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!!