Angular velocity and acceleration for catapult arm

[yohak]

Homework Statement

This isn't a homework problem, but I'm trying to do the calculations for a catapult I built to see if its actual performance matches the theoretical performance. I'm stuck trying to map the trajectory of the projectile. Here is the picture for reference http://farm6.static.flickr.com/5054/5475843963_b8aaea4c8a.jpg

The mass of the arm is 0.0426 kg. The mass of the projectile is 0.0091 kg.

The left end of the arm has a spring attached, and the spring is attached to the bottom of the dotted line. I've calculated that the spring constant is 55.8 N/m, its stretched length is .2867 m and its unstretched length is .0783 m.

The pivot point in the actual catapult is a metal rod with a Delrin sleeve, so you can assume the pivot is frictionless.

Initially, the angle alpha is 46.2 degrees, beta is 133.8 and gamma is 35.14 degrees. I believe that the projectile will launch when beta is roughly 45 degrees.

I'm getting confused on how to find the angular velocity or acceleration at the right end of the arm (where the projectile goes).

Really what I want to know is how fast the projectile is launched. Once I have its initial velocity and launch angle, I can easily figure out the projectile motion by myself.

Homework Equations

F=kx
L=I*omega
tau=I*alpha
kinematic equations
I_(com)=1/12*mL^2
I=I_(com)*md^2
PE_spring = 1/2*kx^2

The Attempt at a Solution

The rod rotates through a point that isn't at the center of mass, so I found the moment of inertia using the parallel axis theorem.
I_(com) = 1/12*(.0426)(.4064m)^2 = 586e-6 kg*m^2
I = 586e-6+(.0426)(.127)^2 = 1.27e-3 kg*m^2

The catapult is driven by a spring. Using F=kx, the force F is 10.418 N. The potential energy of the spring is 1/2*55.8*0.2084^2 = 1.21 N*m. I don't know how to find the acceleration on this end due to the spring.

I don't know if I should view the arm in two pieces (left and right) or try to analyze it as one piece. Is conservation of angular momentum the way to go? I'm so confused on where to even start this. Can you give me a push in the right direction, please?

 

[Beaker87]

I would recommend against force dynamics. I think the problem would be much easier for you to solve using energy dynamics.

The formula you would use to do this is:
KE_{arm}=\frac{1}{2}I\omega^2



Now, this next thing is important:
Are you throwing the projectile with a sling, or is it mounted directly to the arm?
In either case DO NOT forget to add in the rotational inertia of the projectile.

 

[yohak]

it is mounted directly to the arm. We had a sling at first, but it was inconsistent. We're firing a washer into a plastic cup 3 m away, and the sling was hard to adjust to make the washer hit the target.

The rotational inertia of the washer is I = 6.47e-6 kg*m^2 since the OD is .035 m and the ID is .014 m and the mass is .0091 kg.

Since
KE_(arm,initial) + PE_(arm,initial) = KE_(arm_final) + PE_(arm,final)
does that mean we can say:
KE_(projectile end, i) + PE_(projectile end, i) + KE_(spring end, i) + PE_(spring end, i) = KE_(projectile end, f) + PE_(projectile end, f) + KE_(spring end, f) + PE_(spring end, f)
?
Is the moment of inertia different for the two ends since the distance from the pivot point is different?

 

[Beaker87]

Do not solve for the left side of the catapult arm separately from the right. Once you have the moment of inertia for the whole arm, use that to find the angular velocity of the whole arm.As for the energy equation you have written, at the beginning, the catapult only has spring potential, and at the end it only has kinetic energy. so your equation will become initial PE of the spring is equal to the final KE of the arm (taken as a whole with the washer)

 

[yohak]

Thanks! I think I got it! :-)