How Do You Calculate Angular Momentum for a Particle in Motion?

[mcnealymt]

Homework Statement

http://session.masteringphysics.com/problemAsset/1070538/5/12.EX46.jpg

a)What is the magnitude of he angular momentum of the 200g particle relative to the origin.

b) What is the direction of the angular momentum relative to the origin of the 200g particle? Into the page or out of the page.

Homework Equations

L= r *p you can then substitute "mv" in place of "p"

The Attempt at a Solution

a) Okay I'm very confused as to how I'm supposed to get the velocity. I understand that radius is found using Pythagorean Theorem, but I believe the velocity needs to be broken up.

I saw that one person said v= 3(cos45-arctan(.5)) I understand the cos45, but I don't get how arctan is relevant.

b) Into the page because of right hand rule? I am still not sure about this rule, its still very confusing. I've read three different methods and they just don't make any sense.

 

[jambaugh]

Your "relevant equation" is a magnitude equation. for r a position vector and p a momentum vector the proper equation is $\vec{L} = \vec{r}\times \vec{p}=m\vec{r}\times\vec{v}$ where $\times$ is the cross product.

Pick a basis, (the standard i,j,k basis will do) express v as a vector and the position of the particle as a vector then take the cross product.

The velocity's magnitude and direction are given in your diagram. You just need to apply some trig.

 

[mcnealymt]

After I take the cross product what do I do? I know that the equation L= *m*v*r*p and there's another one ABsin(beta)K-hat.

 

[jambaugh]

What is the question asking?

 

[Nickie]

I would approach this problem by first defining the terms and concepts involved. Angular momentum is a vector quantity that describes the rotational motion of an object around a fixed point or axis. It is defined as the cross product of the position vector and the linear momentum vector. In this case, the position vector is the distance from the origin to the particle, and the linear momentum vector is the product of the mass and velocity of the particle.

a) To find the magnitude of the angular momentum, we can use the formula L = r x p, where r is the position vector and p is the linear momentum vector. In this case, the position vector is given as 0.2m in the x-direction and 0.3m in the y-direction. The mass of the particle is 0.2kg, and we can find the velocity by using the given information that the particle is moving at a speed of 3m/s at an angle of 45 degrees. We can use trigonometry to find the x and y components of the velocity, which are vx = 3cos45 and vy = 3sin45. Substituting these values into the formula for linear momentum (p = mv), we get p = (0.2kg)(3cos45i + 3sin45j), where i and j are unit vectors in the x and y directions, respectively. Therefore, the angular momentum can be calculated as L = (0.2m)(3cos45i + 3sin45j) x (0.2kg)(3cos45i + 3sin45j). Using the cross product formula, we get L = (0.2m)(0.2kg)(3cos45)(3sin45)k, where k is the unit vector in the z-direction. Simplifying, we get L = 0.018kgm^2/s in the z-direction (into the page).

b) The direction of the angular momentum can be determined using the right-hand rule. This rule states that if we curl the fingers of our right hand in the direction of rotation, the thumb will point in the direction of the angular momentum vector. In this case, the particle is rotating in a counterclockwise direction, so the angular momentum vector points into the page. Alternatively, we can use the cross product formula to determine the direction, which in this case is given as k (z-direction