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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.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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