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**Unformatted text preview: **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 ω = = (b) The minute hand of the smoothly running watch turns through 2 π radians during 3600 s . Thus, ω = = × 2 3600 175 10 3 π . r a d / s . (c) The hour hand of the smoothly running 12-hour watch turns through 2 π radians during 43200 s. Thus, ω = = × 2 43200 145 10 4 π . r a d / s . 2. The problem asks us to assume v com and ω 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 . m i n . During that time, the angular displacement of a point on the balls surface is θ ω = = t 1800 0 00802 14 rev min rev . b g b g . m i n 3. We have ω = 10 π rad/s. Since α = 0, Eq. 10-13 gives θ = ω 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. (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 = ω t = (10 π rad/s) n (0.050 s) © § ¹ · 1 2 π = 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 = ω t = (10 π rad/s) n (0.040 s) © § ¹ · 1 2 π = 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 problemletting these units be understood. Also, in our manipulations we will generally not display the coefficients with their proper number of significant figures. ...

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