Version 040/AACCA – midterm 02 – Turner – (58220)
1
This
printout
should
have
19
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001
10.0 points
A curve of radius 51
.
5 m is banked so that
a car traveling with uniform speed 63 km
/
hr
can round the curve without relying on fric
tion to keep it from slipping to its left or right.
The acceleration of gravity is 9
.
8 m
/
s
2
.
2
.
9 Mg
μ
≈
0
θ
What is
θ
?
1. 22.0794
2. 22.7692
3. 28.3155
4. 31.2492
5. 27.0
6. 34.1362
7. 23.9453
8. 16.5845
9. 29.315
10. 25.8513
Correct answer: 31
.
2492
◦
.
Explanation:
Let :
m
= 2900 kg
,
v
= 63 km
/
hr
,
r
= 51
.
5 m
,
and
μ
≈
0
.
Basic Concepts:
Consider the free body
diagram for the car.
The forces acting on
the car are the normal force, the force due to
gravity, and possibly friction.
μ
N
N
N
cos
θ
m g
N
sin
θ
x
y
To keep an object moving in a circle re
quires a force directed toward the center of
the circle; the magnitude of the force is
F
c
=
m a
c
=
m
v
2
r
.
Also remember,
vector
F
=
summationdisplay
i
vector
F
i
.
Using the freebody diagram, we have
summationdisplay
i
F
x
N
sin
θ
−
μ
N
cos
θ
=
m
v
2
r
(1)
summationdisplay
i
F
y
N
cos
θ
+
μ
N
sin
θ
=
m g
(2)
(
m g
)
bardbl
=
m g
sin
θ
(3)
m a
bardbl
=
m
v
2
r
cos
θ
(4)
and
,
if
μ
= 0
,
we have
tan
θ
=
v
2
g r
(5)
Solution:
Solution in an Inertial Frame:
Watching from the Point of View of Some
one Standing on the Ground.
The car is performing circular motion with
a constant speed, so its acceleration is just
the centripetal acceleration,
a
c
=
v
2
r
.
The
net force on the car is
F
net
=
m a
c
=
m
v
2
r
The component of this force parallel to the
incline is
summationdisplay
vector
F
bardbl
=
m g
sin
θ
=
F
net
cos
θ
=
m
v
2
r
cos
θ .
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Version 040/AACCA – midterm 02 – Turner – (58220)
2
In this reference frame, the car is at rest,
which means that the net force on the car
(taking in consideration the centrifugal force)
is zero. Thus the component of the net “real”
force parallel to the incline is equal to the
component of the centrifugal force along that
direction.
Now, the magnitude of the cen
trifugal force is equal to
F
c
=
m
v
2
r
, so
F
bardbl
=
F
net
cos
θ
=
F
c
cos
θ
=
m
v
2
r
cos
θ
F
bardbl
is the component of the weight of the car
parallel to the incline. Thus
m g
sin
θ
=
F
bardbl
=
m
v
2
r
cos
θ
tan
θ
=
v
2
g r
=
(63 km
/
hr)
2
(9
.
8 m
/
s
2
) (51
.
5 m)
×
parenleftbigg
1000 m
km
parenrightbigg
2
parenleftbigg
hr
3600 s
parenrightbigg
2
= 0
.
606796
θ
= arctan(0
.
606796)
= (0
.
545402 rad)
bracketleftbigg
180 deg
π
rad
bracketrightbigg
= 31
.
2492
◦
.
002
10.0 points
An
80 kg boat that is 8
.
6 m in length is
initially 8
.
4 m from the pier. A
56 kg child
stands at the end of the boat closest to the
pier. The child then notices a turtle on a rock
at the far end of the boat and proceeds to
walk to the far end of the boat to observe the
turtle.
8
.
4 m
8
.
6 m
How far is the child from the pier when she
reaches the far end of the boat? Assume there
is no friction between boat and water.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Turner
 Physics, Force, Correct Answer, kg, Version 040/AACCA

Click to edit the document details