HW5_Sol_F10

# HW5_Sol_F10 - Sept 27 2010 PHYS 2101 HW#5 WileyPlus Problem Solutions 1 HW#5 Solutions – or at least most of them…(USING BOOK VALUES 9 th

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Unformatted text preview: Sept. 27, 2010 PHYS 2101 HW#5 WileyPlus Problem Solutions 1 HW #5 Solutions – or at least most of them… (USING BOOK VALUES 9 th Edition in problems; problem # indicated by ) 2) [10-6] If we make the units explicit, the function is θ = 4.0 rad/s ( ) t − 3.0 rad/s 2 ( ) t 2 + 1.0 rad/s 3 ( ) t 3 but generally we will proceed as shown in the problem—letting these units be understood. Also, in our manipulations we will generally not display the coefficients with their proper number of significant figures. (a) Equation 10-6 leads to ω = d dt 4 t − 3 t 2 + t 3 ( ) = 4 − 6 t + 3 t 2 . Evaluating this at t = 2 s yields ω 2 = 4.0 rad/s. (b) Evaluating the expression in part (a) at t = 4 s gives ω 4 = 28 rad/s. (c) Consequently, Eq. 10-7 gives (d) And Eq. 10-8 gives α = d ω dt = d dt 4 − 6 t + 3 t 2 ( ) = − 6 + 6 t . Evaluating this at t = 2 s produces α 2 = 6.0 rad/s 2 . (e) Evaluating the expression in part (d) at t = 4 s yields α 4 = 18 rad/s 2 . We note that our answer for α avg does turn out to be the arithmetic average of α 2 and α 4 but point out that this will not always be the case. 3) [10-13] The wheel has angular velocity ω = +1.5 rad/s = +0.239 rev/s at t = 0, and has constant value of angular acceleration α < 0, which indicates our choice for positive sense of rotation. At t 1 its angular displacement (relative to its orientation at t = 0) is θ 1 = +20 rev, and at t 2 its angular displacement is θ 2 = +40 rev and its angular velocity is . (a) We obtain t 2 using Eq. 10-15: which we round off to ....
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## This note was uploaded on 01/26/2011 for the course PHYS 2101 taught by Professor Grouptest during the Spring '07 term at LSU.

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HW5_Sol_F10 - Sept 27 2010 PHYS 2101 HW#5 WileyPlus Problem Solutions 1 HW#5 Solutions – or at least most of them…(USING BOOK VALUES 9 th

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