Angular Momentum Conservation for Projectile-Rod Collision | Homework Solution

[jdboucher]

Homework Statement

A projectile of mass m = 1.49 kg moves to
the right with speed v0 = 19.6 m/s. The pro-
jectile strikes and sticks to the end of a sta-
tionary rod of mass M = 6.55 kg and length
d = 2.29 m that is pivoted about a frictionless
axle through its center.
Find the angular speed of the system right
after the collision.
Answer in units of rad/s.

Homework Equations

Icom rod = (1/12)ML^2

3. The Attempt at a Solution [/b

I figured I would use conservation of angular momentum to determine the final angular speed. I'm unsure how to find the initial angular momentum. I can find the final moment of inertia.

(1/12)(1.49 + 6.55)(2.29)^2 = 3.51

HELP!

 

[Winzer]

I think viewing it from an energy perspective will be easier.

 

[jdboucher]

I think viewing it from an energy perspective will be easier.

Does translational (i think that's what its called) kinetic energy relate to rotational kinetic energy?

 

[Winzer]

Does translational (i think that's what its called) kinetic energy relate to rotational kinetic energy?

What happens in the problem? The energy gets converted from ____ to _____.
It comes from the conversion of motion. From Linear to Rotational right?

 

[jdboucher]

What happens in the problem? The energy gets converted from ____ to _____.
It comes from the conversion of motion. From Linear to Rotational right?

Is it kinetic to potential?

 

[jdboucher]

What happens in the problem? The energy gets converted from ____ to _____.
It comes from the conversion of motion. From Linear to Rotational right?

so can I do this:

0.5mv^2 = Iw

 

[Winzer]

Is it kinetic to potential?

I was looking for the energy gets converted from linear kinetic to rotational energy, right?
Use conservation of energy:
$$KE_{Linear}=KE_{Rotational}$$

 

[Winzer]

so can I do this:

0.5mv^2 = Iw

It should be :
$$\frac{1}{2}mv^2 = \frac{1}{2}I \omega^2$$

 

[jdboucher]

It should be :
$$\frac{1}{2}mv^2 = \frac{1}{2}I \omega^2$$

0.5(1.49)(19.6)^2 = 0.5(1/12)(1.49+6.55)(2.29)^2w^2

I solved this and got w = 12.76

The answer is wrong though. I checked my work. Did I do something wrong?

 

[vela]

The two bodies stick together, so this is an inelastic collision. Kinetic energy isn't conserved. You have to use angular momentum.

Remember $\vec{L}=\vec{r}\times\vec{p}$.

 

[jdboucher]

The two bodies stick together, so this is an inelastic collision. Kinetic energy isn't conserved. You have to use angular momentum.

Remember $\vec{L}=\vec{r}\times\vec{p}$.

So then I need to determine r. Is that d/2?

 

[danielatha4]

Yes, r would be d/2. It's the distance to the pivot point.

 

[jdboucher]

Yes, r would be d/2. It's the distance to the pivot point.

I assume the angle is 90. So L = mvrsin90. If I do that, set it equal to Iw, I get that w= 9.52. According to the online site I'm using that's wrong. Is it my angle?

 

[vela]

Your $I_{final}$ is wrong.

 

[jdboucher]

Your $I_{final}$ is wrong.

Ifinal = (1/12)ML^2 = (1/12)(1.49 + 6.55)(2.29)^2

Ifinal = 3.51

Is that wrong?

 

[vela]

Yes, it's wrong. That would be the moment of inertia for a uniform rod of mass 8.04 kg. You have a total mass of 8.04 kg, but it's not distributed as a uniform rod.

 

[jdboucher]

Yes, it's wrong. That would be the moment of inertia for a uniform rod of mass 8.04 kg. You have a total mass of 8.04 kg, but it's not distributed as a uniform rod.

(1/12)(6.55+1.49)(2.29)^2 + (1.49)(2.29/2)^2 = 6.12

Its wrong again. Am I not supposed to use the parallel axis theorm?

 

[jdboucher]

(1/12)(6.55+1.49)(2.29)^2 + (1.49)(2.29/2)^2 = 6.12

Its wrong again. Am I not supposed to use the parallel axis theorm?

I mean that I assumed the bullet was a point and used mr^2 to solve for its moment of inertia. Thats wrong obviously

 

[vela]

No, the parallel axis theorem let's you calculate the moment of inertia of an object when it's rotating about an axis that doesn't go through its center of mass, like if you were rotating the rod by its end rather than through its center. That's not the case here, so the theorem doesn't apply.

In this problem, the total rotational mass is simply that of the rod (by itself) plus that of the projectile. You have the formula for the rod; just make sure you use the mass of the rod alone since the projectile isn't part of the rod. The projectile is a point mass at a distance $r$ from the rotation axis. Its rotational mass is given by $mr^2$, where $m$ is the mass.

 

[vela]

I mean that I assumed the bullet was a point and used mr^2 to solve for its moment of inertia. That's wrong obviously.

No, that's right. You just plugged the wrong mass in for the rod.

 

[jdboucher]

No, that's right. You just plugged the wrong mass in for the rod.

Yes you're right. Silly me. I got it now. Thank you for your help.