Unformatted text preview: ons made by the larger bicycle
wheel, which has a radius of r2 = r1 + 0.012 m. SOLUTION Because 1 km = 1000 m, the total distance traveled during the race is
D = (4520 km)[(1000 m)/(1 km)] = 4520×103 m. From Equation (1), then, the radius r1 of
the smaller bicycle wheel is
r1 = D
4520 × 103 m
=
= 0.330 m
2π N1 2π 2.18 × 106 ( ) The larger wheel, then, has a radius r2 = 0.330 m + 0.012 m = 0.342 m. Over the same
distance D, this wheel would make N2 revolutions, where, by Equation (1),
N2 = D
4520 × 103 m
=
= 2.10 × 106
2π r2 2π ( 0.342 m ) Chapter 8 Problems 425 61. REASONING As a pennyfarthing moves, both of its wheels roll without slipping. This
means that the axle for each wheel moves through a linear distance (the distance through
which the bicycle moves) that equals the circular arc length measured along the outer edge
of the wheel. Since both axles move through the same linear distance, the circular arc length
measured along the outer edge of the large front wheel must equal the circular arc length
measured along the outer edge of the small rear wheel. In each case the arc length s is equal
to the number n of revolutions times the circumference 2π r of the wheel (r = radius).
SOLUTION Since the circular arc length measured along the outer edge of the large front
wheel must equal the circular arc length measured along the outer edge of the small rear
wheel, we have
nRear 2π rRear = nFront 2π rFront
14 3
24
14243
Arc length for rear
wheel Solving for nRear gives
nRear = nFront rFront
rRear = Arc length for front
wheel 276 (1.20 m )
= 974 rev
0.340 m 62. REASONING While the ball is in the air, its angular speed ω is constant, and thus its
angular displacement is given by θ = ω t (Equation 8.2). The angular speed of the ball is
found by considering its rolling motion on the table top, because its angular speed does not
change after it leaves the table. For rolling motion, the angular speed ω is related to the
linear speed v by ω = v r (Equation 8.12), where r is the radius of the ball. In order to
determine the time t the ball spends in the air, we treat it as a projectile launched
horizontally. The vertical displacement of the ball is then given by y = v0 yt + 1 ayt 2
2
(Equation 3.5b), which we will use to determine the time t that elapses while the ball is in
the air.
SOLUTION Since the ball is launched horizontally in the projectile motion, its initial
velocity v0 has no vertical component: v0y = 0 m/s. Solving y = v0 yt + 1 ayt 2 (Equation 3.5b)
2
for the elapsed time t, we obtain y = ( 0 m/s ) t + 1 a y t 2 = 1 a y t 2
2
2 or t= 2y
ay (1) 426 ROTATIONAL KINEMATICS Substituting Equation (1) and ω = v
(Equation 8.12) into θ = ω t (Equation 8.2) yields
r v 2y θ = r ay
{{
ωt (2) We choose the upward direction to be positive. Once the ball leaves the table, it is in free
fall, so the vertical acceleration of the ball is that due to gravity: ay = −9.80 m/s2. Further,
the ball’s vertical displacement is negative as the ball falls to the floor: y = −2.10 m. The
ball’s angular displacement while it is in the air is, from Equation (2), θ= 63. v
r 2y
3.60 m/s
=
a y 0.200 m/s SSM REASONING AND SOLUTION
"axle" or the center of the moving quarter is 2 ( −2.10 m )
−9.80 m/s 2 = 11.8 rad By inspection, the distance traveled by the d = 2 π ( 2 r ) = 4π r
where r is the radius of the quarter. The distance d traveled by the "axle" of the moving
quarter must be equal to the circular arc length s along the outer edge of the quarter. This arc
length is s = rθ , where θ is the angle through which the quarter rotates. Thus, 4 π r = rθ
so that θ = 4π rad . This is equivalent to 1 rev = 2 revolutions
(4π rad) 2π rad 64. REASONING AND SOLUTION
According to Equation 8.2, ω = ∆θ / ∆t . Since the angular speed of the sun is constant,
ω = ω . Solving for ∆ t , we have
∆t = ∆θ 2π rad
1y 1 h 1 day 8
= = 1.8 ×10 y −15
ω 1.1× 10 rad/s 3600 s 24 h 365.25 day Chapter 8 Problems 65. 427 SSM REASONING The tangential acceleration aT of the speedboat can be found by
using Newton's second law, FT = maT , where FT is the net tangential force. Once the
tangential acceleration of the boat is known, Equation 2.4 can be used to find the tangential
speed of the boat 2.0 s into the turn. With the tangential speed and the radius of the turn
known, Equation 5.2 can then be used to find the centripetal acceleration of the boat.
SOLUTION
a. From Newton's second law, we obtain aT = FT 550 N
=
= 2.5 m/s 2
m 220 kg b. The tangential speed of the boat 2.0 s into the turn is, according to Equation 2.4,
2 1 v T = v 0 T + aT t = 5.0 m/s + (2.5 m/s )(2.0 s) = 1.0 × 10 m/s The centripetal acceleration of the boat is then
v2
T (1.0 × 10 1 m/s) 2
ac =
=
= 3.1 m/s 2
r
32 m 66. REASONING AND SOLUTION From Equation 8.6, θ = (ω 0 + ω ) t . Solving for t gives
1
2 t= 67. 2θ
2(85.1 rad)
=
= 5.22 s
ω 0 + ω 18.5 rad/s +14.1 rad/s SSM REASONING AND SOLUTION Since the angular speed of the...
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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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