At time t = 0, a particle of mass m has position vector (r^=4i-2j^ ) relative to the origin, in meters. For t ≥ 0, its velocity is given by v^=-6t^2i^+4j^.Find expressions of the z component of (a) the particle's angular momentum and (b) the torque acting on the particle at time t.

Question

t is in s and v in m/s
someone pls help ;)

Answer 1

angular momentum L = R cross P = R cross m V
R(0) = 4 i - 2 j at t = 0
R(t) = R(0) + integral V dt from t = 0 to t = t
R(t) = 4 i - 2j -6 integral t^2 dt i^+4 integral dt j
R(t) = (4-2t^3)i + 4 t j
here P = m V = m [ -6 t^2 i + 4 j]
so L =R cross mV
L =m [ 4(4-2t^3)+24t^3] k = m [16-8t^3+24 t^3]
L =16 m[ 1-t^3]
and then
torque = d/dt L
= m[-48 t^2 ]
once again check my arithmetic, I stand behind method only

Answer 2

Are you at MIT? (That sort of problem)

Answer 3

Thank you for help, I’m not at MIT I’m studying engineering tho.