Question

At the instant of Figure, a 2.0kg particle Phas a position vector $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{r}}$ of magnitude 3.0mand angle ${\mathit{\theta }}_{\mathbf{1}}\mathbf{=}{\mathbf{45}}^{\mathbf{o}}$ and a velocity vector$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}$ of magnitude 4.0m/sand angle${\mathit{\theta }}_{\mathbf{2}}\mathbf{=}{\mathbf{30}}^{\mathbf{0}}$. Force$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{F}}$, of magnitude 2.0Nand angle${\mathit{\theta }}_{\mathbf{3}}\mathbf{=}{\mathbf{30}}^{\mathbf{o}}$acts on P. All three vectors lie in the xy plane. About the origin, (a) What is the magnitude of the angular momentum of P? (b) What is the direction of the angular momentum of P? (c) What is the magnitude of the torque acting on P? (d)What is the direction of the torque acting on P?

Short answer

1. The magnitude of angular momentum is 12.0kg.m²/s,
2. The direction of angular momentumis out of plane.
3. Magnitude of torque is 3.0 N.m.
4. The direction of torque is out of plane.

Step-by-step solution

Identification of given data

Magnitude of position vector r = 3.0m

Force magnitude F = 2.0N

Mass of particle m = 2.0kg

Velocity vector v = 4.0m/s

To understand the concept

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, $\tau =rF\sin \theta$

Where, F is the force, $\theta$is the angle between the radial vector and force.

(a) Determining the magnitude of angular momentum

The equation is

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

(b) Determining the direction of angular momentum

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.

(c) Determining the magnitude of torque

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

(d) Determining the direction of torque

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.