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How it Works »New Instructable »I am not a big fan of doing a science fair project, but I would love to do something big and fun with minimal actual work. Does anyone have an idea?

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active| newest | oldest(1) a projectile (the thing being thrown)

(2) an accelerator (some means of giving the projectile a controlled momentum, or equivalently some

initial velocity)(3) a target (some location where you want the projectile to arrive)

Obviously the particulars can be as complicated as you like. This class of problem includes everything from throwing a basketball at a hoop, to putting a lander onto the lunar surface, to dropping white phosphorous on innocent civies.

However, you specifically asked for simplicity, or "minimal actual work" as you phrased it, and I agree it's good to start with something simple, and get that to work before building a trebuchet and declaring war on the neighborhood.

One of the simplest (er most simple) physics-class ballistics demos I have ever seen looks like the illustration below.

The projectile, a steel ball bearing, starts its journey at point A. Then it rolls down the ramp AB, picking up m*g*h

_{1}of kinetic energy by the time it reaches the bottom of the ramp. Note that some of this KE will be rotational, some translational. Then it rolls across the table, the horizontal path BC. At point C it takes flight and falls ballistically through the air until it reaches the target D, somewhere on the floor, which lies a vertical distance h_{2}below the surface of the table.For the target you can use a bottle cap, or a pie pan, depending on how good your math is.

;-)

Obviously a bottle cap would be more impressive, since it's a smaller target.

The trick to predicting where the ball is going to land, as it moves ballistically from C to D, lies in treating the horizontal (x) and vertical (y) motions separately. Because the table top is horizontal, you expect the initial motion of the ball to also to be horizontal. Thus the vertical part of its velocity (v

_{y}) is zero initially. As a result the vertical motion is just free-fall.v

_{y}= -gt with T_{fall}= (2*h_{2}/g)^{1/2}and

v

_{x}= constantx(T

_{fall}) = v_{x}*T_{fall}where v

_{x}depends on the initial height of the ramp h_{1}. The math for this is based on energy conservation. Expect 0.5*m*v_{x}^{2}for the translational KE, while the rotational KE is a formula I don't recall at the moment. (Yet nottoo hardto figure out.)Essentially you can control the position on the floor, v

_{x}*T_{fall}, where the ball hits, depending on what height h_{1}you start the ball on the ramp.As mentioned before, this is a classic physics demo. So somewhere on the web there is probably a superior description of the same thing, compared to the one I've given here.

Anyway, when describing this idea before, I sort of did some "hand waving" regarding the effects of rotational KE on the ball's speed v

_{x}after it falls through h_{1}. For the sake of completeness, I thought I'd go ahead and explain that part for anyone interested.The equation for the energy balance, from PE to KE, is:

m*g*h

_{1}= KE_{translation}+ KE_{rotation}m*g*h

_{1}= (1/2)*m*v_{x}^{2}+ (1/2)*J*ω^{2}Radially symmetric objects, like rings, discs, balls, etc, rotating about their center, have moment of inertia of the form J = k*m*R

^{2}where k is a dimensionless constant that depends on the shape (ring, disc, ball, etc). Also, v_{x}=ω*R, since the object is rolling with its center R from the point where its edge touches the ramp. Substitute these into the energy balance equation to get:m*g*h

_{1}= (1/2)*m*v_{x}^{2}*(1+k)then

v

_{x}= (1/(1+ k))^{1/2}* (2*g*h_{1})^{1/2}where

k=0 for sliding (without rolling, like a wet ice cube)

k=2/5 for a rolling solid ball (like a ball bearing)

k=1/2 for a rolling disc (like a coin, or manhole cover)

k=1 for a rolling ring (like a ring, or section of pipe, soup can, etc)

These give adjustments to v

_{x}of {1, 0.845, 0.817, 0.707}, for sliding, rolling ball, rolling disc, rolling ring, respectively.So if you had a "race" between a ring (0.707) and ball (0.845), you'd expect the ball to win, i.e. reach the bottom of the ramp first, and be moving faster when it did.

Source for moment of inertia formulas:

http://en.wikipedia.org/wiki/List_of_moments_of_inertia

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