Black Friday Lazer Curry Shootout

After stuffing ourselves silly on Thanksgiving Day, we decided to work off those extra calories with a Black Friday Lazer Curry Shootout. We gathered at [Mom McMains->]’ house, gobbled a couple of delicious curries that [Becky->] had put together with a bit of help from [Kathy->] and my cousin Tanya, and then lumbered off to the neighborhood park for some Lazer Tag.

Thanks to some additions from [Chris->], we had nine guns, all of which were pressed into service as we leaped over creekbeds, pushed through forests, and climbed up and around huge oaks. Alternating between King of the Hill, Capture the Flag, and Free-for-All games, we had a superb time plugging away at each other and getting some good exercise to boot.

Once everyone got a bit tuckered out, we retired to the playground for conversation and some less running-intensive play. Barry provides a really nice writeup of that time here, focusing on the fun that was had with using the merry-go-round and other playground equipment as a large-scale physics lab. (I spent about an hour working on figuring out how quickly we’d have to spin the merry-go-round to get one gravity of centrifugal force, but still haven’t been able to get the math to work out!)

Update: Jason provides the math. A few notes in reply, since his comment system is being grouchy:

Hehe. Excellent. I had gone down this road a bit, but I found contradictory formulas, and my math wasn’t working out. Thanks for delving in!

One can simplify this formula a bit by solving for acceleration, rather than force. (You need the force to figure out how strong to make the merry-go-round, but only the acceleration to figure out how to get up to 1G.) Once you do that, the mass doesn’t figure in any more.

Further, to get a pure 1G lateral acceleration in the presence of the normal 1G downward acceleration, you’d need infinite speed, I’m afraid — the combined vector will approach horizontal asymptotically, but since there’s always that downward element (however infinitesimal it may get by comparison to the outward force), it will never become purely horizontal.