Download Centripetal Acceleration

Centripetal Acceleration
13 Examples with full solutions
Example 1
A 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 – Step 1 (Free Body Diagram)
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).
FN
Acceleration
direction
Ff
ac
FG
+y
+x
Example 1 - Step 2 (Sum of Vector Components)
FN
+y
ac
Ff
FG
Vertical Components
F
y
0
Horizontal Components
F
F N  FG  0
x
 ma c
F f  ma c
 s FN  mac
FN  mg  0
FN  mg
Static
Friction
 v2 
 s FN  m  
R
F R
v2  s N
m
  mg  R
 s
m
v  s  g  R
We have an acceleration in
x-direction
From Vertical
Component
+x
Example 1 -
Step 3 (Insert values)
FN
+y
ac
Ff
FG
v  s  g  R

 0.50   9.8
m
  35m 
s2 

m
 13.0958
s
m 1km 3600 s
 13.0958 

s 1000m
1h
km
 47
h
+x
Example 2
A 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 – Step 1 (Free Body Diagram)
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).
FN
Acceleration
direction
Ff
ac
FG
+y
+x
Example 2 - Step 2 (Sum of Vector Components)
FN
+y
ac
Ff
FG
Vertical Components
Fy  0
Horizontal Components
F
x
 ma c
F f  ma c
F N  FG  0
 s FN  mac
FN  mg  0
FN  mg
Static
Friction
From Vertical
Component
 v2 
 s FN  m  
R
m  v2 
s 
 
FN  R 
m  v2 

 
mg  R 
v2

gR
We have an acceleration in
x-direction
+x
Example 2 -
Step 3 (Insert values)
FN
+y
ac
Ff
FG
v2
s 
gR
We require the
minimum value
v2
s 
gR
2
 m
 25 
 s

m

9.8
120m 

2 
s 

 0.53
+x
Example 3
An 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 – Step 1 (Free Body Diagram)
The normal
to the road
Acceleration
First the car
direction
FN
18
FN sin 18
ac
Now for
gravity
+y
FG
+x
FN cos 18
Components of Normal
force
along that
axis we
(we
Notice
ensured
haveone
no axis
staticwas
along
acceleration
friction
force in
direction
this
example
(question did not
require one)
Example 3 - Step 2 (Sum of Vector Components)
FN
18
ac
FN cos 18
+y
FN sin 18 
+x
FG
Horizontal Components
Vertical Components
F
y
F
0
FN 
mg
cos 18 
 ma c
F Nx  ma c
F Ny  F G  0
FN cos 18   mg  0
x
FN sin 18   mac
 mg 
 v2 

 sin 18   m  
cos
18

 
R

v2
g tan 18  
R
From Vertical
Component
We have an
acceleration
in the
x-direction
v  Rg tan 18 
Example 3 -
Step 3 (Insert values)
FN
18
ac
FN sin 18 
v
 200m   9.8
m
tan 18 
2 
s 

m
 25.2357
s
m 1km 3600 s
 25.2357 

s 1000m
1h
km
 91
h
+y
+x
FG
v  Rg tan 18 
FN cos 18
Example 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 – Step 1 (Free Body Diagram)
The normal
to the road
Acceleration
First the car
direction
FN

FN sin   
ac
Now for
gravity
+y
FG
+x
FN cos   
Components of Normal
force
along that
axis we
(we
Notice
ensured
haveone
no axis
staticwas
along
acceleration
friction
force in
direction
this
example
(question did not
require one)
Example 4 - Step 2 (Sum of Vector Components)
FN
FN cos   

FN sin   
ac
+y
+x
FG
Vertical Components
F
y
0
F Ny  F G  0
FN cos     mg  0
FN 
mg
cos   
Horizontal Components
F
x
 ma c
F Nx  ma c
FN sin     mac
 mg 
 v2 

 sin     m  
c
os



R


v2
g tan    
R
2
From Vertical
1  v 
  tan 
Component

 Rg 
We have an
acceleration
in the
x-direction
Example 4 -
Step 3 (Insert values)
FN
FN cos   

FN sin   
ac
+y
+x
FG
 v2 
  tan 

Rg


1
  km 1000m
1h 


  88

h
km
3600 s 
1  
 tan

m

230
m
9.8



2 
s



 15
2






Don’t forget to place in
metres per second
Example 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 – Step 1 (Free Body Diagram)
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)
Bottom
Top
FT
ac
FT
FG
ac
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.
FG
In both cases, gravity is
pulling down
Example 5 - Step 2 (Sum of Vector Components)
FT
ac
ac
FT
+y
+x
FG
FG
Top
F
y
Bottom
 ma c
F
F T  F G  ma c
 FT  mg  mac
FT  mg  mac
 m  ac  g 
 v2

 m  g 
 r

y
 ma c
F T  F G  ma c
Note:
acceleration
is down (-)
FT  mg  mac
FT  mac  mg
 m  g  ac 

v2 
 m g  
r 

Example 5 -
Step 3 (Insert values)
FT
ac
ac
FT
FG
+y
+x
FG
Top
 v2

FT  m   g 
 r

  m 2

 12 

m
s
  9.8 
  2kg   
 3m
s2 




 76.4 N
Bottom

v2 
FT  m  g  
r 

2

 m 
12  


m
s 
  2kg   9.8 2  

s
3m 




 115.6 N
Example 6
A 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 – Step 1 (Free Body Diagram)
Let’s decompose
our Tension Force
into vertical and
horizontal
components
FT
FT cos  25
25
ac
FT sin  25
+y
+x
It’s easier to
make the x axis
positive to the
left
FG
Example 6 - Step 2 (Sum of Vector Components)
FT
FT cos  25
25
ac
FT sin  25
+y
+x
FG
Horizontal
Vertical
F
F
x
 ma c
FTx  mac
FT sin  25   mac
v2
FT sin  25   m
r
y
0
F Ty  F G  0
FT cos  25   mg  0
FT 
mg
cos  25 
Example 6 -
Step 3 (Insert values for velocity)
FT
FT cos  25
25
ac
+y
FT sin  25
+x
FG
v2
FT sin  25   m
r
mg
v2
sin  25   m
cos  25 
r
v2
g tan  25  
r
v  gr tan  25 
m

v   9.8 2  1.25m  sin  25   tan  25 
s 

m
 1.554
s
m
 1.55
s
The speed of the bob is about 1.55 m/s
Example 6 -
Step 3 (Insert values for frequency)
FT
FT cos  25
25
ac
+y
FT sin  25
+x
FT sin  25   mac
f 
FT sin  25   m4 2 rf 2
mg
sin  25   4 2 rf 2
cos  25 
g tan  25   4 2 rf 2
f 
g tan  25 
4 2 r
g tan  25 
4 2 r
m

 9.8 2  tan  25 
s 


4 2 1.25m  sin  25  
 0.46811Hz
 0.468Hz
The frequency of the bob is about 0.468Hz
FG
Example 7
A 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 – Step 1 (Free Body Diagram)
FT
FT cos  30 30 30
ac
FT sin  30
+y
+x
FG
Example 7 - Step 2 (Sum of Vector Components)
FT
FT cos  30 30 30
ac
FT sin  30
+y
+x
FG
Horizontal
F
x
 ma c
FTx  mac
FT sin  30   mac
4 2 r
FT sin  30   m 2
T
Vertical
F
y
0
F Ty  F G  0
FT cos  30   mg  0
FT 
mg
cos  30 
Example 7 - Step 2 (Sum of Vector Components)
FT
FT cos  30 30 30
ac
FT sin  30
+y
3.00 m 5.00 m sin  30.0
4 r
FT sin  30   m 2
T
mg
4 2 r
sin  30   m 2
cos  30 
T
+x
2
4 2 r
g tan  30   2
T
T
4 2 r
g tan  30 
FG
T
4 2  3.00m  5.00m sin  30  
m

9.8
tan  30 

2 
s 

 6.1948s
 6.19s
The period is 6.19s
Example 8
A 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 – Step 1 (Free Body Diagram)
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)
Bottom
Top
FN
ac
FN
FG
When the car is
at the top of
the curve, the
normal force is
“pushing”
down.
ac
When the ball
is at the
bottom of the
curve, the
normal force is
“pushing” up.
FG
In both cases, gravity is
pulling down
Example 8 - Step 2 (Sum of Vector Components)
FN
ac
ac
FN
+y
+x
FG
FG
Top
F
y
Bottom
 ma c
F
F N  F G  ma c
 FN  mg   mac
FN  mg  mac
 m  ac  g 
 v2

 m  g 
 r

y
 ma c
F N  F G  ma c
Note:
acceleration
is down (-)
FN  mg  mac
FN  mac  mg
 m  g  ac 

v2 
 m g  
r 

Example 8 -
Step 3 (Insert values)
FN
ac
ac
FN
FG
+y
+x
FG
Top
 v2

FN  m   g 
 r

2


m
 12.0 

m
s
  9.8 
 1.60kg   
 5.00m
s2 




 30.4 N
Bottom

v2 
FT  m  g  
r 

2

m 

12.0  


m
s 
 1.60kg   9.8 2  

s
5.00m 




 61.8 N
Example 9
A 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 – Step 1 (Free Body Diagram)
FN
Fc
Fs
FG
Example 9 - Step 2 (Sum of Vector Components)
FN
Fc
Fs
FG
a.
 F  ma
c
Fs  m  4 2 rf 2 
Convert to
correct units
2


1
rev
1min


4
2
  2.0 10 kg    4   0.12m   33

 

3
min
6
0
s

 

 2.9 104 N
Example 9 - Step 2 (Sum of Vector Components)
FN
Fc
Fs
b.
 F  ma
FG
c
Fs  m  4 2 rf 2 
 s mg  2.9 104 N
2.9 104 N
s 
mg
2.9 104 N

m

4
2.0

10
kg
9.8

  s 2 
 0.15
Example 10
A 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 – Step 1 (Free Body Diagram)
F T1
1.00m
1.25m
  53.1
sin   
ac
53.1
+y
53.1
+x
FT 2
FG
Example 10 - Step 2 (Sum of Vector Components)
F T1
ac
53.1
+y
53.1
+
FT 2
Horizontal
F
x
 ma c
F T 1x  F T 2 x  ma cx
FT 1 cos  53.1   FT 2 cos  53.1   mac
FG
Vertical
F
y
0
F T1  F T1  F G  0
FT 1 sin  53.1   FT 2 sin  53.1   mg  0
Example 10 - Step 2 (Sum of Vector Components)
F T1
ac
53.1
+y
53.1
+
FT 2
a)

80.0 N  sin  53.1   FT 2 sin  53.1    4.00kg   9.8

FG
N
0
kg 

N
  4.00kg   9.8   80.0 N  sin  53.1 
kg 

 FT 2
sin  53.1 
FT 2  30.98 N
 31N
Example 10 - Step 2 (Sum of Vector Components)
F T1
ac
53.1
+y
53.1
+
FT 2
b) FT 1 cos  53.1   FT 2 cos  53.1   mac
FT 1 cos  53.1   FT 2 cos  53.1   m  4 2 rf 2 
f2
f 

F
T1
cos  53.1   FT 2 cos  53.1  
4 2 rm
F
T1
cos  53.1   FT 2 cos  53.1  
4 2 rm
 80.0 N  cos  53.1   30.98 N  cos  53.1  
4 1.25m  cos  53.1    4.00kg 
2
FG
rev
s
rev 60 s
 0.7498

s m
rev
 44.99
min
rev
 45
min
 0.7498
Example 11
The 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 – Step 1 (Free Body Diagram)
FG
Example 11 - Step 2 (Sum of Vector Components)
FG
Horizontal
F
x
 mm a c
F a  mm a c
GM E mm
 mm ac
2
r
GM E mm
 mm ac
r2
GM E
 ac
2
r
ac r 2
ME 
G
4 2 rr 2

GT 2
4 2 r 3

GT 2

4 2  3.8 108 m 
3
2

24h 3600 s 
11 N  m  
6.67

10
27
d




2 
kg
1
d
1h 



 5.97 1024 kg
2
The mass of the Earth is about 6.0x1024 kg
Example 12
Cassiopia takes a ride on child’s Ferris Wheel. This ride has no retaining bar, so that she
only sides on the seat as the ride moves.
a) Determine the Normal Force she would experience from the bottom of the seat
when she is at the lowest point on the ride.
b) Determine the Normal Force she would experience from the bottom of the seat
when she is at the highest point on the ride.
c) Determine the Net Force she would experience from the bottom of the seat when
she is at the mid-point on the ride with her height equal to the axis.
Cassiopia takes a ride on child’s Ferris Wheel. This ride has no retaining bar, so
that she only sides on the seat as the ride moves.
a) Determine the Normal Force she would experience from the bottom of the
seat when she is at the lowest point on the ride.
With this ride, gravity always points down, the normal (seat) force always
points up, and the centripetal acceleration is always toward the centre.
F
c
 mac
 FN  Fg  mac
FN  Fg  mac
 W  mac
Cassiopia takes a ride on child’s Ferris Wheel. This ride has no retaining bar, so
that she only sides on the seat as the ride moves.
Determine the Normal Force she would experience from the bottom of the
seat when she is at the highest point on the ride.
With this ride, gravity always points down, the normal (seat) force always
points up, and the centripetal acceleration is always toward the centre.
F
c
 mac
 FN  Fg  mac
FN  Fg  mac
 W  mac
Example 12
Cassiopia takes a ride on child’s Ferris Wheel. This ride has no retaining bar, so that she
only sides on the seat as the ride moves.
Determine the Net Force (force of seat) she would experience from the bottom of
the seat when she is at the mid-point on the ride with her height equal to the axis.
Fc  mac
FN  mg
F Net  F c  F N
FNet 

 Fc    FN 
2
2
 mac    mg 
2
 mg
  tan 
 mac
1
2

1  g 
  tan  

 ac 
Example 13 (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 13 – Step 1 (Free Body Diagram)
The normal
to the road
Acceleration
First the car
direction
ac
FN

FN sin   
Ff
Friction
+y
FG
+x
FN cos   

Components of Normal
force along axis (we
ensured one axis was
along acceleration
direction
We have friction going
down by assuming car
Now forwants to slide up. This
gravity
will provide an equations
for the maximum
velocity
Example 13 – Step 2 (Components)F
N
FN cos   


FN sin   
Ff
FG
Horizontal Components
Vertical Components
F
y
0
F
x
 ma c
F Ny  F G  F f  0
F Nx  F fx  ma c
FN cos     mg  Ff sin     0
FN sin     Ff cos     mac
FN cos     mg  uFN sin     0
FN sin      FN cos     mac
From Vertical
m
FN cos      FN sin   
g
ac
+y
+
x
Example 13 – Step 2 (Components)F
N
ac
FN cos   


FN sin   
+y
Ff
+
x
FG
From Vertical
m
FN cos      FN sin   
g
Solve for v
v 
2
v
Sub into Horizontal
FN sin      FN cos     mac
 FN cos      FN sin     v 2
FN sin      FN cos     

g

R
gR  FN sin      FN cos    
Minimum velocity (slides down)
gR  sin      cos    
v
FN cos      FN sin   
cos      sin   
gR  sin      cos    
cos      sin   
π
Flash