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Unformatted text preview: ncreases from 5.0 to 8.0 s, the angular
velocity increases from +6.0 rad/s to
6 rad/s + 3×(3.0 rad/s) = +15 rad/s. A graph of
the angular velocity from 0 to 8.0 s is shown at
–9.0 rad/ s
the right. The average angular velocity during
this time is equal to one half the sum of the initial and final angular velocities: Time (s)
8.0 s ω = 1 (ω 0 + ω ) = 1 ( −9.0 rad/s + 15 rad/s ) = + 3.0 rad/s
2
2
The angular displacement of the wheel from 0 to 8.0 s is
∆θ = ω ∆t = ( +3.0 rad/s ) ( 8.0 s ) = +24 rad 73. REASONING The time required for the change in the angular velocity to occur can be
found by solving Equation 8.4 for t. In order to use Equation 8.4, however, we must know
the initial angular velocity ω 0 . Equation 8.6 can be used to find the initial angular velocity. Chapter 8 Problems 431 SOLUTION From Equation 8.6 we have
1
2 θ = (ω 0 + ω )t
Solving for ω0 gives ω0 = 2θ
−ω
t Since the angular displacement θ is zero, ω0 = –ω. Solving ω = ω0 + α t (Equation 8.4) for
t and using the fact that ω0 = –ω give t= 2ω α = 2(− 25.0 rad/s) 2= −4.00 rad/s 12.5 s ____________________________________________________________________________________________ 74. REASONING The drawing shows a top view of
the race car as it travels around the circular turn.
Its acceleration a has two perpendicular
components: a centripetal acceleration ac that
arises because the car is moving on a circular path
and a tangential acceleration aT due to the fact that
the car has an angular acceleration and its angular
velocity is increasing. We can determine the
magnitude of the centripetal acceleration from
Equation 8.11 as ac = rω2, since both r and ω are
given in the statement of the problem. As the
drawing shows, we can use trigonometry to
determine the magnitude a of the total
acceleration, since the angle (35.0°) between a and ac is given. aT a
35.0° ac Race car SOLUTION Since the vectors ac and a are one side and the hypotenuse of a right triangle,
we have that
ac
a=
cos 35.0° The magnitude of the centripetal acceleration is given by Equation 8.11 as ac = rω , so the
magnitude of the total acceleration is
2 ( 23.5 m )( 0.571 rad /s )
rω2
a=
=
=
= 9.35 m /s 2
cos 35.0° cos 35.0°
cos 35.0°
ac 2 432 ROTATIONAL KINEMATICS 75. REASONING The golf ball must travel a distance equal to its diameter in a maximum
time equal to the time required for one blade to move into the position of the previous blade.
SOLUTION The time required for the golf ball to pass through the opening between two
blades is given by ∆t = ∆θ / ω , with ω = 1.25 rad/s and ∆θ = ( 2π rad)/16 = 0.393 rad .
Therefore, the ball must pass between two blades in a maximum time of ∆t = 0.393 rad
= 0.314 s
1.25 rad/s The minimum speed of the ball is v= ∆x 4.50 × 10 –2 m
=
= 1.43 × 10–1 m/s
∆t
0.314 s 76. REASONING The wheels on both sides of the car have the same radius r = 0.350 m and
undergo rolling motion, so we will use v = rω (Equation 8.12) to calculate their individual
angular speeds: ωleft = vleft
r and ωright = vright (1) r In Equations (1) the linear speeds vleft and vright at which the wheels on opposite sides of the
car travel around the track differ. This is because the wheels on one side of the car are closer
to the center of the track than are the wheels on the other side. As the car makes one
complete lap of the track, therefore, both sets of wheels follow circular paths of different
radii Rleft and Rright. The linear speed of each wheel is the circumference of its circular path
divided by the elapsed time t, which is the same for both sets of wheels:
vleft = 2π Rleft
t and vright = 2π Rright (2) t Substituting Equations (2) into Equations (1), we obtain ωleft 2π Rleft
2π Rleft
t
=
=
r
rt 2π Rright
and ωright = t
r = 2π Rright
rt (3) SOLUTION We will assume that the wheels on the left side of the car are closer to the
center of the track than the wheels on the right side. We do not know the radii of the circular
paths of either set of wheels, but the difference between them is Rright − Rleft = 1.60 m . We Chapter 8 Problems 433 can now calculate the difference between the angular speeds of the wheels on the left and
right sides of the car by subtracting ωleft from ωright [see Equations (3)]: ωright − ωleft = 2π Rright
rt − 2π Rleft
rt = ( 2π Rright − Rleft
rt )= 2π (1.60 m ) ( 0.350 m )(19.5 s ) = 1.47 rad/s CHAPTER 9 ROTATIONAL DYNAMICS
ANSWERS TO FOCUS ON CONCEPTS QUESTIONS
___________________________________________________________________________________________ 1. (d) A rigid body is in equilibrium if it has zero translational acceleration and zero angular
acceleration. A body, such as a bicycle wheel, can be moving, but the translational and
2
2
angular accelerations must be zero (a = 0 m/s and α = 0 rad/s ). 2. (e) As discussed in Section 9.2, a body is in equilibrium if the sum of the externally applied
forces is zero and the sum of the externally applied torques is zero. 3. (b) The torque τ3...
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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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