Physics Solution Manual for 1100 and 2101

Its acceleration a has two perpendicular components a

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Unformatted text preview: fan decreases, the sign of the angular acceleration must be opposite to the sign for the angular velocity. Taking the angular velocity to be positive, the angular acceleration, therefore, must be a negative quantity. Using Equation 8.4 we obtain ω 0 = ω − α t = 83.8 rad/s – (–42.0 rad/s 2 )(1.75 s) = 157.3 rad/s 68. REASONING The top of the racket has both tangential and centripetal acceleration components given by Equations 8.10 and 8.11, respectively: aT = rα and a c = r ω 2 . The total acceleration of the top of the racket is the resultant of these two components. Since these acceleration components are mutually perpendicular, their resultant can be found by using the Pythagorean theorem. 428 ROTATIONAL KINEMATICS SOLUTION Employing the Pythagorean theorem, we obtain a= 2 aT + a 2 = ( r α ) 2 + ( r ω 2 )2 = r α 2 + ω 4 c Therefore, a = (1.5 m) (160 rad/s2 )2 + (14 rad/s)4 = 380 m/s2 69. REASONING The length of tape that passes around the reel is just the average tangential speed of the tape times the time t. The average tangential speed vT is given by Equation 8.9 ( vT = rω ) as the radius r times the average angular speed ω in rad/s. SOLUTION The length L of tape that passes around the reel in t = 13 s is L = vT t . Using Equation 8.9 to express the tangential speed, we find L = vT t = rω t = ( 0.014 m ) ( 3.4 rad/s ) (13 s ) = 0.62 m 70. REASONING a. Since the angular velocity of the fan blade is changing, there are simultaneously a tangential acceleration aT and a centripetal acceleration ac that are oriented at right angles to each other. The drawing shows these two accelerations for a point on the tip of one of the blades (for clarity, the blade itself is not shown). The blade is rotating in the counterclockwise (positive) direction. The a= magnitude 2 ac 2 + aT of the total acceleration a aT φ ac is , according to the Pythagorean theorem. The magnitude ac of the centripetal acceleration can be evaluated from ac = rω 2 (Equation 8.11), where ω is the final angular velocity. The final angular velocity can be determined from Equation 8.4 as ω = ω0 + α t . The magnitude aT of the tangential acceleration follows from aT = rα (Equation 8.10). b. From the drawing we see that the angle φ can be obtained by using trigonometry, φ = tan −1 ( aT / ac ) . Chapter 8 Problems 429 SOLUTION 2 2 a. Substituting ac = rω 2 (Equation 8.11) and aT = rα (Equation 8.10) into a = ac + aT gives 2 2 a = ac + aT = ( rω 2 )2 + ( rα )2 = r ω4 + α 2 The final angular velocity ω is related to the initial angular velocity ω0 by ω = ω0 + α t (see Equation 8.4). Thus, the magnitude of the total acceleration is a = r ω4 +α 2 = r (ω0 + α t )4 + α 2 = ( 0.380 m ) 1.50 rad/s + ( 2.00 rad/s 2 ) ( 0.500 s ) + ( 2.00 rad/s 2 ) = 2.49 m/s 2 4 2 b. The angle φ between the total acceleration a and the centripetal acceleration ac is (see the drawing above) aT ac φ = tan −1 α −1 r α = tan −1 = tan 2 2 rω (ω0 + α t ) 2.00 rad/s 2 = 17.7° = tan −1 2 2 )( 1.50 rad/s + ( 2.00 rad/s 0.500 s ) where we have used the same substitutions for aT, ac, and ω as in part (a). ______________________________________________________________________________ 71. SSM REASONING The tangential speed vT of a point on the “equator” of the baseball is given by Equation 8.9 as vT = rω, where r is the radius of the baseball and ω is its angular speed. The radius is given in the statement of the problem. The (constant) angular speed is related to that angle θ through which the ball rotates by Equation 8.2 as ω = θ /t, where we have assumed for convenience that θ0 = 0 rad when t0 = 0 s. Thus, the tangential speed of the ball is θ vT = r ω = r t The time t that the ball is in the air is equal to the distance x it travels divided by its linear speed v, t = x/v, so the tangential speed can be written as 430 ROTATIONAL KINEMATICS θ θ rθ v vT = r = r = x x t v SOLUTION The tangential speed of a point on the equator of the baseball is −2 r θ v ( 3.67 × 10 m ) ( 49.0 rad ) ( 42.5 m/s ) vT = = = 4.63 m/s x 16.5 m 72. REASONING The average angular velocity is defined as the angular displacement divided by the elapsed time (Equation 8.2). Therefore, the angular displacement is equal to the product of the average angular velocity and the elapsed time The elapsed time is given, so we need to determine the average angular velocity. We can do this by using the graph of angular velocity versus time that accompanies the problem. Angular velocity SOLUTION The angular displacement ∆θ is +15 rad/ s related to the average angular velocity ω and the elapsed time ∆t by Equation 8.2, ∆θ = ω ∆t . The elapsed time is given as 8.0 s. To obtain the average angular velocity, we need to extend the graph that accompanies this problem from a ω time of 5.0 s to 8.0 s. It can be seen from the +3.0 rad/s graph that the angular velocity increases by 0 +3.0 rad/s during each second. Therefore, when 3.0 s the time i...
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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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