Physics Solution Manual for 1100 and 2101

In each case the arc length s is equal to the number

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Unformatted text preview: wheel v = rω . It would then travel a distance 60 s x = v t = r ω t = ( 0.45 m )( 9.1 rad/s )( 35 min ) = 1 min 8.6 ×103 m ____________________________________________________________________________________________ 55. REASONING The angular displacement θ of each wheel is given by Equation 8.7 (θ = ω0t + 12 α t 2 ) , which is one of the equations of rotational kinematics. In this expression ω0 is the initial angular velocity, and α is the angular acceleration, neither of which is given directly. Instead the initial linear velocity v0 and the linear acceleration a are given. However, we can relate these linear quantities to their analogous angular counterparts by means of the assumption that the wheels are rolling and not slipping. Then, according to Equation 8.12 (v0 = rω0), we know that ω0 = v0/r, where r is the radius of the wheels. Likewise, according to Equation 8.13 (a = rα), we know that α = a/r. Both Equations 8.12 and 8.13 are only valid if used with radian measure. Therefore, when we substitute the expressions for ω0 and α into Equation 8.7, the resulting value for the angular displacement θ will be in radians. 422 ROTATIONAL KINEMATICS SOLUTION Substituting ω0 from Equation 8.12 and α from Equation 8.13 into Equation 8.7, we find that v a θ = ω 0t + 1 α t 2 = 0 t + 1 t 2 2 2 r r 2 2 20.0 m/s 1 1.50 m/s = ( 8.00 s ) = 693 rad ( 8.00 s ) + 2 0.300 m 0.300 m 56. REASONING AND SOLUTION a. If the wheel does not slip, a point on the rim rotates about the axle with a speed vT = v = 15.0 m/s For a point on the rim ω = vT/r = (15.0 m/s)/(0.330 m) = 45.5 rad/s vT = rω = (0.175 m)(45.5 rad/s) = 7.96 m/s b. 57. REASONING a. The constant angular acceleration α of the wheel is defined by Equation 8.4 as the change in the angular velocity, ω −ω0, divided by the elapsed time t, or α = (ω − ω0 ) / t . The time is known. Since the wheel rotates in the positive direction, its angular velocity is the same as its angular speed. However, the angular speed is related to the linear speed v of a wheel and its radius r by v = rω (Equation 8.12). Thus, ω = v / r , and we can write for the angular acceleration that v v0 ω − ω0 r − r v − v0 = = α= t t rt b. The angular displacement θ of each wheel can be obtained from Equation 8.7 of the equations of kinematics: θ = ω0t + 1 α t 2 , where ω0 = v0/r and α can be obtained as 2 discussed in part (a). SOLUTION a. The angular acceleration of each wheel is α= v − v0 rt = 2.1 m/s − 6.6 m/s = −1.4 rad/s 2 ( 0.65 m )( 5.0 s ) Chapter 8 Problems 423 b. The angular displacement of each wheel is v0 1 2 t + α t r 2 θ = ω0t + 1 α t 2 = 2 2 6.6 m/s ( 2 1 = 5.0 s ) + 2 ( −1.4 rad/s ) ( 5.0 s ) = +33 rad 0.65 m ______________________________________________________________________________ 58. REASONING For a wheel that rolls without slipping, the relationship between its linear speed v and its angular speed ω is given by Equation 8.12 as v = rω, where r is the radius of a wheel. For a wheel that rolls without slipping, the relationship between the magnitude a of its linear acceleration and the magnitude α of the angular acceleration is given by Equation 8.13 as a = rα, where r is the radius of a wheel. The linear acceleration can be obtained using the equations of kinematics for linear motion, in particular, Equation 2.9. SOLUTION a. From Equation 8.12 we have that v = r ω = ( 0.320 m ) ( 288 rad /s ) = 92.2 m /s b. The magnitude of the angular acceleration is given by Equation 8.13 as α = a /r. The linear acceleration a is related to the initial and final linear speeds and the displacement x by 2 v 2 − v0 Equation 2.9 from the equations of kinematics for linear motion; a = . Thus, the 2x magnitude of the angular acceleration is ( ) v − v0 / ( 2 x ) a α= = r r 2 = 2 2 v − v0 2x r 2 ( 92.2 m /s )2 − ( 0 m /s )2 = 2 ( 384 m )( 0.320 m ) = 34.6 rad /s 2 59. REASONING AND SOLUTION a. If the rope is not slipping on the cylinder, then the tangential speed of the teeth on the larger gear (gear 1) is 2.50 m/s. The angular speed of gear 1 is then ω1 = v/r1 = (2.50 m/s)/(0.300 m) = 8.33 rad/s 424 ROTATIONAL KINEMATICS The direction of the larger gear is counterclockwise . b. The gears are in contact and do not slip. This requires that the teeth on both gears move with the same tangential speed. vT1 = vT2 or ω1r1 = ω2r2 So r 0.300 m ω 2 = 1 ω1 = ( 8.33 rad/s ) = 14.7 rad/s 0.170 m r2 The direction of the smaller gear is clockwise . 60. REASONING The distance d traveled by the axle of a rolling wheel (radius = r) during one complete revolution is equal to the circumference (2π r) of the wheel: d = 2π r. Therefore, when the bicycle travels a total distance D in a race, and the wheel makes N revolutions, the total distance D is N times the circumference of the wheel: D = Nd = N ( 2π r ) (1) We will apply Equation (1) first to the smaller bicycle wheel to determine its radius r1. Equation (1) will then also determine the number of revoluti...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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