2.101
A bicycle is placed on a service rack with its wheels hanging free. As part of a bearing test, the front wheel
is spun at a rate of N=45 rev/min. Assume that this rate is constant and determine the speed v and the
magnitude a of the acceleration of point A.
Given: Diameter of wheel. Angle to point A on rim. Angular speed of wheel.
Find: The speed and scalar acceleration of point A.
Solution:
Since we are just looking for the speed and the magnitude of the acceleration the angle is not needed.
v = rω or with the book’s notation v = ρβ̇. Need to experess the the angular velocity in radians.
rad
rev rad
2π
= 282.7
min rev min
rad
v = (13.5 inches) 282.7
min
inches
inches
ft
v = 3820
= 63.6
= 5.30
min
s
s
ω = 45
Since the wheel is spinning at a constant speed there is no tangential acceleration.
2
3820 inches
v2
inches
min
a=
=
= 1.08 × 106
r
13.5 inches
min2
2
63.6 inches
inches
s
=
= 300 2
13.5 inches
s
2
5.30 fts
ft
= 25.0 2
= 13.5
s
12 ft
2.104
The two cars A and B enter an unbanked and level turn. They cross the line C − C simultaneously, and
each car has the speed corresponding to a maximum normal acceleration of 0.9g in the turn. Determine the
elapsed time for each car between its two crossings of line C − C What is the relative position of the two
cars as the second exits the turn? Assume no speed changes throughout.
Given: Radius of curve for each car. Maximum normal acceleration.
Find: The time it takes each car to complete the curve and how far down the track the faster car is when
the slower car leaves the curve.
Solution:
First finding the speed of each car
v2
= 0.9g
r
p
v = 0.9gr
r m
m
vA = 0.9 9.81 2 (72 m) = 25.2
s
s
r m
m
vB = 0.9 9.81 2 (88 m) = 27.9
s
s
Now the distance each car needs to cover:
a=
sA = π (72 m) = 226.2 m
sB = π (88 m) = 276.5 m
and now the time
226.2 m
sA
=
= 8.97 s
vA
25.2 m
s
276.5 m
sB
= 9.92 s
=
tB =
vB
27.9 m
s
tA =
Car A finishes ahead of car B by
tB − tA = 0.95 s
so Car A is 0.95 s 25.2 m
s = 24 m ahead as car B leaves the curve.
2.117
For the football of the previous problem, determine the radius of curvature ρ of the path and the time rate
of change v̇ of the speed at times t = 1 sec and t = 2 sec, there t = 0 is the time of release from the
quarterback’s hand.
Given: Initial speed and angle of the ball.
Find: The radius of curvature and scalar acceleration of the ball one and two seconds after the throw.
Solution:
To do this we need to break the acceleration into two components one, at , in the direction of motion of the
ball, the other an normal to the direction of travel of the ball. Let theta be the angle between the x-axis
and the velocity vector. Then,
vy
vx
at = v̇ = −g sin θ
tan θ =
v2
v2
= g cos θ ⇒ ρ =
ρ
g cos θ
q
v = vx2 + vy2
an =
vx = vx,o = vo cos (θo )
vy = vy,o − gt = vo sin (θo )
θo = 35◦
ft
vo = 80
s
Plugging in t=1,
Plugging in t=2,
ft
ft
cos (35◦ ) = 65.5
vx = 80
s
s
ft
ft
ft
◦
vy = 80
sin (35 ) − 32.2 2 (1 s) = 13.7
s
s
s
!
ft
13.7 s
vy
= 11.8◦
arctan
θ = arctan
vx
65.5 fts
ft
ft
v̇ = −g sin θ = − 32.2 2 sin (11.8◦ ) = −6.58 2
s
s
2
2
vx2 + vy2
65.5 fts + 13.7 fts
v2
= 142 ft
=
=
ρ=
g cos θ
g cos θ
32.2 sft2 cos (11.8◦ )
ft
ft
80
cos (35◦ ) = 65.5
s
s
ft
ft
ft
sin (35◦ ) − 32.2 2 (1 s) = −18.5
vy = 80
s
s
s
!
ft
−18.5 s
vy
= −15.8◦
θ = arctan
arctan
vx
65.5 fts
ft
ft
v̇ = −g sin θ = − 32.2 2 sin (−15.8◦) = 8.75 2
s
s
2
2
vx2 + vy2
65.5 fts + −18.5 fts
v2
=
=
= 150 ft
ρ=
g cos θ
g cos θ
32.2 sft2 cos (−15.8◦)
vx =