The angular momentum of a particle relative to a certain point O varies with time as `M=a+bt^2`, where a and b are constant vectors, with `a_|_b`. Find the force moment N relative to the point O acting on the particle when the angle between the vectors N and M equals `45^@`.

Explanation

Force moment relative to point O,
`vecN=(dvecM)/(dt)=2vec(bt)`
Let the angle between `vecM` and `vecN`,
`alpha=45^@` at `t=t_0`,
Then `1/sqrt2=(vecM*vecN)/(|vecM||vecN|)=((veca+vecbt_0^2)*(2vecbt_0))/(sqrt(a^2+b^2t_0^4)2bt_0)`
`=(2b^2t_0^3)/(sqrt(a^2+b^2t_0^4)2bt_0)=(bt_0^2)/(sqrt(a^2+b^2t_0^4))`
So, `2b^2t_0^4=a^2+b^2t_0^4`, or, `t_0=sqrt(a/b)` (as `t_0` cannot be negative)
It is also obvious from the figure that the angle `alpha` is equal to `45^@` at the moment `t_0`, when `a=bt_0^2`, i.e. `t_0=sqrt(a//b)` and `vecN=2sqrt(a/b)vecb`.