What is the Angular Momentum After a Totally Inelastic Collision?

[jalapenojam]

A 2.6 kg particle that is moving horizontally over a floor with velocity (-3.00 m/s)j undergoes a totally inelastic collision with a 4.00 kg particle that is moving horizontally over the floor with velocity (4.20 m/s)i. The collision occurs at xy coordinates (-0.50 m, -0.10 m). After the collision, what is the angular momentum of the stuck-together particles with respect to the origin using vector notation?

I first tried using conservation of momentum, which gives me 9 as the x-momentum. This is wrong. I know that angular momentum (L) = Iw, but I'm not sure how to calculate I or w.

I'm also stuck on a similar problem:
At the instant the displacement of a 2.00 kg object relative to the origin is d = (2.00 m)i + (4.00 m)j - (3.00 m)k, its velocity is v = - (6.40 m/s)i + (2.90 m/s)j + (2.70 m/s)k, and it is subject to a force F = (6.40 N)i - (7.80 N)j + (4.10 N)k. Find the angular momentum of the object and the torque acting on the object about the origin using vector notation.

To find momentum, I again didn't know how to find Iw, so I just tried multiplying the given mass times the velocity vector. I'm guessing this is wrong because that's for linear momentum.

Any help would greatly be appreciated!

 

[Fermat]

The angular momentum of a particle about an origin is given by,

L = rp

where r is position vector of the particle and p is the linear momentum

 

[jalapenojam]

how do you determine which way the velocity of stuck-together particles are going? would i just use pythagorean theorem to find the hypotenuse of the 4.2 velocity and the -3 velocity? and from there, how do i figure out which way the momentum is going? how can there be momentum in the 'k' (vector notation) direction?

 

[NateTG]

A 2.6 kg particle that is moving horizontally over a floor with velocity (-3.00 m/s)j undergoes a totally inelastic collision with a 4.00 kg particle that is moving horizontally over the floor with velocity (4.20 m/s)i. The collision occurs at xy coordinates (-0.50 m, -0.10 m). After the collision, what is the angular momentum of the stuck-together particles with respect to the origin using vector notation?

You need to specify (and keep track of) the direction of the velocity and momentum.

I first tried using conservation of momentum, which gives me 9 as the x-momentum. This is wrong. I know that angular momentum (L) = Iw, but I'm not sure how to calculate I or w.

For particles you can calculate the angular momentum as:
$$\vec{L}=\vec{r} \times \vec{p}$$
where $\times$ is the cross product, $\vec{r}$ is the vector from the axis of rotation, and $\vec{p}$ is the momentum of the particle.
The magnitude of $\vec{L}$ will be equal to
$$|r p \sin \theta|$$
Where $\theta$ is the angle between $\vec{r}$ and $\vec{p}$.

I'm also stuck on a similar problem:
At the instant the displacement of a 2.00 kg object relative to the origin is d = (2.00 m)i + (4.00 m)j - (3.00 m)k, its velocity is v = - (6.40 m/s)i + (2.90 m/s)j + (2.70 m/s)k, and it is subject to a force F = (6.40 N)i - (7.80 N)j + (4.10 N)k. Find the angular momentum of the object and the torque acting on the object about the origin using vector notation.
To find momentum, I again didn't know how to find Iw, so I just tried multiplying the given mass times the velocity vector. I'm guessing this is wrong because that's for linear momentum.

For forces that apply at a single point, torque can be calculated by
$$\vec{\tau}=\vec{r} \times \vec{F}$$

 

[jalapenojam]

how do i determine r? it isn't just the displacement from the origin, right? because that doesn't work

 

[Doc Al]

Sure it works. (Use the coordinates of the collision to figure out the perpendicular distance between the momentum vectors and the origin.)

 

[jalapenojam]

sorry i was referring to the second problem. but on the first problem, how do i know how fast the stuck-together objects are going in the x or y directions? would they both be moving -3j down and 4.2i across? if so, would i determine the angular momentum in the j direction by doing

(-.5)(2.6 kg)(-3.00 m/s) = 3.9 kg*m^2/s?

 

[Doc Al]

For the first problem, realize that angular momentum is conserved during the collision. (Find the total angular momentum of the particles before the collision.)

For the second problem, calculate $\vec{L}=\vec{r} \times \vec{p}$. (I assume you know how to take the cross product of vectors in cartesian coordinates.)

 

[jalapenojam]

i.. actually don't know how to take the cross product of vectors in cartesian coordinates :( but anyhow, after i find the angular momentum, how do i figure out which direction it's going? because i have to give the angular momentum in vector notation.

 

[Doc Al]

Once you learn how to take the cross product in cartesian coordinates, you'll see that the answer is automatically in (unit) vector notation. See here: http://www.netcomuk.co.uk/~jenolive/vect8.html