Dynamics II
Motion in a Plane
Review Problems
Problem 1
A 500 g model rocket is on a cart that is rolling
to the right at a speed of 3.0 m/s. The rocket
engine, when it is fired , exerts an 8.0 N thrust
on the rocket. Your goal is to have the rocket
pass through a small horizontal loop that is 20
m above the launch point. At what horizontal
point to the left of the hoop should you
launch?
2
AP Physics C
Problem 1
con’t
3
AP Physics C
Problem 1
con’t
Analysis
: Essentially a projectile motion problem with
positive vertical acceleration and constant horizontal
velocity. Assume a particle model. Compute the time
required to rise 20 m and use that to determine the
horizontal distance.
4
AP Physics C
T
F  mg=ma
2
s
8N.5kg(9.8m/s )
a = =6.2m/s
.5kg
2
1
20 = 6.2 t
2
t =2.54sec
d=2.54s(3m/s) =7.62m
Problem 2
A highway curve of radius 500 m is designed for traffic
moving at speeds of 90
kph
. What is the correct
banking angle of the road?
5
AP Physics C
θ
θ
Ncos
θ =mg
2
mv
Nsin
θ =
r
2
v
tan
θ =
rg
2
25
= 0.128 7.3
500(9.8)
Analysis
: N
r
provides the centripetal acceleration and
N
z
counters the
gravitational force.
Problem 3
A 30 g ball rolls around a 40 cm diameter horizontal
track at 60 rpm. What is the magnitude of the net force
that the track exerts on the ball? Neglect rolling
friction.
6
AP Physics C
Analysis
: Force against the vertical
element of the track provides centripetal
acceleration and the horizontal supports
the weight.
F
r
F
z
2
r
2
60 rev 2
π rad 1min
F =mr = 0.030 kg 0.20 m × × =0.24 N
min 1 rev 60 s
Problem 4
A student has 65 cm arm length. What is the minimum angular velocity,
in rpm, for swinging bucket of water in a vertical circle without spilling
any? The distance from the handle to the bottom of the bucket is 35 cm.
7
AP Physics C
Analysis
: The minimum angular velocity for swinging a
bucket of water in a vertical circle without spilling any
water corresponds to the case when the speed of the
bucket is critical. In this case, n = 0 N when the bucket is
in the top position of the circular motion.
2
2
c
rG c
mv
F =0 N+mg= =mr
ω
r
2
c
9.8 m/s
ω = g/r = =3.13 rad/s = 30 rpm
1.00 m
Problem 5
A car is tested on a 200 m diameter track. If the car speeds up at a
steady 1.5 m/s
2
, how long after starting is the magnitude of its
centripetal accelerations equal to the tangential acceleration?
8
AP Physics C
Analysis
: NUCM; we know a
t
= 1.5 m/s
2
from that we should be able to
determine the angular velocity and acceleration,
ω
and
α
, and then the time.
2 2
r
a =
ωr =1.5 m/s
2
2 2
t
a 1.5 m/s
α = = =1.5
×10 s.
r 100 m
i
ω=ω +αΔt
1 1
i
2
ω ω 0.122 s  0 s
Δt = = =8.2 s
α 0.015 s
2 2
1.5 m/s 1.5 m/s
ω= = =0.122 rad/s
r 100 m
Problem 6
A popular pastime is to see who ca push an object closest to the
edge of a table without its going off. You push the 100 g object and
release it 2.0 m from the table edge. Unfortunately you push a little
too hard. The object slides across, sails off the edge, falls 1.0 m to
the floor, and lands 30 cm from the edge of the table. If the
coefficient of kinetic friction is 0.50, what was the object’s speed as
you released it?
9
AP Physics C
Analysis
: After release velocity is
affected by friction only. When it
reaches the edge of the table it
becomes a projectile motion
problem with horizontal velocity
constant. Best approaches appear
to be to work backward from the
projectile motion to the release
velocity.
Problem 6
con’t
AP Physics C
10
x k x k
F =f =ma =
μ mg
2 2
x k
a =
μ g= 0.50 9.8 m/s =4.9 m/s
2
1
2 1 1y 2 1 2 1
2
y = y +v t  t + g t  t
2
2 1
g
0 m=1.0 m+0 m t  t
2
2 1
t  t =0.4518 s
2 1 1x 2 1
x = x +v t  t
1x
2.30 m=2.0 m+v 0.4518 s
1x
v =0.664 m/s
Δ
y = 1 m and v
1y
= 0
Similarly in the x direction
Given v
1x
we can find v
0x
given the acceleration.
2 2
1x 0x x 1 0
v = v +2a x  x:
2
2 2
0x 0x
0.664 m/s = v +2 4.9 m/s 2.0 m v = 4.5 m/s
Problem 7
A motorcycle daredevil plans to ride up a 2.0 m high, 20
°
ramp, sail
across a 10 m wide pool filled with crocodiles, and land at ground
level on the other side. Unfortunately, the motorcycle engine dies
just as he starts up the ramp. He is going 11 m/s at that instant, and
the coefficient of rolling friction is 0.02. Does he make it?
11
AP Physics C
Analysis
: Two part problem: First is a
ramp problem to determine velocity at
end of ramp. Second part is a
projectile motion problem determining
how far right he will travel while
dropping 2.0 m. Different sets of axes
will be required for each part.
Problem 7
con’t
AP Physics C
12
net  r
(F ) =f  mgsin20°
r
=
μn mgsin20°
r 0
=
μmgcos20° mgsin20° =ma
0 r
a =g(
μ cos20° +sin20°)
2 2
=(9.8 m/s )((0.02)cos20° +sin20°) =3.536
m/s
Dividing by M and solving for a
0
gives:
First find the acceleration parallel to the ramp:
The length of the ramp is 2 m/ sin 20
°
= 5.85 m
2 2
1 0 0 1 0
2 2
1
Using v = v +2a (s  s )
v = (11.0 m/s) +2(3.536 m/s )(5.85 m) =8.92
m/s
Problem 7
con’t
AP Physics C
13
Now we have the velocity at the beginning
of projectile motion. Axes are now
horizontal and vertical. Velocity
components are:
m
s
1x 1
v = v cos20° =8.38
m
S
1y 1
v = v sin20° =3.05
2
1
2 1 1y 1y 2 1
2
2 2
y =0 m= y +v t + a t For
Δt = t  t
=2.0 m+(3.05 m/s) t  (4.90 m/s ) t
t =1.021s
Time in the air is given by:
X
2
is given by: 8.38 m/s (
Δ
t) = 8.56 m.
Looks like crocodile food!!
Maximum speed is when the static friction force
reaches its maximum value
Substituting this into the above equations gives:
Problem 8
A concrete highway curve of radius 70 m is banked at a 15
°
angle.
What is the maximum speed with which a 1500 kg rubber tired car can
take this curve without sliding?
14
AP Physics C
2
r k
mv
F = f cos
θ+nsinθ =
r
z G
F =ncos
θ f sinθ F =0
k
k k
max
f =
μ n.
2
s
mv
n
μ cos15° +sin15° =
r
s
n cos15° 
μ sin15° =mg
Dividing the bottom into the top gives:
2
s s
m
s
s s
μ +tan15° μ +tan15°
v
= v = gr =34
1
μ tan15° gr 1 μ tan15°
The summation of r and z forces gives:
Problem 9
A conical pendulum is formed by attaching a 500g ball
to a 1.0 m long string, then allowing the mass to move
in a horizontal circle of radius 20 cm. a. Find the
tension is the string and b. the angular speed of the
ball in rpm.
15
AP Physics C
Analysis
: The mass moves in a
horizontal
circle of radius
The acceleration and the net force vector point to the
center of the circle,
not
along the string. The only two
forces are the string tension, which does point along the
string, and the gravitational force. These are shown in
the free

body diagram. Newton’s second law for circular
motion is
z G
F = Tcos
θ F = Tcosθ mg=0 N
2
r r
mv
F =Tsin
θ =ma =
r
2
0.500 kg 9.8 m/s
mg
T = = =5.00 N
cos
θ cos11.54°
From F
Z
:
Problem 10
A 500 g ball moves in a vertical circle on a 102
cm long string. If the speed at the top of the
4.0m/s, then the speed at the bottom will be 7.5
m/s. Find (a) the gravitational force acting on
the ball, (b) The tension in the string at the top
of the circle and (c) the tension in the string at
the bottom of the circle.
16
AP Physics C
2
G
F =mg= 0.500 kg 9.8 m/s = 4.9 N.
(a) The gravitational force is:
(b) The tension in the string at the top of the circle is:
2
2
2
1
4.0 m/s
v
T =m  g = 0.500 kg  9.8 m/s =2.9N
r 1.02 m
Problem 10
con’t
AP Physics C
17
(c)
Similarily
at the bottom of the circle the tension is:
2
r 2 G
mv
F =T  F =
r
2
2
2
2
7.5 m/s
v
T =m g+ = 0.500 kg 9.8 m/s + =32 N
r 1.02 m
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