Comments on: physics…10pts to best/correct answer? http://antiquerecordplayers.info/physics10pts-to-bestcorrect-answer.htm Wed, 08 Feb 2012 05:06:42 +0000 http://wordpress.org/?v=2.7.1 hourly 1 By: Schmedley http://antiquerecordplayers.info/physics10pts-to-bestcorrect-answer.htm/comment-page-1#comment-3220 Schmedley Sat, 09 Jan 2010 07:32:57 +0000 http://antiquerecordplayers.info/physics10pts-to-bestcorrect-answer.htm#comment-3220 This is probably too late, but here's my approach anyway A) First, find the angular acceleration α, from initial and final angular velocities ωo and ωf: ωf = ωo + αt; substituting, (33.33 rpm) = 0 + α (2.45 s); α = 13.6 rpm/s Converting revolutions per minute per second to degrees per square second: (13.6 rpm/s) (360° /rev) / (60 sec/min) = 81.6°/s² Now calculate angular distance θ: θ = ωot + (1/2)αt², substituting, θ = 0 + (1/2) (81.6°/s²) (2.45 s)²; θ = 245°, or 4.28 radians (there are π radians per 180°). B) s = rθ, were s is the linear distance, r is the radius, and θ is the angle in radians; substituting, s = (0.5 ft) (4.28 radians) = 2.14 ft. Don't be confused by the radian unit; since it is defined by a ratio, it is actually a dimensionless number. This is probably too late, but here’s my approach anyway

A) First, find the angular acceleration α, from initial and final angular velocities ωo and ωf:
ωf = ωo + αt; substituting,
(33.33 rpm) = 0 + α (2.45 s);
α = 13.6 rpm/s
Converting revolutions per minute per second to degrees per square second:
(13.6 rpm/s) (360° /rev) / (60 sec/min) = 81.6°/s²
Now calculate angular distance θ:
θ = ωot + (1/2)αt², substituting,
θ = 0 + (1/2) (81.6°/s²) (2.45 s)²;
θ = 245°, or 4.28 radians (there are π radians per 180°).

B) s = rθ, were s is the linear distance, r is the radius, and θ is the angle in radians; substituting,
s = (0.5 ft) (4.28 radians) = 2.14 ft.
Don’t be confused by the radian unit; since it is defined by a ratio, it is actually a dimensionless number.

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