Angular speed of rod after projectile collides into it

In summary, the conversation discusses the calculation of the angular speed of a rod after a projectile collides into it. The correct formula for the moment of inertia of a thin rod is 1/12mr^2, where r is the length of the rod. The question also involves calculating the change in kinetic energy before and after the collision.
  • #1
StrawHat
33
0
[SOVLED] Angular speed of rod after projectile collides into it

Homework Statement


dcnzX.png



Homework Equations


L = r x p
Iparallel = ICM + md2


The Attempt at a Solution


L = r x p = mvir = mvid/2
mvi(d/2) = (Iparallel + md2
mvi(d/2) = (1/12M(d/2)^2 + m(d/2)^2)ω

Plugging and solving eventually leads me to...
ω = (mvi)/(1/24Md+1/2md)
...Which is apparently incorrect.
 
Last edited:
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  • #2
What's the moment of inertia of a thin rod about its center?
 
  • #3
1/12mr2
 
  • #4
And what should r be in this problem? Hint: I don't think it's d/2.

As a matter of interest, how good is the submission system at algebra? Will it understand that a/(1/24b+1/2c)=24a/(b+12c), or will it only recognise expressions that your prof entered? The only one I've ever used could only do exact string comparison, so if the answer was 0.3 and you typed .3, that was marked wrong...
 
  • #5
Why wouldn't it be d/2? If the diameter of the circular motion is d, then wouldn't the radius be d/2?
 
  • #6
You have the correct functional form for the second moment of inertia of the rod, but are misinterpreting the thing you called r. It's the length of the rod (d, in this case). You can easily look it up, or derive it from first principles by integrating [itex]\mu r^2dr[/itex] (where [itex]\mu[/itex] is the mass per unit length of the rod) over the range -d/2 to d/2.
 
  • #7
Oh okay, I thought r was supposed to be the distance between the rotation point and the end of the rod.

I got the right answer of ω = (mvi)/(1/6Md+1/2md)

Thanks for your help!
 
  • #8
Okay, I've tried part B of the question and this is what I've done so far.

Kfinal = 1/2Iω2
Kinitial = 1/2mv2

I divided both of those to get (Iω2)/(mvinitial)

Calculating that further, and I get this:
D2LsZ.png


I'm guessing that it's not Kfinal/Kinitial, but (Kinitial - Kfinal)/Kfinal?

EDIT: Okay, I got the answer. Thank you all!
 
Last edited:
  • #9
StrawHat said:
I'm guessing that it's not Kfinal/Kinitial, but (Kinitial - Kfinal)/Kfinal?

Almost. After subtracting, divide by Kinitial rather than Kfinal. (It's like one of those percent decrease problems in algebra.)
 

Related to Angular speed of rod after projectile collides into it

What is angular speed?

Angular speed, also known as angular velocity, is a measure of how quickly an object is rotating or moving in a circular path. It is usually represented by the symbol ω and is measured in radians per second.

How is angular speed calculated?

The formula for calculating angular speed is ω = Δθ/Δt, where ω is the angular speed, Δθ is the change in angle, and Δt is the change in time.

What factors affect the angular speed of a rotating object?

The angular speed of a rotating object can be affected by various factors, such as the mass and shape of the object, the force applied to it, and the distance from the object's axis of rotation.

How does a projectile colliding into a rod affect its angular speed?

When a projectile collides into a rod, it transfers its momentum and energy to the rod, causing it to rotate. The resulting angular speed will depend on the mass and velocity of the projectile, as well as the mass and moment of inertia of the rod.

Can the angular speed of a rod change after a projectile collision?

Yes, the angular speed of a rod can change after a projectile collision. This change will depend on the factors mentioned above, as well as the conservation of angular momentum and energy principles.

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