Question

A particle of mass m= 2.5 kg has velocity of v = 2 i m/s, when it is at the origin (0,0). Determine the z- component of the angular momentum of the particle about each of the following reference points. The coordinates of the reference points have units of (meters, meters). Assume the positive z-axis is directed out of the screen. Randomized Variables m= 2.5 kg v = 2 i m/sPart (a) Determine the angular momentum in kg.m-/s of the particle about (0,1). Part (b) Determine the angular momentum in kgm/s of the particle about (0, -1). Part (c) Determine the angular momentum in kg.m-/s of the particle about (1,0).Part (d) Determine the angular momentum in kg.m/s of the particle about (-1,0). Part (e) Determine the angular momentum in kg. m²/s of the particle about (1,1). Part (f) Determine the angular momentum in kg ·mº/s of the particle about (-1,1). Part (g) Determine the angular momentum in kg.mʻls of the particle about (-1,-1).Part (h) Determine the angular momentum in kg.m-/s of the particle about (1,-1). Part (i) Determine the angular momentum in kg.m/s of the particle about (0,0).

Answer 1

Answer:

please read the answer below

Explanation:

The angular momentum is given by

$|\vec{L}|=|\vec{r}\ X \ \vec{p}|=m(rvsin\theta)$

By taking into account the angles between the vectors r and v in each case we obtain:

a)

v=(2,0)

r=(0,1)

angle = 90°

$L=(2.5kg)(1)(2\frac{m}{s})sin90\°=5.0kg\frac{m}{s}$

b)

r=(0,-1)

angle = 90°

$L=(2.5kg)(1)(2\frac{m}{s})sin90\°=5.0kg\frac{m}{s}$

c)

r=(1,0)

angle = 0°

r and v are parallel

L = 0kgm/s

d)

r=(-1,0)

angle = 180°

r and v are parallel

L = 0kgm/s

e)

r=(1,1)

angle = 45°

$L = (2.5kg)(2\frac{m}{s})(\sqrt{2})sin45\°=5kg\frac{m}{s}$

f)

r=(-1,1)

angle = 45°

the same as e):

L = 5kgm/s

g)

r=(-1,-1)

angle = 135°

$L=(2.5kg)(2\frac{m}{s})(\sqrt{2})sin135\°=5kg\frac{m}{s}$

h)

r=(1,-1)

angle = 135°

the same as g):

L = 5kgm/s

hope this helps!!

Answer 2

The determinant method allows to find the z component of the angular momentum for the particle moving in the x axis are:

     a) L_z = - 5.0 kg m / s

     b) L_z = 5 kg m/s

     c)  L_z = 0

     d)  L_z = 0

     e) L_z = -5.0 kg m / s

     f)  L_z = 5.0 kg / ms

     g)  L_z = 5.0 kg m / s

The angular momentum is defined by the vector product of the position and the linear momentum, therefore it is a vector quantity.

      L = r x p

Where the bold letters represent vectors, L the angular momentum, r the position and p the linear moment

Momentum is defined by.

     p = m v

where m is the mass.

The modulus of angular momentum is:

    L = m r v sin θ

One way to find the angular momentum in various dimensions is to solve for the determinant method.

    $L = m \ \left[\begin{array}{ccc}i&j&k\\x&y&z\\v_x&v_y&v_z\end{array}\right]$  

   

The component z of the angular momentum is

     $L_z = m ( x v_y - y v_x)$  

Indicate that the mass of the body is m = 2.5 kg and it moves in the direction x, v = 2 i m/s, let's calculate the angular momentum in direction z for each case:

The velocity is v = 2i

a) point r = (0,1)

    $L_z$ = - 2,5 1 2

    L_z = - 5.0 kg m / s

b) point r = (0, -1)

    L_z = -2.5 (-1) 2

    L_z = 5 kg m/s

c) point r = (1,0)

   L_z = 2,5 0 2

   L_z = 0

velocity is parallel to position

d) point r = (-1,0)

   L_z = 0

Velocity is parallel to position

e) point r = (1,1)

   L_z= 2.5 (1 0 - 2 1)

   L_z = -5.0 kg m / s

f) point r = (- 1, -1)

   L_z = 2.5 (1 0 - 2 (-1))

   L_z = 5.0 kg / ms

g) point r = (1, -1)

    L_z = 2,5 (1 0 - 2 (-1))

    L_z = 5.0 kg m / s

In conclusion using the determinant method we can find the z component of the angular momentum for the particle moving in the x axis are:

     a) L_z = - 5.0 kg m / s

     b) L_z = 5 kg m/s

     c)  L_z = 0

     d)  L_z = 0

     e) L_z = -5.0 kg m / s

     f)  L_z = 5.0 kg / ms

     g)  L_z = 5.0 kg m / s

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