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

Known:

·      ΔΘ = 2π

·      ω_f = 0 rad/s

·      ω_o = (1rev/s)(2π/rev) = 6.28rad/s

Use ω_f² = ω_o² +2αΔΘ to find α = -ω_o²/2ΔΘ = -(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)

Known:

·      ΔΘ = 3π

·      ω_f = 0 rad/s

·      ω_o = (1.5rev/1.5s)(2π/rev) = 6.28rad/s

α = -ω_o²/2ΔΘ = -(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