At the instant of the figure, a 6.50 kg particle P has a position vect

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At the instant of the figure, a 6.50 kg particle P has a position vector r→ of magnitude 1.30 m and angle θ1 = 45.0∘ and a velocity vector v→ of magnitude 6.90 m/s and angle θ2 = 31.0∘. Force F→, of magnitude 3.60 N and angle θ3 = 31.0∘ acts on P. All three vectors lie in the xy plane. About the origin, what are the magnitude of (a) the angular momentum of the particle and (b) the torque acting on the particle?

Image text

At the instant of the figure, a

6.50 k g

particle

P

has a position vector

\vec{r}

of magnitude

1.30 m

and angle

θ_{1} = {45.0}^{\circ}

and a velocity vector

\vec{v}

of magnitude

6.90 m / s

and angle

θ_{2} = {31.0}^{\circ}

. Force

\vec{F}

, of magnitude

3.60 N

and angle

θ_{3} = {31.0}^{\circ}

acts on

P

. All three vectors lie in the xy plane. About the origin, what are the magnitude of (a) the angular momentum of the particle and (b) the torque acting on the particle?