A particle of mass m is shot with an initial velocity vector v i and makes an angle θ with the horizontal, as shown below. The particle moves in the gravitational field of the Earth. (makes a shape of a parabola)Find the angular momentum of the particle about the origin when the particle is at the following locations. (Use v_i for vi, theta for θ, m, and g as appropriate in your equations.)
Known equations:
L=r x p ====> r x mv
My answers: (which are wrong)
a)m(v_i)(sin(theta))
b) ?
c)v_i^2( sin2(theta))/g * v_i sin(theta)
Thank you I appreciate your help!
You do not provide or describe the figure.
You do not identify "the following locations"
You do not tell us what parts (b) and (c) of the question are.
Your answer to (a) does not have the correct dimensions for angular momentum, and therefore cannot be right.
It is not possible to provided assistance because of these omissions
My apologies
The "following locations"
a) at the origin
b)at the highest point of its trajectory
c)just before it hits the ground
the dimensions of angular momentum: ? kg*m^(2)/s
the picture is simply a parabola, a projectile launched with some angle theta with respect to the horizontal
and for part a I realized that it is 0 because the particle is at the origin and I have to find the angular momentum of the particle about the origin:
a = vi x= r
r = 0
emath:
Thank you for taking the time to clarify and complete your question. Your answer to part (a) is correct.
There is a shortcut you can take for calculating the angular momentum with respect to the origin. Write the position in vector form as r = x i + y j, and the momentum as
p = m Vx i + m Vy j. The cross product is then
r = p x r = m(Vx*y - Vy*x) k
(k is a unit vector perpendicular to i and j . 'i' is horizontal along the path and 'j' is vertical.
Next, write down the equations for x, y, Vx and Vy in terms of t
x = Vi cos theta * t
y = Vi sin theta * t - (g/2) t^2
Vx = Vi cos theta
Vy = Vi sin theta - gt
L = m [Vi^2 sincos theta *t -Vi cos theta *t - Vi^2 sincos theta *t + Vi cos theta * g t/2]
At t=0, L obviously = 0, in agreement with your result
For parts (b) and (c) of your question, calculate the corresponding t and then calculate L using the equation above.
For example, at the top of the trajectory, t = (Vi sin theta)/g
At the end of the trajectory, t is twice that amount.
The resulting equation will look a bit messy. I leave the final steps to you
Hey drwls, can you go ahead and finish the rest of the equations, please? I'm having much difficulty with the final equation.
To find the angular momentum of the particle about the origin at different locations, we need to understand the concept of angular momentum.
Angular momentum is a vector quantity defined as the cross product of the position vector and linear momentum of an object about a certain reference point or axis. Mathematically, it can be represented as:
L = r x p
where L is the angular momentum vector, r is the position vector of the object relative to the reference point, and p is the linear momentum vector of the object.
Now, let's solve the problem step by step:
a) To find the angular momentum of the particle about the origin when it is at location A, we need to determine the position vector (r) and the linear momentum vector (p) at that location.
Since the particle is at the origin, the position vector (r) will be zero. The linear momentum vector (p) can be calculated by multiplying the mass of the particle (m) with its velocity vector (v):
p = m * v
So, the angular momentum (L) at location A will be:
L = r x p = 0 x m * v = 0
Therefore, the angular momentum of the particle about the origin at location A is zero.
b) To find the angular momentum of the particle about the origin when it is at location B, we use the same concept. The position vector (r) will again be determined relative to the origin.
Since the particle moves in the gravitational field of the Earth, we can consider its motion as projectile motion. At location B, the particle will have a horizontal and vertical displacement. The position vector (r) will depend on both displacements.
The linear momentum vector (p) can be calculated as before:
p = m * v
To determine the position vector (r), we can consider the horizontal and vertical components of the displacement. Let's denote the horizontal displacement as dx and the vertical displacement as dy.
So, the position vector (r) at location B will be:
r = <dx, dy>
Now, we can calculate the angular momentum (L) at location B:
L = r x p = <dx, dy> x m * v
Since we don't know the exact values of dx, dy, v, we cannot provide a specific formula at this stage, but you can calculate it based on the given values using the cross product formula.
c) To find the angular momentum of the particle about the origin when it is at location C, we again use the same concept of angular momentum.
At location C, the particle will have both horizontal and vertical displacements from the origin. The linear momentum vector (p) can be calculated as:
p = m * v
The position vector (r) will again depend on the horizontal and vertical displacements. Let's denote the horizontal displacement as dx and the vertical displacement as dy.
So, the position vector (r) at location C will be:
r = <dx, dy>
Now, we can calculate the angular momentum (L) at location C:
L = r x p = <dx, dy> x m * v
Similar to location B, we cannot provide a specific formula without knowing the exact values of dx, dy, v, but you can calculate it using the cross product formula.
In summary, the angular momentum of the particle about the origin depends on the position vector (r) and linear momentum vector (p) at each location. By calculating the cross product of these vectors, you can find the angular momentum at the desired locations.