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Unformatted text preview: Chapter 10 1. (a) The second hand of the smoothly running watch turns through 2 π radians during 60 s . Thus, 2 0.105 rad/s. 60 π ϖ = = (b) The minute hand of the smoothly running watch turns through 2 π radians during 3600 s . Thus, ϖ = = × 2 3600 175 10 3 π . rad / s. (c) The hour hand of the smoothly running 12hour watch turns through 2 π radians during 43200 s. Thus, ϖ = = × 2 43200 145 10 4 π . rad / s. 2. The problem asks us to assume v com and ϖ are constant. For consistency of units, we write v com mi h ft mi 60min h ft min = F H G I K J = 85 5280 7480 b g . Thus, with ∆ x = 60ft , the time of flight is t x v = = = ∆ com 60 7480 0 00802 . min . During that time, the angular displacement of a point on the ball’s surface is θ ϖ = = ≈ t 1800 0 00802 14 rev min rev . b gb g . min 3. We have ϖ = 10 π rad/s. Since α = 0, Eq. 1013 gives ∆ θ = ϖ t = (10 π rad/s)( n ∆ t ), for n = 1, 2, 3, 4, 5, …. For ∆ t = 0.20 s, we always get an integer multiple of 2 π (and 2 π radians corresponds to 1 revolution). (a) At f 1 ∆ θ = 2 π rad the dot appears at the “12:00” (straight up) position. 441 CHAPTER 10 (b) At f 2 , ∆ θ = 4 π rad and the dot appears at the “12:00” position. ∆ t = 0.050 s, and we explicitly include the 1/2 π conversion (to revolutions) in this calculation: ∆ θ = ϖ t = (10 π rad/s) n (0.050 s) = ¼ , ½ , ¾ , 1, … (revs) (c) At f 1 (n=1) , ∆ θ = 1/4 rev and the dot appears at the “3:00” position. (d) At f 2 (n=2) , ∆ θ = 1/2 rev and the dot appears at the “6:00” position. (e) At f 3 (n=3) , ∆ θ = 3/4 rev and the dot appears at the “9:00” position. (f) At f 4 (n=4) , ∆ θ = 1 rev and the dot appears at the “12:00” position. Now ∆ t = 0.040 s, and we have ∆ θ = ϖ t = (10 π rad/s) n (0.040 s) = 0.2 , 0.4 , 0.6 , 0.8, 1, … (revs) Note that 20% of 12 hours is 2.4 h = 2 h and 24 min. (g) At f 1 (n=1) , ∆ θ = 0.2 rev and the dot appears at the “2:24” position. (h) At f 2 (n=2) , ∆ θ = 0.4 rev and the dot appears at the “4:48” position. (i) At f 3 (n=3) , ∆ θ = 0.6 rev and the dot appears at the “7:12” position. (j) At f 4 (n=4) , ∆ θ = 0.8 rev and the dot appears at the “9:36” position. (k) At f 5 (n=5) , ∆ θ = 1.0 rev and the dot appears at the “12:00” position. 4. If we make the units explicit, the function is θ = + 4 0 30 10 3 . . . rad / s rad / s rad / s 2 2 3 b g c h c h t t t 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) Eq. 106 leads to ϖ = + = + d dt t t t t t 4 3 4 6 3 2 3 2 c h ....
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This note was uploaded on 04/19/2008 for the course PHYS 1100, 1200 taught by Professor Many during the Fall '08 term at Rensselaer Polytechnic Institute.
 Fall '08
 Many
 Physics

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