Big idea for Feb. 1, 2010

Today's main topic was applying energy to the motion of a roller coaster car as it goes along the track and does a loop-de-loop. Mr. Burk presented us with a problem requiring us to apply energy transferring, centripetal motion, and FBD presentation to solve.

The drawing for the problem is this:



The problem that Mr. Burk presented us with is as follows:
1) Find the speed of the car at points A, B, C, and D.
2) Find the minimum speed to make the loop.
3) Find the minimum height the car was dropped from to make the loop.
4) Construct an FBD for points A, B, C, and D.

The answers are as follows:
1) a. V=0 because the object is at the top of the course.
b. all of the Ug the car started with has now transfered into K
so mgh=.5*mv^2
v=square root(2gh)
c. the equation is the same except the height is now h-R
so v=square root(2gh-2gR)
d. the equation is the same except the height is now h-2R
so v=square root(2gh-4gR)
2) the centripetal acceleration is v^2/R and acceleration is Fnet/m.
so Vmin^2/R=mg/m
Vmin^2/R=g
Vmin=square root(Rg)
3) for the car to make the loop, the velocity at D squared is equal to mg/g*R because of the fact that the centripetal acceleration equals the def. of acceleration (see #2)
so 2gHmin-4gR=gR
2gHmin=5gR
Hmin=2.5R
4) the FBDs and diagrams for the car are below:

That is the entire insightful problem. HW: e-mail proposal to Mr. Burk and work more on the energy packet. Next Scribe: Timothy!!