Lecture 14: Circular motion and force

Dynamics
Applying Newton’s Laws
Circular Motion
Lana Sheridan
De Anza College
April 27, 2015
Overview
• Circular motion and force
• Centripetal force
• Examples
• Non-uniform circular motion
Circular Motion - Now with Force
If an object moves in a uniform circle, its velocity must always be
changing. ⇒ It is accelerating.
1
Figures from Serway & Jewett.
Circular Motion - Now with Force
If an object moves in a uniform circle, its velocity must always be
changing. ⇒ It is accelerating.
F = ma ⇒ F 6= 0
1
Figures from Serway & Jewett.
Circular Motion - Now with Force
If an object moves in a uniform circle, its velocity must always be
changing. ⇒ It is accelerating.
F = ma ⇒ F 6= 0
n’s Laws
Any object moving in a circular (or curved) path must be
experiencing a force.
nsion)We call this the centripetal force.
S
SJOHPGDPOTUBOUMFOHUI
wirled in a circle
PSDFBDUJOHPOBQMBOFU
the Sun in a perfectly
apter 13)
FBDUJOHPOBDIBSHFE
1
Figures from Serway & Jewett.
!F
S
ac
r
S
v
Uniform Circular Motion
For an object moving in a uniform circle, a = ac =
This gives the expression for centripetal force:
F = ma
so,
Fc =
mv 2
r
v2
r .
Centripetal Force
Something must provide this force:
6.1 Extending the Particle in
S
A force Fr , directed
toward the center
of the circle, keeps
the puck moving
in its circular path.
S
Fr
r
m
When the
string breaks,
the puck
moves in the
direction
tangent
to the circle.
r
S
Fr
Figure 6.1
It could be tension
An overhead view of a
puck moving in a circular path in a
inhorizontal
a rope.plane.
Figure 6.2
The string holding th
puck in its circular path breaks.
S
ction of the velocity vector will be smaller in a given time
Centripetal Force
red tension in the string is smaller. As a result, the string
r radius.
Something must provide this force:
the Car?
AM
S
fs
a curve as shown
efficient of static
maximum speed
a large circle so
a
It could be friction.
we model the car
on. The car is not
um in the vertical
S
n
Centripetal Force
Consider the example of a string constraining the motion
a puck:the Parti
6.1 of
Extending
S
A force Fr , directed
toward the center
of the circle, keeps
the puck moving
in its circular path.
S
Fr
r
m
When the
string breaks,
the puck
moves in the
direction
tangent
to the circle.
r
S
Fr
Figure 6.1 An overhead view of a
puck moving in a circular path in a
horizontal plane.
Figure 6.2 The string hold
puck in its circular path brea
Centripetal Force
Question. What will the puck do if the string breaks?
(A) Fly radially outward.
(B) Continue along the circle.
(C) Move tangentially to the circle.
Centripetal Force
Question. What will the puck do if the string breaks?
(A) Fly radially outward.
(B) Continue along the circle.
6.1 Extending the Particle in Uniform Circular Motion Model
←
(C) Move tangentially to the circle.
S
A force Fr , directed
toward the center
of the circle, keeps
the puck moving
in its circular path.
S
Fr
r
m
When the
string breaks,
the puck
moves in the
direction
tangent
to the circle.
S
S
Fr
Figure 6.1
An overhead view of a
puck moving in a circular path in a
horizontal plane.
r
v
Figure 6.2
The string holding the
puck in its circular path breaks.
15
Determine (a) the astronaut’s orbital speed and (b) the
of the orbit.
Findingperiod
Centripetal
Force Example
odel tutorial available in
WebAssign 3. In the Bohr model of the hydrogen atom, an electron
moves in a circular path around a proton. The speed
of the electron is approximately 2.20 3 10 6 m/s. Find
utorial available
in Enhanced
(a)
the force acting on the electron as it revolves in a
210
BIO
circular
Page
169,in#orbit
4 of radius 0.529 3 10 m and (b) the
deo solution
available
Q/C
centripetal acceleration
of the electron.
blem
WebAssign
4. A curve in a road forms part of a horizontal circle. As a
car goes around Sit at constant speed 14.0 m/s, the total
horizontal force on the driver has magnitude 130 N.
What is the total horizontal force on the driver if the
speed on the same curve is 18.0 m/s instead?
5. In a cyclotron (one type of particle accelerator), a
deuteron (of mass 2.00 u) reaches a final speed of
10.0% of the speed of light while moving in a circular
path of radius 0.480 m. What magnitude of magnetic
force is required to maintain the deuteron in a circular path?
6. A car initially traveling
y
a
7. A
d
3
w
r
t
8. C
W o
t
m
Finding Centripetal Force Example
Page 169, # 4
Fc =
mv 2
r
Finding Centripetal Force Example
Page 169, # 4
Fc =
m
r
mv 2
r
=
Fc
v2
=
130 N
(14.0m/s)2
= 0.6633 kg/m
Finding Centripetal Force Example
Page 169, # 4
Fc =
m
r
mv 2
r
=
Fc
v2
=
130 N
(14.0m/s)2
= 0.6633 kg/m
Fc =
m 2
v
r
= (0.6633 kg/m)(18.0 m/s)2
= 215 N
angle u. Therefore, if g is smaller, as it is on Mars, the speed v with which the roa
er.
Ferris Wheel Forces
A Ferris wheel is a ride you tend to see at fairs and theme parks.
e Ferris Wheel
AM
Top
S
v
erris wheel as shown
in a vertical circle of
of 3.00 m/s.
S
nbot
S
nt
R
by the seat on the
xpress your answer in
mg.
S
v
Figure 6.6a. Based
S
S
mg
mg
on a Ferris wheel or
Bottom
c
way, you would expect
a
b
path. During
Similarly,
the you
ride the speed, v , is constant.
Figure 6.6 (Example 6.5) (a) A child rides on a Ferris whee
e bottom of the path.
Ferris Wheel Forces
Quick Quiz 6.11 You are riding on a Ferris wheel that is rotating
with constant speed. The car in which you are riding always
maintains its correct upward orientation; it does not invert.
(i) What is the direction of the normal force on you from the seat
when you are at the top of the wheel?
(A) upward
(B) downward
(C) impossible to determine
1
Page 153, Serway & Jewett
Ferris Wheel Forces
Quick Quiz 6.11 You are riding on a Ferris wheel that is rotating
with constant speed. The car in which you are riding always
maintains its correct upward orientation; it does not invert.
(i) What is the direction of the normal force on you from the seat
when you are at the top of the wheel?
(A) upward
←
(B) downward
(C) impossible to determine
1
Page 153, Serway & Jewett
Ferris Wheel Forces
Quick Quiz 6.11 You are riding on a Ferris wheel that is rotating
with constant speed. The car in which you are riding always
maintains its correct upward orientation; it does not invert.
(ii) From the same choices, what is the direction of the net force
on you when you are at the top of the wheel?
(A) upward
(B) downward
(C) impossible to determine
1
Page 153, Serway & Jewett
Ferris Wheel Forces
Quick Quiz 6.11 You are riding on a Ferris wheel that is rotating
with constant speed. The car in which you are riding always
maintains its correct upward orientation; it does not invert.
(ii) From the same choices, what is the direction of the net force
on you when you are at the top of the wheel?
(A) upward
(B) downward
←
(C) impossible to determine
1
Page 153, Serway & Jewett
1 35.0 m 2 1 9.80 m/s 2
dius r banked at a fixed angle u. Therefore, if g is smaller, as it is on Mars, the speed v with which the roadway
s traveled
that the
speed
ly
is also
smaller. v is proportional to the square root of g for a roadway
Finalize Equation (3) shows that the banking angle is independent of the mass of the vehicle negotiating the curve. If a
ore,
g is the
smaller,
it is
speed vacceleration
with which
theTherefore,
roadway
carif
rounds
curve at aas
speed
lesson
thanMars,
13.4 m/s,the
the centripetal
decreases.
the normal force,
which is unchanged, is sufficient to cause two accelerations: the lower centripetal acceleration and an acceleration of the
car down the inclined roadway. Consequently, an additional friction force parallel to the roadway and upward is needed
AM
Top
6.5 to keepRiding
the carthe
fromFerris
slidingWheel
down the bank (to the left in Fig.Sv 6.5). Similarly,
a driver attempting
to negotiate the curve
top
net
at a speed greater than 13.4 m/s has to depend on friction to keep from sliding up the bank (to the right in Fig. 6.5).
mass m rides on a Ferris wheel as shown
S
nbot
H ATchild
I F ? moves
Imagine
this circle
same of
roadway were built on Mars in the future to connect different
colony centers.
S
.6a. W
The
in a that
vertical
ntop
it be speed
traveled
them/s.
same speed?
m atCould
a constant
of at
3.00
Ferris Wheel
Assume the speed, v , is constant.
n
< mg : F
points down
R
el
mine Answer
theAM
forceThe
exerted
by the
seat onTop
theforce on Mars would mean that the car is not pressed as tightly to the roadway. The
reduced
gravitational
S
bottom
of the
ride. Express
your answer
inv component of the normal force toward the center of the circle. This smaller
reduced
normal
force results
in a smaller
e weight
of the child,
component
wouldmg.
not be sufficient to provide the centripetal acceleration associated with the original speed. The censhown
S the speed v.
tripetal acceleration must be reduced, which can be done by reducing
nbot
N
Mathematically, notice that Equation (3) shows that the speed v is proportional
to the square root of g for a roadway
S
rcleofof
S
fixed
radius
r
banked
at
a
fixed
angle
u.
Therefore,
if
g
is
smaller,
as
it is on Mars, the speed vnwith
which the roadway
v
top
ize Look carefully at Figure 6.6a. Based
S
S
mg
mg
can
be
safely
traveled
is
also
smaller.
nces you may have had on a Ferris wheel or
Bottom
c
r small hills on a roadway, you would expect R a
b
he at the top of the path. Similarly, you
hter
Figure
6.6
(Example 6.5) (a) A child rides on a Ferris wheel.
ct to feel
of the path.
net
bot
wer
in heavier6.5at the bottom
(b) The forces
at
the bottom of the path.
AMacting on the child
Top
Example
e bottom
of the path and Riding
the top,the
the Ferris
nor- Wheel
S top of the path.
(c) The forces acting on the child at the
v
avitational forces on the child act in opposite
childsum
of mass
m rides
on a Ferris
S
TheAvector
of these
two forces
gives awheel as shown
nbot
inmagnitude
Figure 6.6a.
The
child
a vertical
circle of path at a constant speed. To yield net force vecnstant
that
keeps
themoves
child in
moving
in a circular
radius
10.0
m
at
a
constant
speed
of
3.00
m/s.
e same magnitude, the normal force at the bottom must be greater than that at the top.
R
continued
(A) Determine the force exerted by the seat on the
S
v
child at the bottom of the ride.
Express your answer in
Based
S
S
terms of the weight of the child, mg.
mg
mg
n
heel or
Bottom
SOLUTION
expect
a
Conceptualize Look carefully at Figure 6.6a. Based
y, you
on experiences you may have had on a Ferris wheel or
Figure 6.6
b
> mg : F
points up
S
ntop
c
S
v
S
Bottom
mg
(Example 6.5) (a) A child rides on a Ferris wheel.
S
mg
A
Turn
1 8.00Banked
m/s 2
2
9.80 m/s 2 1 35.0 m 2
2
5 0.187
Sharp turns in roads are often banked inwards to assist cars in
for the dry road.
making the turn: the centripetal force comes from the normal
force, not friction.
M
n
x
ay in Example 6.3 in such a way
the curve without skidding. In
negotiate the curve even when
ked, which means that the roadhe opening photograph for this
to be 13.4 m/s (30.0 mi/h) and
d the curve be banked?
S
n
u
ny
←
e and Example 6.3 is that the
.5 shows the banked roadway,
o the left of the figure. Notice
articipates in causing the car’s
as a particle in equilibrium in
u
S
Fg
Figure 6.5 (Example 6.4) A car
moves into the page and is round-
1 8.00 m/s 2 2
5 0.187
9.80 m/s2 2 1 35.0 m 2
A Banked Turn
for the dry road.
A turn has a radius r . What should the angle θ be so that a car
traveling at speed v can turn the corner without relying on friction?
M
ay in Example 6.3 in such a way
the curve without skidding. In
negotiate the curve even when
ked, which means that the roadhe opening photograph for this
to be 13.4 m/s (30.0 mi/h) and
d the curve be banked?
e and Example 6.3 is that the
.5 shows the banked roadway,
o the left of the figure. Notice
articipates in causing the car’s
as a particle in equilibrium in
ular motion in the horizontal
nx
S
n
u
u
ny
S
Fg
Figure 6.5 (Example 6.4) A car
moves into the page and is rounding a curve on a road banked at an
angle u to the horizontal. When
1 8.00 m/s 2 2
5 0.187
9.80 m/s2 2 1 35.0 m 2
A Banked Turn
for the dry road.
A turn has a radius r . What should the angle θ be so that a car
traveling at speed v can turn the corner without relying on friction?
M
ay in Example 6.3 in such a way
the curve without skidding. In
negotiate the curve even when
ked, which means that the roadhe opening photograph for this
to be 13.4 m/s (30.0 mi/h) and
d the curve be banked?
e and Example 6.3 is that the
.5 shows the banked roadway,
o the left of the figure. Notice
articipates in causing the car’s
as a particle in equilibrium in
Hint: consider what
ular motion in the horizontal
nx
S
n
u
u
ny
S
Fg
Figure 6.5 (Example 6.4) A car
moves into the page and is round-
the
vector
ing anet
curveforce
on a road
bankedmust
at an be in this case.
angle u to the horizontal. When
A Banked Turn
5 0.187
such a way
idding. In
even when
t the roadph for this
mi/h) and
d?
s that the
roadway,
re. Notice
g the car’s
librium in
horizontal
l acceleran the prewever, the
nx
S
n
u
y -direction (vertical):
ny
Fy ,net = 0
ny − mg
u
S
Fg
Figure 6.5 (Example 6.4) A car
moves into the page and is rounding a curve on a road banked at an
angle u to the horizontal. When
friction is neglected, the force that
causes the centripetal acceleration and keeps the car moving in
its circular path is the horizontal
component of the normal force.
= 0
A Banked Turn
5 0.187
such a way
idding. In
even when
t the roadph for this
mi/h) and
d?
s that the
roadway,
re. Notice
g the car’s
librium in
horizontal
l acceleran the prewever, the
nx
S
n
u
y -direction (vertical):
ny
Fy ,net = 0
ny − mg
= 0
n cos θ = mg
n =
u
S
Fg
Figure 6.5 (Example 6.4) A car
moves into the page and is rounding a curve on a road banked at an
angle u to the horizontal. When
friction is neglected, the force that
causes the centripetal acceleration and keeps the car moving in
its circular path is the horizontal
component of the normal force.
mg
cos θ
A Banked Turn
0.187
ch a way
ding. In
en when
he roadh for this
/h) and
that the
oadway,
. Notice
he car’s
brium in
rizontal
ccelerathe preever, the
x-direction (horizontal):
nx
S
n
u
u
ny
S
Fg
Figure 6.5 (Example 6.4) A car
moves into the page and is rounding a curve on a road banked at an
angle u to the horizontal. When
friction is neglected, the force that
causes the centripetal acceleration and keeps the car moving in
its circular path is the horizontal
component of the normal force.
Fx,net = Fc
mv 2
nx =
r
A Banked Turn
0.187
ch a way
ding. In
en when
he roadh for this
/h) and
that the
oadway,
. Notice
he car’s
brium in
rizontal
ccelerathe preever, the
x-direction (horizontal):
nx
S
n
u
u
ny
S
Fg
Figure 6.5 (Example 6.4) A car
moves into the page and is rounding a curve on a road banked at an
angle u to the horizontal. When
friction is neglected, the force that
causes the centripetal acceleration and keeps the car moving in
its circular path is the horizontal
component of the normal force.
Fx,net = Fc
mv 2
nx =
r
mv 2
n sin θ =
r
mg
sin θ =
cos θ
tan θ =
mv 2
r
v2
⇒ θ = tan−1
rg
v2
rg
Non-Uniform Circular Motion
If the speed is changing the net force will not be pointed into the
center of the circle.
The net force exerted on
the particle is the vector
sum of the radial force
and the tangential force.
S
!F
S
! Fr
S
! Ft
Figure 6.7 When the net f
ticle moving in a circular pa
component o F t , the particle
at points !, ", and #. (b) Suppose the
nstant tangential acceleration as it moves
presenting the force on the bead at points
#
Non-Uniform Circular Motion
Example
Figure 6.8 (Quick
Quiz 6.2) A
bead slides along a curved wire.
Consider a ball of mass m swinging in a vertical circle, radius R.
What is the tension T when the string makes an angle θ with the
AMif the speed at that instant is v ?
on the Ball
vertical
e end of a cord of length
about a fixed point O as
tangential acceleration
d at any instant when the
akes an angle u with the
S
vtop
S
Ttop
S
mg
R
O
the sphere in Figure 6.9
ssociated with Example
ath. Unlike the child in
sphere is not uniform in
ong the path, a tangenm the gravitational force
rticle under a net force and
particle in uniform circuues discussed in this sec-
S
T
mg cos u
u
u
S
Tbot
S
vbot
mg sin u
S
mg
S
mg
Figure 6.9 (Example 6.6) The forces acting on a
sphere of mass m connected to a cord of length R and
Non-Uniform Circular Motion Example
In the radial direction:
Fr ,net =
mv 2
R
Non-Uniform Circular Motion Example
In the radial direction:
Fr ,net =
mv 2
R
T − mg cos θ =
mv 2
R
T
T
mv 2
+ mg cos θ
R
2
v
= mg
+ cos θ
Rg
=
NBHOFUJDGPSDFBDUJOHPOBDIBSHFE
icle moving in a uniform magnetic field (Chapter 29)
One More Example: Conical pendulum
MFDUSJDGPSDFBDUJOHPOBOFMFDUSPOJOPSCJUBSPVOEB
eus in the Bohr model of the hydrogen atom (Chapter 42)
A small ball of mass m is suspended from a string of length L. The
ball revolves with constant speed v in a horizontal circle of radius r
as shown in the figure. Find an expression for v in terms of the
geometry in the figure.
h L. The ball revolves
shown in Figure 6.3.
system is known as a
eometry in Figure 6.3.
L
u
S
T cos u
T
u
r
T sin u
3a and convince yours in a horizontal circle.
rtically. Therefore, we
. It experiences a cenodeled as a particle in
S
S
mg
mg
a
b
Figure 6.3
(Example 6.1) (a) A
conical pendulum. The path of the
1
Serway & Jewett, ball
pageis 152.
a horizontal circle. (b) The
One More Example: Conical pendulum
x-direction:
Fnet,x =
mv 2
= T sin θ
r
(1)
y-direction:
Fnet,y = 0 = T cos θ − mg
T cos θ = mg
(2)
One More Example: Conical pendulum
x-direction:
Fnet,x =
mv 2
= T sin θ
r
(1)
y-direction:
Fnet,y = 0 = T cos θ − mg
T cos θ = mg
Divide (1) by (2):
tan θ =
v2
rg
Replace r = L sin θ and rearrange:
p
v = Lg sin θ tan θ
(2)
Summary
• Uniform circular motion with forces
• Centripetal force
• Banked turns
• Non-uniform circular motion
(Uncollected) Homework Serway & Jewett,
• Ch 6, onward from page 169. Questions: Section Qs 1, 5, 7,
9, 15, 17
`