Angular momentum of a bullet problem

[Ocsycleen]

Homework Statement

A 4 g bullet traveling at 500 m/s strikes a disk of mass 1 kg and
radius 10 cm that is free to rotate around an axis passing through its
center. The bullet’s incoming path is 5 cm above the rotation axis and
the bullet comes to rest in the position shown in the figure. At how
many revolutions per second is the disk is rotating afterwards?
(Ignore the mass of the bullet after it hits the disk.)

https://gyazo.com/7ba8193726c76f57aa6c3ffa9e0c2930

the answer is 3.18 but i can't seem to figure out how to do this.

Homework Equations

L(initial) = L (final)

LDisk = 1/2MR^2[/B]

The Attempt at a Solution

[/B]
Here is what i got so far using conservation of angular momentum

R* Mass of Bullet * Velocity of bullet * sin([PLAIN]http://physics-help.info/physicsguide/appendices/si_units_images/image002.gif[/I]) = IDisk * ωFinal
0.1m * (0.004kg)(500)sin([PLAIN]http://physics-help.info/physicsguide/appendices/si_units_images/image002.gif) = (1/2) (1kg) (0.1)^2 * ωFinal

But i can't seem to figure what the theta is in this case and how to incorporate 5cm above the rotation axis into the problem.

 

[TSny]

Hello, and welcome to PF!

You have the right approach using conservation of angular momentum.

For a refresher on the meaning of θ, look at second paragraph here:
https://cluster31-files.instructure...iles/apb11o/resources/guides/G11-4.angmom.htm

This discussion shows that you don't really need to worry about finding θ in this problem.

Why did you not include the contribution of the bullet to the final moment of inertia $I$?

 

[Ocsycleen]

Thanks.

I did not include the bullet's contribution to the inertia because the problem said to Ignore the mass of the bullet after it hits the disk. I am still confused about how the 5cm above the rotation axis should be used in this problem.

 

[CWatters]

It says to ignore the mass of the bullet after impact so no effect on the moment of inertia.

 

[TSny]

I did not include the bullet's contribution to the inertia because the problem said to Ignore the mass of the bullet after it hits the disk.

OK, great. I somehow overlooked that part of the statement of the problem. If you had included the bullet, you would have found that it does not change the answer to 3 significant figures.

I am still confused about how the 5cm above the rotation axis should be used in this problem.

Do you understand why you can write $L = p r_{\perp}$ for the bullet before the collision?

 

[Ocsycleen]

Because the disk is stationary before the bullet hits so right before the moment of impact, the only thing angular momentum that should exist should be for the bullet which is the cross product of r and p.

 

[TSny]

Because the disk is stationary before the bullet hits so right before the moment of impact, the only thing angular momentum that should exist should be for the bullet which is the cross product of r and p.

OK. If you understand the meaning of a cross product of two vectors, then you should understand the meaning of θ in the formula
L = r p sinθ.

From the link I gave, you see the following diagram (which I slightly modified)

This shows a particle moving with constant speed along a straight line. Note how θ is indicated for the three positions A, B, and C. Hopefully, that agrees with your understanding of θ. Using trig, can you relate r_A, θ_A, and r_B. Note r_B is the perpendicular distance from the origin O to the line of travel of the particle.

 

[Ocsycleen]

sin(180-θA) = RB / RA

 

[TSny]

OK. Can you simplify sin(180-θ_A) using a trig identity? Then solve your equation for R_B in terms of R_A and sin θ_A.

(Note, there is a formatting toolbar that you can use for subscripts, etc.)

 

[Ocsycleen]

RB = sin(θA) * RA

 

[TSny]

Good. So, how can you express the angular momentum of the particle at point A in terms of R_B?

 

[Ocsycleen]

L = P * RB

 

[TSny]

Yes. Note that R_B is the perpendicular distance from the origin to the line along which the particle is traveling. So, you can see now that no matter where the particle is along the line, the angular momentum may be calculated using L = R$_\perp$ p.

What is the value of R$_\perp$ in your problem?

 

[Ocsycleen]

is it 10cm?

 

[TSny]

No. Draw a line representing the line along which the bullet is moving. You can extend that line as far as you like. R$_{\perp}$ is the shortest distance from the origin (center of disk) to this line. See R_B in the previously posted figure.

 

[Ocsycleen]

No. Draw a line representing the line along which the bullet is moving. You can extend that line as far as you like. R$_{\perp}$ is the shortest distance from the origin (center of disk) to this line. See R_B in the previously posted figure.

Oh it's 5cm

 

[TSny]

Yes. Good.

 

[Ocsycleen]

So the left hand side of the equation should be
R⊥ * P = IDisk * ωFinal

But is there something else i am missing here?

 

[TSny]

##r

So the left hand side of the equation should be
R⊥ * P = IDisk * ωFinal

Yes, that's it.

But is there something else i am missing here?

I don't think so. Why do you feel unsure?

 

[TSny]

Note, however, that the question asks for the number of revolutions per second. So, make sure you express your answer in those units.

 

[Ocsycleen]

Ok yea. i think i did a little miscalculation and converted my final answer to rpm instead. Thanks for the help.

 

[TSny]

Good work.