Question

At the instant the displacement of a 2.00kg object relative to the origin is $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{d}}\mathbf{=}\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{m}\mathbf{)}\hat{\mathbf{i}}\mathbf{+}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{m}\mathbf{)}\hat{\mathbf{j}}\mathbf{-}\mathbf{(}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{m}\mathbf{)}\hat{\mathbf{k}}$, its velocity is

$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}\mathbf{=}\mathbf{(}\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{)}\hat{\mathbf{i}}\mathbf{+}\mathbf{(}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{)}\hat{\mathbf{j}}\mathbf{-}\mathbf{(}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{)}\hat{\mathbf{k}}$ and it is subject to a force$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{F}}\mathbf{=}\mathbf{(}\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{N}\mathbf{)}\hat{\mathbf{i}}\mathbf{-}\mathbf{(}\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{N}\mathbf{)}\hat{\mathbf{j}}\mathbf{+}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{N}\mathbf{)}\hat{\mathbf{k}}$(a) Find the acceleration of the object. (b) Find the angular momentum of the object about the origin. (c) Find the torque about the origin acting on the object. (d) Find the angle between the velocity of the object and the force acting on the object.

Short answer

1. The acceleration of the object is$\stackrel{\rightharpoonup }{a}=\left(3.00m/s^2\right)\hat{i}-\left(4.00m/s^2\right)\hat{j}+\left(2.00m/s^2\right)\hat{k}$
2. The angular momentum of the object is
   $\stackrel{\rightharpoonup }{a}=\left(42.0kg\frac{m^2}{s}\right)\hat{i}+\left(24.0kg\frac{m^2}{s}\right)\hat{j}+\left(60.0kg\frac{m^2}{s}\right)\hat{k}$
3. The torque acting on the object is $\stackrel{\rightharpoonup }{t}=\left(-8.00N.m\right)\hat{i}-\left(26.0N.m\right)\hat{j}-\left(40.0N.m\right)\hat{k}$
4. The angle between the velocity and force acting on the object is$\theta ={127}^o$

Step-by-step solution

Identification of given data

$$\begin{gathered} m=2.00kg \\ \stackrel{\rightharpoonup }{d}=\left(2.00m\right)\hat{i}+\left(4.00m\right)\hat{j}-\left(3.00m\right)\hat{k} \\ \stackrel{\rightharpoonup }{v}=\left(-6.00m/s\right)\hat{i}+\left(3.00m/s\right)\hat{j}-\left(3.00m/s\right)\hat{k} \\ \stackrel{\rightharpoonup }{F}=\left(6.00N\right)\hat{i}-\left(8.00N\right)\hat{j}+\left(4.00N\right)\hat{k} \end{gathered}$$

To understand the concept

The problem deals with the calculation of angular momentum. The angular momentum of a rigid object is product of the moment of inertia and the angular velocity. It is analogous to linear momentum. Also it involves Newton’s second law of motion. It states that the acceleration of an object is dependent upon the net force acting upon the object and the mass of the object.Use the expression of acceleration, angular momentum, and torque acting on the object and find their values. By using dot product of velocity vector and force vector, find the angle between them.

Formulae:

$\begin{array}{rcl}\stackrel{\rightharpoonup }{F}& =& m\stackrel{\rightharpoonup }{a}\\ \stackrel{\rightharpoonup }{l}& =& m\left(\stackrel{\rightharpoonup }{r}\times \stackrel{\rightharpoonup }{v}\right)\\ \stackrel{\rightharpoonup }{t}& =& \stackrel{\rightharpoonup }{r}\times \stackrel{\rightharpoonup }{F}\\ \stackrel{\rightharpoonup }{v}.\stackrel{\rightharpoonup }{F}& =& vF\cos \theta \end{array}$

(a) Determining the acceleration of the object

The expression of the force acting on the object is,

$\begin{array}{rcl}\stackrel{\rightharpoonup }{F}& =& m\stackrel{\rightharpoonup }{a}\\ \stackrel{\rightharpoonup }{a}& =& \frac{\stackrel{\rightharpoonup }{F}}{m}\\ & =& \frac{\left(6.00N\right)\hat{i}-\left(8.00N\right)\hat{j}+\left(4.00N\right)\hat{k}}{2.00kg}\\ & =& \left(3.00m/s^2\right)\hat{i}-\left(4.00m/s^2\right)\hat{j}+\left(2.00m/s\right)\hat{k}\end{array}$

(b) Determining the angular momentum of the object

Let position vector be $\stackrel{\rightharpoonup }{r}=x\hat{i}+y\hat{j}+z\hat{k}$and velocity vector be $\stackrel{\rightharpoonup }{v}=v_x\hat{i}+v_y\hat{j}+v_z\hat{k}$.

The cross product of the position vector and velocity vector is,

$\stackrel{\rightharpoonup }{r}\times \stackrel{\rightharpoonup }{v}=\left(y{v}_z-z{v}_y\right)\hat{i}+\left(z{v}_x-x{v}_z\right)\hat{j}+\left(x{v}_y-y{v}_x\right)\hat{k}$

The expression of the angular momentum of the object is

role="math" localid="1661343441140" width="410" height="188">l⇀=mr⇀×v⇀=myvz-zvyi^+zvx-xvzj^+xvy-yvxk^=2.004.00m×3.00m/s--3.00m×3.00m/si^+-3.00m×-6.00m/s-2.00m×3.00m/sj^+2.00m×3.00m/s-4.00m×-6.00m/sk^=42.0kgm2si^+24.0kgm2sj^+60.0kgm2sk^

(c) Determining the torque acting on the object

To find torque acting on the object, force acting on it will be $\stackrel{\rightharpoonup }{F}={\stackrel{\rightharpoonup }{F}}_x\hat{i}+{\stackrel{\rightharpoonup }{F}}_y\hat{j}+{\stackrel{\rightharpoonup }{F}}_z\hat{k}$.

The expression of torque is

role="math" localid="1661343902025" width="638" height="140">τ⇀=r⇀×F⇀=yFz-zFyi^+zFx-xFzj^+xFy-yFxk^=4.00m×3.00N--3.00m×8.00Ni^+-3.00m×-6.00N-2.00m×4.00Nj^+2.00m×3.00m/s-4.00m×-6.00m/sk^=-8.00N.mi^-26.0N.mj^-40.0N.mk^

(d) Determinng the angle between the velocity and force acting on the object

To find the angle between the velocity and force acting on the object, we can use dot product of two vectors.

The magnitude of force is

$\begin{array}{rcl}F& =& \sqrt{F_x^2+F_y^2+F_z^2}\\ & =& \sqrt{{\left(6.00N\right)}^2+{\left(-8.00N\right)}^2+{\left(4.00N\right)}^2}\\ & =& 10.8N\end{array}$

The magnitude of velocity is

$\begin{array}{rcl}v& =& \sqrt{v_x^2+v_y^2+v_z^2}\\ & =& \sqrt{{\left(-6.00m/s\right)}^2+{\left(3.00m/s\right)}^2+{\left(3.00m/s\right)}^2}\\ & =& 7.35m/s\end{array}$

According to the property of scalar dot product

$\begin{array}{rcl}\stackrel{\rightharpoonup }{v}.\stackrel{\rightharpoonup }{F}& =& v_x{F}_x+v_y{F}_y+v_z{F}_z\\ & =& \left(-6.00m/s\right)\left(6.00N\right)+\left(3.00m/s\right)\left(-8.00N\right)+\left(3.00m/s\right)\left(4.00N\right)\\ & =& -48Nm/s\end{array}$

The dot product of two vectors is

$\begin{array}{rcl}\stackrel{\rightharpoonup }{v}.\stackrel{\rightharpoonup }{F}& =& vF\cos \theta \\ \cos \theta & =& \frac{\stackrel{\rightharpoonup }{v}.\stackrel{\rightharpoonup }{F}}{vF}\\ & =& \frac{-48N.m/s}{7.35{\displaystyle \frac{m}{s}}\times 10.8N}\end{array}$

$\Rightarrow \theta ={127}^o$