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Centripetal AccelerationCentripetal Acceleration
12 Examples with full solutions
Example 1Example 1A 1500 kg car is moving on a flat road and negotiates a curve whose radius is 35m. If the coefficient of static friction between the tires and the road is 0.5, determine the maximum speed the car can have in order to successfully make the turn.
35m
Example 1 – Example 1 – Step 1 Step 1 ((Free Body Free Body DiagramDiagram))
GF--------------
NF--------------
fF--------------
+y
+x
This static friction is the only horizontal force
keeping the car moving toward the centre of the arc (else the car
will drive off the road).
ca
Acceleration direction
Example 1 -Example 1 - Step 2 Step 2 ((Sum of Vector Sum of Vector
ComponentsComponents))
GF--------------
NF--------------
fF--------------
ca +y
+x
0
0
0
y
N G
N
N
F
F F
F mg
F mg
--------------
----------------------------
Vertical Components
Horizontal Components
2
2
x c
f c
s N c
s
s
s
s
N
N
F ma
F ma
F ma
vF m
F
mg
R
Rv
mR
m
v g R
----------------------------
---------------------------- We have an acceleration in x-
direction
Static Frictio
nFrom Vertical Component
Example 1 -Example 1 - Step 3 Step 3 ((Insert valuesInsert values))
GF--------------
NF--------------
fF--------------
ca +y
+x
20.50 9.8 35
13.0958
1 360013.0958
1000 1
47
sv g R
mm
s
m
sm km s
s m hkm
h
Example 2Example 2A car is travelling at 25m/s around a level curve of radius 120m. What is the minimum value of the coefficient of static friction between the tires and the road to prevent the car from skidding?
120 m
Example 2 – Example 2 – Step 1 Step 1 ((Free Body Free Body DiagramDiagram))
GF--------------
NF--------------
fF--------------
+y
+x
This static friction is the only horizontal force
keeping the car moving toward the centre of the arc (else the car
will drive off the road).
ca
Acceleration direction
Example 2 -Example 2 - Step 2 Step 2 ((Sum of Vector Sum of Vector
ComponentsComponents))
GF--------------
NF--------------
fF--------------
ca +y
+x
0
0
0
y
N G
N
N
F
F F
F mg
F mg
--------------
----------------------------
Vertical Components
Horizontal Components
2
2
2
2
x c
f c
s N c
s N
sN
F ma
F ma
F ma
vF m
R
m v
R
m v
R
v
mg
R
F
g
----------------------------
---------------------------- We have an acceleration in x-
direction
Static Frictio
n
From Vertical Component
Example 2 -Example 2 - Step 3 Step 3 ((Insert valuesInsert values))
GF--------------
NF--------------
fF--------------
ca +y
+x
2
2
2
2
25
9.8 120
0.53
s
s
v
gR
v
gR
ms
mm
s
We require the minimum value
Example 3Example 3An engineer has design a banked corner with a radius of 200m and an angle of 180. What should the maximum speed be so that any vehicle can manage the corner even if there is no friction?
180
200 m
Example 3 – Example 3 – Step 1 Step 1 ((Free Body Free Body DiagramDiagram))
First the car
Now for gravity
GF--------------
The normal to the road NF
--------------
18
Components of Normal force along axis (we ensured
one axis was along acceleration
direction
sin 18NF
+y
+x
ca
Acceleration
direction
cos 18NF
Notice that we have no static friction force
in this example
(question did not require
one)
Example 3 -Example 3 - Step 2 Step 2 ((Sum of Vector Sum of Vector
ComponentsComponents))
ca +y
+x
0
0
cos 18 0
cos 18
y
Ny G
N
N
F
F F
F mg
mgF
--------------
----------------------------
Vertical Components
Horizontal Components
2
2
cos
sin 18
sin 18
tan 18
tan 18
18
x c
N
Nx c
c
F ma
F ma
ma
vm
R
vg
R
g
R
m
v g
F
----------------------------
----------------------------We have
an acceleration in the
x-direction
From Vertical Component
GF--------------
NF--------------
18
sin 18NF
cos 18NF
Example 3 -Example 3 - Step 3 Step 3 ((Insert valuesInsert values))
ca +y
+x
GF--------------
NF--------------
18
sin 18NF
cos 18NF
2
tan 18
200 9.8 tan 18
25.2357
1 360025.2357
1000 1
91
v Rg
mv m
s
m
sm km s
s m hkm
h
Example 4Example 4
An engineer has design a banked corner with a radius of 230m and the bank must handle speeds of 88 km/h. What bank angle should the engineer design to handle the road if it completely ices up?
230 m
?
Example 4 – Example 4 – Step 1 Step 1 ((Free Body Free Body DiagramDiagram))
First the car
Now for gravity
GF--------------
The normal to the road NF
--------------
Components of Normal force along axis (we ensured
one axis was along acceleration
direction
sinNF
+y
+x
ca
Acceleration
direction
cosNF
Notice that we have no static friction force
in this example
(question did not require
one)
Example 4 -Example 4 - Step 2 Step 2 ((Sum of Vector Sum of Vector
ComponentsComponents))
ca +y
+x
0
0
cos 0
cos
y
Ny G
N
N
F
F F
F mg
mgF
--------------
----------------------------
Vertical Components
Horizontal Components
2
2
21
sin
sc
in
tan
ta
os
n
x c
N
N
x c
cF
m
F ma
F ma
ma
vm
R
vg
R
g
v
Rg
----------------------------
---------------------------- We have an
acceleration in the
x-direction
From Vertical Component
GF--------------
NF--------------
sinNF
cosNF
ca +y
+x
21
2
1
2
tan
1000 188
3600tan
230 9.8
15
v
Rg
km m hh km s
mm
s
GF--------------
NF--------------
sinNF
cosNF
Example 4 -Example 4 - Step 3 Step 3 ((Insert valuesInsert values))
Don’t forget to place in metres per
second
Example 5Example 5
A 2kg ball is rotated in a vertical direction. The ball is attached to a light string of length 3m and the ball is kept moving at a constant speed of 12 m/s. Determine the tension is the string at the highest and lowest points.
Example 5 – Example 5 – Step 1 Step 1 ((Free Body Free Body DiagramDiagram))
Top Bottom
TF--------------
GF--------------
TF--------------
GF--------------
When the ball is at the
top of the curve, the string is “pulling”
down.
When the ball is at the
bottom of the curve,
the string is “pulling” up.
In both cases, gravity is pulling down
ca
ca
Note: any vertical motion problems that do not include a solid
attachment to the centre, do not maintain a constant speed, v and
thus (except at top and bottom) have an acceleration that does not point
toward the centre. (it is better to use energy conservation techniques)
Example 5 -Example 5 - Step 2 Step 2 ((Sum of Vector Sum of Vector
ComponentsComponents))
TF--------------
GF--------------
TF--------------
GF--------------
Top Bottom
2
y c
T G c
T c
T c
c
F ma
F F ma
F mg ma
F mg ma
m a g
vm g
r
----------------------------
------------------------------------------
2
y c
T G c
T c
T c
c
F ma
F F ma
F mg ma
F ma mg
m g a
vm g
r
----------------------------
------------------------------------------
+y
+x
ca
ca
Note: acceleration is down
(-)
Example 5 -Example 5 - Step 3 Step 3 ((Insert valuesInsert values))
TF--------------
GF--------------
TF--------------
GF--------------
+y
+x
ca
ca
Top Bottom
2
2
2
122 9.8
3
76.4
T
vF m g
r
mms
kgm s
N
2
2
2
122 9.8
3
115.6
T
vF m g
r
mm s
kgs m
N
Example 6Example 6A conical pendulum consists of a mass (the pendulum bob) that travels in a circle on the end of a string tracing out a cone. If the mass of the bob is 1.7 kg, and the length of the string is 1.25 m, and the angle the string makes with the vertical is 25o.
Determine:
a) the speed of the bob b) the frequency of the bob
Example 6 – Example 6 – Step 1 Step 1 ((Free Body Free Body DiagramDiagram))
ca
TF--------------
GF--------------
25 cos 25TF
sin 25TF
Let’s decompose our Tension Force
into vertical and horizontal
components
+y
+x
It’s easier to make the x axis positive
to the left
ca
TF--------------
GF--------------
25 cos 25TF
sin 25TF +y
+x
Example 6 -Example 6 - Step 2 Step 2 ((Sum of Vector Sum of Vector
ComponentsComponents))
2
sin 25
sin 25
x
x c
T c
T c
T
F ma
F ma
F ma
vF m
r
----------------------------
0
0
cos 25 0
cos 25
y
Ty G
T
T
F
F F
F mg
mgF
--------------
----------------------------
Horizontal Vertical
Example 6 -Example 6 - Step 3 Step 3 ((Insert values for Insert values for velocityvelocity))
ca
TF--------------
GF--------------
25 cos 25TF
sin 25TF +y
+x
2
2
2
sin 25
sin 25
tan 25
ta
co
n 2
s 2
5
5
T
vmr
vmr
v
F
mg
gr
v gr
29.8 1.25 sin 25 tan 25
1.554
1.55
mv m
s
m
sm
s
The speed of the bob is about 1.55 m/s
Example 6 -Example 6 - Step 3 Step 3 ((Insert values for Insert values for frequencyfrequency))
ca
TF--------------
GF--------------
25 cos 25TF
sin 25TF +y
+x
2 2
2 2
2 2
2
sin 25
sin 25 4
sincos 25
25 4
tan 25 4
tan 25
4
T c
T
ma
m rf
rf
g rf
gf
F
F
mg
r
2
2
2
tan 25
4
9.8 tan 25
4 1.25 sin 25
0.46811
0.468
gf
r
msm
Hz
Hz
The frequency of the bob is about 0.468Hz
Example 7Example 7A swing at an amusement park consists of a vertical central shaft with a number of horizontal arms. Each arm supports a seat suspended from a cable 5.00m long. The upper end of the cable is attached to the arm 3.00 m from the central shaft.
Determine the time for one revolution of the swing if the cable makes an angle of 300 with the vertical
Example 7 – Example 7 – Step 1 Step 1 ((Free Body Free Body DiagramDiagram))
GFC
TF--------------
30
+y
+x
ca
cos 30TF
sin 30TF
30
Example 7 -Example 7 - Step 2 Step 2 ((Sum of Vector Sum of Vector
ComponentsComponents))
GFC
TF--------------
30 cos 30TF
sin 30TF
30
+y
+x
ca
2
2
sin 30
4sin 30
x
x c
T c
T c
T
F ma
F ma
F ma
rF m
T
----------------------------
0
0
cos 30 0
cos 30
y
Ty G
T
T
F
F F
F mg
mgF
--------------
----------------------------
Horizontal Vertical
Example 7 -Example 7 - Step 2 Step 2 ((Sum of Vector Sum of Vector
ComponentsComponents))
GFC
TF--------------
30 cos 30TF
sin 30TF
30
+y
+x
ca
2
2
2
2
2
2
2
4sin 30
4sin 30
4tan 30
cos 3
4
t 0
0
an 3
TF
mg
rmT
rmT
rg
T
rT
g
2
2
4
9.8
3
tan 30
6.1948
6.19
.00 5.00 sin 30T
ms
s
s
m m
The period is 6.19s
3.00 5.00 sin 30.0
m m
Example 8Example 8A toy car with a mass of 1.60 kg moves at a constant speed of 12.0 m/s in a vertical circle inside a metal cylinder that has a radius of 5.00 m. What is the magnitude of the normal force exerted by the walls of the cylinder at A the bottom of the circle and at B the top of the circle
Example 8 – Example 8 – Step 1 Step 1 ((Free Body Free Body DiagramDiagram))
Top Bottom
NF--------------
GF--------------
NF--------------
GF--------------
When the car is at the top of the curve, the
normal force is “pushing”
down.
When the ball is at the
bottom of the curve, the normal
force is “pushing”
up.
In both cases, gravity is pulling down
ca
ca
Note: any vertical motion problems that do not include a solid
attachment to the centre, do not maintain a constant speed, v and
thus (except at top and bottom) have an acceleration that does not point
toward the centre. (it is better to use energy conservation techniques)
Example 8 -Example 8 - Step 2 Step 2 ((Sum of Vector Sum of Vector
ComponentsComponents))
NF--------------
GF--------------
NF--------------
GF--------------
Top Bottom
2
y c
N G c
N c
N c
c
F ma
F F ma
F mg ma
F mg ma
m a g
vm g
r
----------------------------
------------------------------------------
2
y c
N G c
N c
N c
c
F ma
F F ma
F mg ma
F ma mg
m g a
vm g
r
----------------------------
------------------------------------------
+y
+x
ca
ca
Note: acceleration is down
(-)
Example 8 -Example 8 - Step 3 Step 3 ((Insert valuesInsert values))
NF--------------
GF--------------
NF--------------
GF--------------
+y
+x
ca
ca
Top Bottom
2
2
2
12.01.60 9.8
5.00
30.4
N
vF m g
r
mms
kgm s
N
2
2
2
12.01.60 9.8
5.00
61.8
T
vF m g
r
mm s
kgs m
N
Example 9Example 9A 0.20g fly sits 12cm from the centre of a phonograph record revolving at 33.33 rpm.
a) What is the magnitude of the centripetal force on the fly?b) What is the minimum static friction between the fly and the record to prevent the fly from sliding off?
Example 8 – Example 8 – Step 1 Step 1 ((Free Body Free Body DiagramDiagram))
cF--------------
GF--------------
sF--------------
NF--------------
Example 9 -Example 9 - Step 2 Step 2 ((Sum of Vector Sum of Vector
ComponentsComponents))
cF--------------
2
4 2
2 2
4
1 1min2.0 10 4 0.12 33
3 min 6
2. 1
4
0
9 0
s
cF m
F m
revk
a
r
g m
N
f
s
GF--------------
sF--------------
NF--------------
a.
Convert to correct units
Example 9 -Example 9 - Step 2 Step 2 ((Sum of Vector Sum of Vector
ComponentsComponents))
cF--------------
4
4
4
4
2
2
2
2.9 10
2.9 10
2.9 10
2.0 10 9.8
0. 5
4
1
c
s
k
k
F m
F m
mg N
N
mg
Nm
r
s
a
kg
f
GF
--------------
sF--------------
NF--------------
b.
Example 10Example 10A 4.00 kg mass is attached to a vertical rod by the means of two 1.25 m strings which are 2.00 m apart. The mass rotates about the vertical shaft producing a tension of 80.0 N in the top string.
a)What is the tension on the lower string?b)How many revolutions per minute does the system make?
Example 10 – Example 10 – Step 1 Step 1 ((Free Body Free Body DiagramDiagram))
1.00sin
1.2553.1
m
m
1TF--------------
2TF--------------
53.153.1
GF--------------
+y
+x
ca
Example 10 -Example 10 - Step 2 Step 2 ((Sum of Vector Sum of Vector
ComponentsComponents))
1 2
1 2cos 53.1 cos 53.1
x c
T x T x cx
T T c
F ma
F F ma
F F ma
----------------------------
------------------------------------------
1 1
1 2
0
0
sin 53.1 sin 53.1 0
y
T T G
T T
F
F F F
F F mg
--------------
------------------------------------------
Horizontal Vertical
1TF--------------
2TF--------------
53.153.1
GF--------------
+y
+
ca
Example 10 -Example 10 - Step 2 Step 2 ((Sum of Vector Sum of Vector
ComponentsComponents))1TF
--------------
2TF--------------
53.153.1
GF--------------
+y
+
ca
2
2
2
80.0 sin 53.1 sin 53.1 4.00 9.8 0
4.00 9.8 80.0 sin 53.1
sin 53.1
30.98
31
T
T
T
NN F kg
kg
Nkg N
kgF
F N
N
a)
Example 10 -Example 10 - Step 2 Step 2 ((Sum of Vector Sum of Vector
ComponentsComponents))1TF
--------------
2TF--------------
53.153.1
GF--------------
+y
+
ca
b)
1 2
2 21 2
1 222
1 2
2
2
cos 53.1 cos 53.1
cos 53.1 cos 53.1 4
cos 53.1 cos 53.1
4
cos 53.1 cos 53.1
4
80.0 cos 53.1 30.98 c
1.25 cos 53.
os 53.1
4 4.001
T T c
T T
T T
T T
F F ma
F F m rf
F Ff
rm
F Ff
m
N N
r
m kg
0.7498
600.7498
44.99min
45min
rev
srev s
s mrev
rev
Example 11Example 11The moon orbits the Earth in an approximately circular path of radius 3.8 x 108 m. It takes about 27 days to complete one orbit. What is the mass of the Earth as obtained by this data?
Example 11 – Example 11 – Step 1 Step 1 ((Free Body Free Body DiagramDiagram))
GF--------------
Example 11 -Example 11 - Step 2 Step 2 ((Sum of Vector Sum of Vector
ComponentsComponents))
GF--------------
2
2
2
2
E
x cm
a cm
E mm c
E mm c
Ec
c
F m a
F m a
GM mm a
rGM m
m arGM
ar
a rM
G
----------------------------
----------------------------
2
2
2 3
2
32 8
2211
2
2
24
4
4 3.8 10
24 36006.67 10 27
1 1
5.97 10
4 r
GT
r
GT
m
N m h sd
kg d h
k
r
g
The mass of the Earth is about 6.0x1024 kg
Horizontal
Example 12Example 12 (Hard Question) (Hard Question)
An engineer has design a banked corner with a radius of R and an angle of β. What is the equation that determines the velocity of the car given that the coefficient of friction is µ ?
Example 12 – Example 12 – Step 1 Step 1 ((Free Body Free Body DiagramDiagram))
First the car
Now for gravity
GF--------------
The normal to the road NF
--------------
Components of Normal force along axis (we ensured
one axis was along acceleration
direction
sinNF
+y
+x
ca
Acceleration
direction
cosNF
fF--------------
Friction
We have friction going down by
assuming car wants to slide up. This will
provide an equations for the
maximum velocity
0
0
cos sin 0
cos sin 0
N f
N G
N
f
N
y
y
F
F
F F F
F
F uF
mg
mg
--------------
------------------------------------------
Vertical Components
Horizontal Components
sin cos
sin cos
x c
Nx fx c
N
cN
cf
N
F F
F ma
F F ma
m
F
a
maF
----------------------------
------------------------------------------
Example 12 – Example 12 – Step 2 Step 2 ((ComponentsComponents))
GF--------------
NF--------------
sinNF +y
+x
ca
cosNF
fF--------------
From Vertical
cos sinN Nmg
F F
Sub into Horizontal
2
sin cos
cos sinsin cos
N N c
N NN N
F F ma
F
R
FF
v
gF
Example 12 – Example 12 – Step 2 Step 2 ((ComponentsComponents))
GF--------------
NF--------------
sinNF +y
+x
ca
cosNF
fF--------------
From Vertical
cos sinN Nmg
F F
Solve for v
2sin cos
cos sin
sin cos
cos sin
N N
N N
gR F Fv
F F
gRv
π
sin cos
cos sin
gRv
Minimum velocity (slides down)
Flash
