Unformatted text preview: REASONING In addition to knowing the initial angular velocity ω0 and the acceleration α,
we know that the final angular velocity ω is 0 rev/s, because the wheel comes to a halt.
With values available for these three variables, the unknown angular displacement θ can be
2
calculated from Equation 8.8 (ω 2 = ω 0 + 2αθ ) .
When using any of the equations of rotational kinematics, it is not necessary to use radian
measure. Any selfconsistent set of units may be used to measure the angular quantities,
such as revolutions for θ, rev/s for ω0 and ω, and rev/s2 for α.
A greater initial angular velocity does not necessarily mean that the wheel will come to a
halt on an angular section labeled with a greater number. It is certainly true that greater Chapter 8 Problems 409 initial angular velocities lead to greater angular displacements for a given deceleration.
However, remember that the angular displacement of the wheel in coming to a halt may
consist of a number of complete revolutions plus a fraction of a revolution. In deciding on
which number the wheel comes to a halt, the number of complete revolutions must be
subtracted from the angular displacement, leaving only the fraction of a revolution
remaining.
2
ω 2 − ω0
SOLUTION Solving Equation 8.8 for the angular displacement gives θ =
.
2α a. We know that ω0 = +1.20 rev/s, ω = 0 rev/s, and α = −0.200 rev/s2, where ω0 is positive
since the rotation is counterclockwise and, therefore, α is negative because the wheel
decelerates. The value obtained for the displacement is ω 2 − ω02 ( 0 rev/s ) − ( +1.20 rev/s )
θ=
=
= +3.60 rev
2α
2 ( −0.200 rev/s 2 )
2 2 To decide where the wheel comes to a halt, we subtract the three complete revolutions from
this result, leaving 0.60 rev. Converting this value into degrees and noting that each angular
section is 30.0º, we find the following number n for the section where the wheel comes to a
halt: 360° 1 angular section n = ( 0.60 rev ) = 7.2
30.0° 1 rev A value of n = 7.2 means that the wheel comes to a halt in the section following number 7.
Thus, it comes to a halt in section 8 .
b. Following the same procedure as in part a, we find that ω 2 − ω02 ( 0 rev/s ) − ( +1.47 rev/s )
θ=
=
= +5.40 rev
2α
2 ( −0.200 rev/s 2 )
2 2 Subtracting the five complete revolutions from this result leaves 0.40 rev. Converting this
value into degrees and noting that each angular section is 30.0º, we find the following
number n for the section where the wheel comes to a halt: 360° 1 angular section n = ( 0.40 rev ) = 4.8
30.0° 1 rev A value of n = 4.8 means that the wheel comes to a halt in the section following number 4.
Thus, it comes to a halt in section 5 . 410 ROTATIONAL KINEMATICS 33. SSM WWW REASONING The angular displacement of the child when he catches the
horse is, from Equation 8.2, θ c = ω c t . In the same time, the angular displacement of the
horse is, from Equation 8.7 with ω 0 = 0 rad/s, θ h = 1 α t 2 . If the child is to catch the horse
2
θ c = θ h + (π / 2) .
SOLUTION Using the above conditions yields
1 αt2
2 or
1 (0.0100
2 − ωc t + 1 π = 0
2 rad/s 2 )t 2 − ( 0.250 rad/s ) t + 1
2 (π rad ) = 0 The quadratic formula yields t = 7.37 s and 42.6 s; therefore, the shortest time needed to
catch the horse is t = 7.37 s . The angular acceleration α gives rise to a tangential acceleration aT
according to aT = rα (Equation 8.10). Moreover, it is given that aT = g, where g is the
magnitude of the acceleration due to gravity. 34. REASONING SOLUTION Let r be the radial distance of the point from the axis of rotation. Then,
according to Equation 8.10, we have
g
{ = rα
aT Thus,
r= g α = 9.80 m /s 2 12.0 rad /s 2 = 0.817 m 35. REASONING AND SOLUTION
a. ωA = v/r = (0.381 m/s)/(0.0508 m) = 7.50 rad/s b. ωB = v/r = (0.381 m/s)(0.114 m) = 3.34 rad/s
α= ω B −ω A
t = 3.34 rad/s − 7.50 rad/s
2.40 × 10 s The angular velocity is decreasing . 3 = − 1.73 × 10 −3 rad/s 2 Chapter 8 Problems 411 36. REASONING The tangential speed vT of a point on a rigid body rotating at an angular speed ω is given by vT = rω (Equation 8.9), where r is the radius of the circle described by the moving point. (In this equation ω must be expressed in rad/s.) Therefore, the angular
speed of the bacterial motor sought in part a is ω = vT r . Since we are considering a point
on the rim, r is the radius of the motor itself. In part b, we seek the elapsed time t for an
angular displacement of one revolution at the constant angular velocity ω found in part a.
We will use θ = ω0 t + 1 α t 2 (Equation 8.7) to calculate the elapsed time.
2
SOLUTION
a. The angular speed of the bacterial motor is, from ω = vT r (Equation 8.9), ω= vT 2.3 × 10−5 m/s
=
= 1500 rad/s
r
1.5 × 10 −8 m b. The bacterial motor is spinning at a constant angular velocity, so it has no angular
acceleration. Substituting α = 0 rad/s2 into θ = ω0t + 1 α t 2 (Equation 8.7), and solving for
2
the elapsed time yields ( ) θ = ω0t + 1 0 rad/s 2 t 2 = ω0t
2 or t=...
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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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