)^j]
=[36(t−1)^i−72t2t2^j]
(iv)→L=→r×→p=[(3t2−6t)^i+(−4t3)^j]×[36(t−1)^i−72t2^j]
=[−72t4+288t3]^k