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# Chapter-10 - Chapter 10 1 The problem asks us to assume...

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439 Chapter 10 1. 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 = 60 ft , the time of flight is com (60 ft) /(7480 ft/min) 0.00802min t x v = ∆ = = . 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 2. (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 12-hour watch turns through 2 π radians during 43200 s. Thus, ϖ = = × - 2 43200 145 10 4 π . rad / s. 3. Applying Eq. 2-15 to the vertical axis (with + y downward) we obtain the free-fall time: 2 0 2 1 2(10 m) 1.4 s. 2 9.8 m/s y y v t gt t = + = = Thus, by Eq. 10-5, the magnitude of the average angular velocity is avg (2.5 rev)(2 rad/rev) 11 rad/s. 1.4 s π ϖ = =

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441 (b) The largest angle (less than 1 revolution) turned for the toast to land butter-side down is max 0.75 rev 3 / 2 rad. θ π = = This corresponds to an angular speed of max max 3 / 2 rad 12.0 rad/s. 0.394 s t θ π ϖ = = = 6. 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 θ = + + but in some places we will proceed as indicated in the problem—by letting these units be understood. (a) We evaluate the function θ at t = 0 to obtain θ 0 = 2.0 rad. (b) The angular velocity as a function of time is given by Eq. 10-6: ( 29 ( 29 2 3 2 8.0 rad/s 6.0 rad/s d t t dt θ ϖ = = + which we evaluate at t = 0 to obtain ϖ 0 = 0.

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Chapter-10 - Chapter 10 1 The problem asks us to assume...

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