Solution to HW#8
CJ5 8.CQ.006. REASONING AND SOLUTION The wheels are rotating with a constant
angular velocity.
a. Since the angular velocity is constant, each wheel has zero angular acceleration, α
=
0 rad/s. Since the tangential acceleration aT is related to the angular acceleration
through Equation 8.10,
acceleration.
aT = rα , every point on the rim has zero tangential
b. Since the particles on the rim of the wheels are moving along a circular path, they
must have a centripetal acceleration. This can be supported by Equation 8.11,
ac = r ω 2 , where ac is the magnitude of the centripetal acceleration. Since ω is
nonzero, ac is nonzero.
CJ5 8.CQ.012. REASONING AND SOLUTION The bicycle wheel has an angular
acceleration. The arrows are perpendicular to the radius of the wheel. The
magnitude of the arrows increases with increasing distance from the center in
accordance with vT = r ω , or aT = rα . The arrows in the picture could represent
either the tangential velocity or the tangential acceleration. The arrows are not
directed radially inward; therefore, they cannot represent the centripetal acceleration.
CJ5 8.P.008. REASONING AND SOLUTION We can use Equation 8.4, assuming, as
usual, that
t0 = 0 s. The time is
t=
ω − ω 0 5.2 rad/s − 0 rad/s
=
= 1.3 s
2
α
4.0 rad/s
CJ5 8.P.011. SSM REASONING The time required for the bullet to travel the
distance d is equal to the time required for the discs to undergo an angular displacement
of 0.240 rad. The time can be found from Equation 8.2; once the time is known, the
speed of the bullet can be found using Equation 2.2.
SOLUTION From the definition of average angular velocity:
ω=
the required time is
∆t =
∆θ
ω
=
∆θ
∆t
0.240 rad
−3
= 2.53 × 10 s
95.0 rad/s
Note that ω = ω because the angular speed is constant. The (constant) speed of
the bullet can then be determined from the definition of average speed:
v=
∆x
d
0.850 m
=
=
=
∆t
∆t
2.53 × 10−3 s
336 m/s
CJ5 8.P.016. REASONING AND SOLUTION
a. Since the skater comes to a stop, her final angular speed is zero. Using
Equation 8.8, ω 2 = ω 02 + 2αθ , and solving for α , we obtain
ω 02
(15 rad/s) 2
2
=−
= –22 rad/s
α =−
2θ
2(5.1 rad)
b. Using Equation 8.4 ( ω = ω 0 + α t ) with ω = 0 rad/s, we have
t =−
ω0
15 rad/s
=−
= 0.68 s
α
–22 rad/s 2
CJ5 8.P.023. SSM 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.
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 Equation 8.4 for t,
and using the fact that ω0 = –ω, gives
2ω 2(−25.0 rad/s)
t=
=
= 12.5 s
α
−4.00 rad/s2
CJ5 8.P.033. REASONING AND SOLUTION The average tangential speed v T of
points on the rim of the reel is given by Equation 8.9,
vT = r ω = (2.5 × 10−2 m)(1.9 rad/s) = 0.048 m/s
where ω is the average angular speed. The average tangential speed must equal
the average linear speed v of the tape. The length L of tape that passes around the reel in
a time ∆t can be found from the definition of average velocity:
∆x
∆t
Solving for ∆x and setting ∆x = L gives
v=
L = v ∆t = (0.048 m/s)(7.1 s) = 0.34 m
CJ5 8.P.039. SSM REASONING The top of the racket has both tangential and
centripetal acceleration components given by Equations 8.10 and 8.11, respectively:
2
aT = rα and a c = rω . 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.
SOLUTION Employing the Pythagorean theorem, we obtain
a=
aT2 + a 2c = (r α ) 2 + (r ω 2 )2 = r α 2 + ω 4
Therefore,
a = (1.5 m) (160 rad/s 2 )2 + (14 rad/s)4 = 380 m/s2
CJ5 8.P.048. REASONING AND SOLUTION
a.
If the wheel does not slip, a point on the rim rotates about the axle with a
speed
vT = v = 15.0 m/s
For a point on the rim
ω = vT/r = (15.0 m/s)/(0.330 m) = 45.5 rad/s
b.
vT = rω = (0.175 m)(45.5 rad/s) = 7.96 m/s