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Unformatted text preview: 427 Chapter 10 1. The problem asks us to assume v com and 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 60 ft , the time of flight is com (60 ft) /(7480 ft/min) 0.00802 min t x v . During that time, the angular displacement of a point on the balls surface is t 1800 0 00802 14 rev min rev . b gb g . min 2. (a) The second hand of the smoothly running watch turns through 2 radians during 60 s . Thus, 2 0.105 rad/s. 60 (b) The minute hand of the smoothly running watch turns through 2 radians during 3600 s . Thus, 2 3600 175 10 3 . rad / s. (c) The hour hand of the smoothly running 12hour watch turns through 2 radians during 43200 s. Thus, 2 43200 145 10 4 . rad / s. 3. The falling is the type of constantacceleration motion you had in Chapter 2. The time it takes for the buttered toast to hit the floor is 2 2 2(0.76 m) 0.394 s. 9.8 m/s h t g (a) The smallest angle turned for the toast to land butterside down is min 0.25 rev / 2 rad. This corresponds to an angular speed of CHAPTER 10 428 min min / 2 rad 4.0 rad/s. 0.394 s t (b) The largest angle (less than 1 revolution) turned for the toast to land butterside down is max 0.75 rev 3 / 2 rad. This corresponds to an angular speed of max max 3 / 2 rad 12.0 rad/s. 0.394 s t 4. If we make the units explicit, the function is 2 2 3 3 2.0 rad 4.0 rad/s 2.0 rad/s t t but in some places we will proceed as indicated in the problemby letting these units be understood. (a) We evaluate the function at t = 0 to obtain = 2.0 rad. (b) The angular velocity as a function of time is given by Eq. 106: 2 3 2 8.0 rad/s 6.0 rad/s d t t dt which we evaluate at t = 0 to obtain = 0. (c) For t = 3.0 s, the function found in the previous part is 4 = (8.0)(3.0) + (6.0)(3.0) 2 = 78 rad/s. (d) The angular acceleration as a function of time is given by Eq. 108: 2 3 8.0 rad/s 12 rad/s d t dt which yields 2 = 8.0 + (12)(4.0) = 56 rad/s 2 at t = 4.0 s. (e) The angular acceleration, given by the function obtained in the previous part, depends on time; it is not constant. 5. Applying Eq. 215 to the vertical axis (with + y downward) we obtain the freefall time: 429 2 2 1 2(10 m) 1.4 s. 2 9.8 m/s y y v t gt t Thus, by Eq. 105, the magnitude of the average angular velocity is avg (2.5 rev)(2 rad/rev) 11 rad/s....
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 Spring '11
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