Friday, November 1, 2013

Spin the Bottle: How to get the bottle to land where you want!

Spin the Bottle: How to get the bottle to land where you want!

You are playing a game of spin the bottle and you want to know how much force to exert in order to get the bottle to A) land back on yourself so you don’t have to participate or B) land on the cutie across from you. The bottle used is an empty 2L coca-cola bottle with a mass of .052kg and a length of .30m. (Source: google) 

Assumptions: the bottle rotates at a speed of 1 revolution/s. The force being exerted on it is perpendicular to the radius and is at the tip of the bottle (r = .15). We are also ignoring friction.  

A) Landing on self
·      ΔΘ = 2π
·      ωf = 0 rad/s
·      ωo = (1rev/s)(2π/rev) = 6.28rad/s
Use ωf2 = ωo2 +2αΔΘ to find α = -ωo2/2ΔΘ = -(6.28rad/s)2/4π = -3.138rad/s2
·      Knowing that τ = Iα = rFsinΘ we can rearrange the relationship to find that
F = Iα/rsinΘ
·      Since force is being exerted perpendicular to the radius, sinΘ=sin90° = 1
Find the moment of inertia (I) for the coca-cola bottle (which we are considering a cylinder in this case), with the axis of rotation through the central diameter:
I = ¼ mr2 + 1/12mr2 = ¼ (.052kg)(.15m)2 + 1/12(.052kg)(.30m)2
= 6.825x10-4kg Ÿm2
F = Iα/rsinΘ = (6.825x10-4kg Ÿm2)(-3.138rad/s2)/.15m = -.0143N = -1.43x10-2N

B) Landing across from you (with 1.5 rotations)
·      ΔΘ = 3π
·      ωf = 0 rad/s
·      ωo = (1.5rev/1.5s)(2π/rev) = 6.28rad/s
α = -ωo2/2ΔΘ = -(6.28rad/s)2/6π = -2.0923rad/s2
I is the same as before so I = 6.825x10-4kg Ÿm2
F = Iα/rsinΘ = (6.825x10-4kg Ÿm2)(-2.0923rad/s2)/.15m = -.00952N = -9.52x10-3N

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