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

Known:

·
ΔΘ = 2π

·
ω

_{f }= 0 rad/s
·
ω

_{o }= (1rev/s)(2π/rev) = 6.28rad/s
Use ω

_{f}^{2}= ω_{o}^{2}+2αΔΘ to find α = -ω_{o}^{2}/2ΔΘ = -(6.28rad/s)^{2}/4π = -3.138rad/s^{2}
·
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 mr

^{2}+ 1/12mr^{2 }= ¼ (.052kg)(.15m)^{2}+ 1/12(.052kg)(.30m)^{2}
= 6.825x10

^{-4}kg Ÿm^{2}
F = Iα/rsinΘ = (6.825x10

^{-4}kg Ÿm^{2})(-3.138rad/s^{2})/.15m = -.0143N = -1.43x10^{-2}N
B) Landing across from you (with 1.5 rotations)

Known:

·
ΔΘ = 3π

·
ω

_{f }= 0 rad/s
·
ω

_{o }= (1.5rev/1.5s)(2π/rev) = 6.28rad/s
α = -ω

_{o}^{2}/2ΔΘ = -(6.28rad/s)^{2}/6π = -2.0923rad/s^{2}
I is the same as before so I = 6.825x10

^{-4}kg Ÿm^{2}
F = Iα/rsinΘ = (6.825x10

^{-4}kg Ÿm^{2})(-2.0923rad/s^{2})/.15m = -.00952N = -9.52x10^{-3}N
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