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ay myuonM G Google D Assignment 06JA... a Amazon.com frunl Plusu Email Ee uM Emall a YouTube & vacation Rentals w Twitter n Facebook Login e crammark Yo Fundamentals of Physics, 10e Assignment Halliday, Practice student/main LUS Gradebeek ounces chapter Problem 007 Incorrect. In the figure, asoldcyfinder ofradius 60 em and mass 9 a from withou Mpping roof? (b) The roofs edge is at adstanoe L 62 m doin a roof that is incined at angle 26a. (a) what s the anguar (a) Numberjaos (b) click if you would like to show work for this question: oprn shon ay accessing this Question Assistance, you wit learn whar rou eam prints based on the Point Potereal set av reur instructor. An Retin ved a Division MacBook Air

Explanation / Answer

(a)
(initial gravitational potential energy) = (final translational kinetic energy) + (final rotational kinetic energy)
mgLsin = mv2/2 + I2/2

moment of inertia of solid cylinder: I = ½ mr2

mgLsin = m(r)2/2 + (½ mr2)(2/2)
gLsin = 3(r)2/4
sqrt{4gLsin/3} / r =
= sqrt{4(9.81)(6.2 m)(sin(26°))/3} / (0.06 m)
= 99.37 rad/s

(b)
vertical motion: H = voy*t + gt2/2
horizontal motion: x = (vox)(t)

voy = rsin = (99.37 s-1)(0.06 m)sin26° = 2.61365 m/s
vox = rcos = (99.37 s-1)(0.06 m)cos26° = 5.35878 m/s


t = {sqrt[(voy)2 + 2gH] - voy} / g
t = (sqrt[(2.61365 )2 + 2(9.81)(6.5 )] - 2.61365) / 9.81 m/s^2
t = 0.91516 s

x = (5.35878 m/s)(0.91516 s)
x = 4.904 m

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