Physics Help Forum Spring Launcher Project

 Energy and Work Energy and Work Physics Help Forum

 Apr 20th 2008, 07:44 AM #1 Junior Member     Join Date: Apr 2008 Posts: 5 Spring Launcher Project OK, so the point of this project is to build a launching pad for a spring launcher that can launch a spring (that will be provided on the day of the launch) anywhere in the the classroom. My class room is about 4m long, 5m or 6m wide and about 2m or 2.5m high, my teacher said the box that you have to get the spring into can be anywhere in the room so you have to do the calculations, calibration and all other preparation before hand. These are the calculations I've done, can somebody check them over if they look right? There might be some algebra mistakes. My design concept is pretty simple, two wooden blocks connected by a door hinge and a protractor attached to measure the angle.I have a ruler on the top block to measure how much I have to stretch the spring. Does anybody know a good way to fix the angle, once I have it lined up? I haven't started building it
Apr 20th 2008, 08:41 AM   #2

Join Date: Apr 2008
Location: On the dance floor, baby!
Posts: 2,730
 Originally Posted by Hasan1 OK, so the point of this project is to build a launching pad for a spring launcher that can launch a spring (that will be provided on the day of the launch) anywhere in the the classroom. My class room is about 4m long, 5m or 6m wide and about 2m or 2.5m high, my teacher said the box that you have to get the spring into can be anywhere in the room so you have to do the calculations, calibration and all other preparation before hand. These are the calculations I've done, can somebody check them over if they look right? There might be some algebra mistakes. My design concept is pretty simple, two wooden blocks connected by a door hinge and a protractor attached to measure the angle.I have a ruler on the top block to measure how much I have to stretch the spring. Does anybody know a good way to fix the angle, once I have it lined up? I haven't started building it
The calculations look good. You'll also have to be able to account for the ceiling. I'd suggest you work out an equation for the maximum angle you can fire at for a given initial spring displacement (in other words, given a spring stretch and an angle of firing, how high will the spring go?)

As far as fixing the angle is concerned, if I have your setup right the spring is attached to a board that rotates, right? How about sticking a vertical prop under it that can slide back and forth?

-Dan
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.

See the forum rules here.

 Apr 21st 2008, 01:57 PM #3 Junior Member     Join Date: Apr 2008 Posts: 5 I'm not really sure how to calculate the max angle...I already know how to calculate the minimum angle, wouldn't it be enough to just calculate the min. angle then aim it a few degrees higher than that? I thought about using a sliding vertical prop to fix the angle in place but my only concern is that I think it's going to minimize my max. or min. angle range depending on how high it is.
Apr 23rd 2008, 10:38 AM   #4

Join Date: Apr 2008
Location: On the dance floor, baby!
Posts: 2,730
 Originally Posted by Hasan1 I'm not really sure how to calculate the max angle...I already know how to calculate the minimum angle, wouldn't it be enough to just calculate the min. angle then aim it a few degrees higher than that? I thought about using a sliding vertical prop to fix the angle in place but my only concern is that I think it's going to minimize my max. or min. angle range depending on how high it is.
Say that you are planning to fire the projectile at an angle theta and will pull the spring back an initial distance x0. Then we know the highest the projectile can go is the height H, which for simplicity I will take as the vertical distance between the launch point and the ceiling. I will call the starting point of the projectile of mass m the 0 point for the GPE and calling the spring constant k.

First we find the speed v0 that the projectile leaves the launcher:
$\displaystyle \frac{1}{2}kx_0^2 = \frac{1}{2}mv_0^2 + mgx_0~sin(\theta)$

So we can find v0.

Now, you know that the velocity v0 will be directed at angle theta. So now do a kinematics problem and see what the max height of the projectile will be. This must be less than H.

-Dan
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.

See the forum rules here.

 Tags launcher, project, spring