91Ó°ÊÓ

Magnitude of position vector r = 3.0m

Force magnitude F = 2.0N

Mass of particle m = 2.0kg

Velocity vector v = 4.0m/s

All the three vectors lie in the same plane. Mass of particle is also given. And by using the conditions given in problem, it can be concluded that velocity vector is perpendicular to position vector. By using given angles, find the value of angular momentum. By using formula for torque, find the value of torque.

Formulae:

The angular momentum of a body, l = rmv

Where, m is the mass of the object, r is the radial distance of the object, v is the velocity of the object.

The torque acting on a body, $\mathrm{τ}=rF\sin \mathrm{θ}$

Where, F is the force, $\mathrm{θ}$is the angle between the radial vector and force.

The equation is

$\begin{array}{rcl}l& =& rmv\\ & =& \left(3.0m\right)\left(2.0kg\right)\left(4.0m/s\sin \mathrm{θ}\right)\\ & =& 12.0kg.m^2/s\end{array}$

If velocity vector and position vector are in the plane, then by assuming right hand rule, the angular momentum is out of paper, that is, perpendicular to the plane of figure.

$\begin{array}{rcl}\mathrm{τ}& =& rFs\sin \mathrm{θ}\\ & =& \left(3.0m\right)\left(2.0N\right)\left(\sin 30\right)\\ & =& 3.0N.m\end{array}$

By using the right-hand thumb rule, if point our fingers in the direction of radial vector then, we curl our fingers in the direction of the force, we can get the direction of torque. Hence, the torque is perpendicular to plane of the paper.