Chapter10 - 1. The problem asks us to assume vcom and are...

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1. 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 bg . Thus, with Δ x = 60ft , the time of flight is com (60 ft)/(7480 ft/min) 0.00802min tx v = = . During that time, the angular displacement of a point on the ball’s surface is θω == t 1800 0 00802 14 rev min rev . b gb g .m i n
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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 ω == (b) The minute hand of the smoothly running watch turns through 2 π radians during 3600 s . Thus, × 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, × 2 43200 145 10 4 π r a d / s .
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3. Applying Eq. 2-15 to the vertical axis (with + y downward) we obtain the free-fall time: 2 0 2 12 ( 1 0 m ) 1.4 s. 29 . 8 m / s y yv t g t t Δ= + ¡ == 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 ω
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4. If we make the units explicit, the function is θ =− + 40 30 10 3 .. . rad / s rad / s rad / s 22 3 b g ch tt 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. 10-6 leads to ω + = + d dt ttt 43 4 63 23 2 c h . Evaluating this at t = 2 s yields 2 = 4.0 rad/s. (b) Evaluating the expression in part (a) at t = 4 s gives 4 = 28 rad/s. (c) Consequently, Eq. 10-7 gives α avg 2 rad / s = = 42 12 . (d) And Eq. 10-8 gives == += + d dt d dt t 46 3 66 2 . c h Evaluating this at t = 2 s produces 2 = 6.0 rad/s 2 . (e) Evaluating the expression in part (d) at t = 4 s yields 4 = 18 rad/s 2 . We note that our answer for avg does turn out to be the arithmetic average of 2 and 4 but point out that this will not always be the case.
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(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. θπ Δ= = This corresponds to an angular speed of max max 3/ 2 r a d 12.0 rad/s. 0.394 s t θ π ω Δ == = Δ 5. The falling is the type of constant-acceleration motion you had in Chapter 2. The time it takes for the buttered toast to hit the floor is 2 22 ( 0 . 7 6 m ) 0.394 s. 9.8 m/s h t g = = (a) The smallest angle turned for the toast to land butter-side down is min 0.25 rev / 2 rad. = This corresponds to an angular speed of min min /2 rad 4.0 rad/s. 0.394 s t Δ = Δ
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6. If we make the units explicit, the function is () 22 33 2.0 rad 4.0 rad/s 2.0 rad/s tt θ =+ + but in some places we will proceed as indicated in the problem—by letting these units be understood. (a) We evaluate the function at t = 0 to obtain 0 = 2.0 rad. (b) The angular velocity as a function of time is given by Eq. 10-6: 23 2 8.0 rad/s 6.0 rad/s d dt ω == + which we evaluate at t = 0 to obtain 0 = 0. (c) For t = 4.0 s, the function found in the previous part is 4 = (8.0)(4.0) + (6.0)(4.0) 2 = 128 rad/s.
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Chapter10 - 1. The problem asks us to assume vcom and are...

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