At time t, r^=4t^2 i^-(2t+6t^2)j^ gives the position of a particle of mass m relative to the origin of an xy coordinate system Find an expression for the (a) x, (b) y, and (c) z components of the torque acting on the particle relative to the origin.

Question

At time t, r^=4t^2 i^-(2t+6t^2)j^  gives the position of a particle of mass m relative to the origin of an xy coordinate system  Find an expression for the (a)  x, (b)  y, and (c)  z components of the torque acting on the particle relative to the origin.

Damon · Accepted Answer

since motion is in x,y plane, torque is about z axis

what is linear acceleration?
dr/dt = 8 t i -( 2+12 t)j
d^2r/dt^2 = 8 i -12 j constant linear acceleration so constant linear force
F = m (8 i - 12 j) = 4m (2 i - 3j)

now what is F cross r , the torque
i j k
rx ry 0
2 -3 0
all times 4 m
4m (-3rx -2 ry)
in k which is z direction
4 m [ -3(4t^2) +2(2t+6t^2)]
4 m [ 12 t^2 - 12 t^2 + 4 t]
16 m t

Damon · Answer

sorry, sign reversed moment = R cross F not F cross R it has been a while check all my arithmetic