Circular motion
1. In the following examples name the force that is providing the centripetal force and draw it
on the diagram.
(a) A runner running round a circular track.
(b) A car on a rollercoaster.
2. A 2kg mass travels in a circle of radius 50cm. If the time for one revolution is 2s calculate:
(a) The angular velocity of the mass
(b) The centripetal acceleration of the mass
(c) The centripetal force of the mass
3. A ball rolls around the inside of a vertical cylinder as shown. Indentify the force that stops it
from falling down.
Formulae
ω=2π/T
F=mv2/r=mω2r
© Chris Hamper, InThinking
www.physics-inthinking.co.uk
1
1
Gravitational field
1. State Newton’s universal law of gravity.
2. Two masses are positioned as shown in the diagram.
Calculate the Force on the red one.
3. Define gravitational field strength.
4. Given that the mass of the moon is about 1/80 of the earth and its radius is ¼ estimate the
acceleration due to gravity on the surface of the moon.
Formulae
F=GMm/r2
G=6.7x10-11 m3kg-1s-2
© Chris Hamper, InThinking
www.physics-inthinking.co.uk
1
2
E N D - O F -TO P I C Q U E S T I O N S
Solutions for Topic 6 – Circular motion and gravitation
ν
1.
α
P
2. friction between the tyres and the road.
mv2 .
3. F = _
r
4. a) (i) The drops are increasingly far apart and so the speed is increasing.
(ii) 5.6 s is 6 time intervals so the distance travelled is 14.4 cm on the scale, or 57.6 m on
the ground.
b) (i) centripetal force; acting towards the centre of the circle
(ii) Passengers are in a rotating frame of reference. Seen from above the passengers would
move in a straight line from Newton’s First Law of motion but friction acts at seat to provide
centripetal force to centre of circle. Passenger interprets the reaction to this force
as being flung outwards.
534 = 0.297 m s–1.
5. a) Circumference is 2π × 85 = 534 m. So linear speed is _
30 × 60
v2
_
m
2
v so fractional change = _
r
v2 _
0.592
_
–4
b) (i) change in weight = m _
r
mg = gr = 9.8 × 170 = 2.1 × 10
(ii) a smaller apparent weight as in passenger frame of reference there is an apparent additional
upward force
c) Capsule turns 2π rad in 30 minutes, so 3.5 mrad s–1.
2π
6. Angular speed of Earth = __
= 7.3 × 10–5 rad s–1
24 × 60 × 60
Linear speed v = ω r cos θ where θ is the latitude.
a) 14’ of arc = 4.1 mrad, linear speed = 7.3 × 10–5 × 6.4 × 106 × 1 = 470 m s–1
b) 46° gives 320 m s–1
c) At the geographical south pole the linear speed is zero.
7. a) Maximum friction force = 6500 × 9.8 × 0.7 = 44.6 kN ___________
v2 = _
6500 v2. So v = __
44600 × 150 = 32 m s–1
Centripetal force required = m _
max
r
150
6500
2
_______
__
v
_
–1
b) m r < mg; v > √ rg . Maximum speed = √75 × 9.8 = 27 m s
√
8. The component of the normal reaction force acting horizontally contributes to the centripetal force
so the faster the cyclist is travelling, the greater the component required and this is achieved by
moving up the slope to a point where the slope angle is greater.
9. Gravitational force on planet provides the centripetal force on the planet (Keplar’s third law)
mS mp
so mp ω2R = G _
R2
m
re-arranging ω2 = G _3S
R
2π
ω = 2πf = _
T
2 3
4π
R
and T2 = _
Gms
3
© Oxford University Press 2014: this may be reproduced for class use solely for the purchaser’s institute
1
E N D - O F -TO P I C Q U E S T I O N S
10. a) Speed is a scalar but velocity is a vector. The direction of the velocity is constantly changing, so
the vector velocity is changing too. Acceleration occurs when velocity changes and so there is
acceleration in this case.
b) Work done = distance travelled × force in direction of distance travelled. The force
(acceleration) acting and the distance travelled are at 90° to each other so in this case no work
is done. OR
Work is done when kinetic energy or potential energy change. The speed is constant so kinetic
energy is constant. Distance from Earth is constant so gravitational potential energy does not
change. So no work is done.
Gm
11. g = _
r2
GmE
GmM
=
2
rE
rM2
Where rE and rM are the distances from the centres of Earth and Moon respectively to the point
where the field strengths are equal (known as the Lagrangian point).
rE2
mE _
6 × 1024 = 82
So r 2 = _
=
mM
7.3 × 1022
M
rE
__
√
rM = 82 = 9.06
9
Therefore the point is th of the way from the Earth to the Moon (3.42 × 108 m).
10
G mM
12. a) (i) force on 1 kg of water = _
= 3.4 × 10–5 N due to Moon
r2
G mS
(ii) force on 1 kg of water = _
= 6.0 × 10–3 N due to Sun
r2
b) When the Moon is overhead there is a gravitational force of attraction (a tide) on objects. So
fluids are able to respond to this by an increase in the water level (tides are also observed in
the rocks). There are two tides because there is a corresponding bulge in the surface on the
opposite side of the Earth.
GM = __
6.7 × 10–11 × 6 × 1024 = 7.3 N kg–1
13. a) g = – _
r2
(7.4 × 106)2
2π =
b) The satellite orbits in 24 hours; orbital time = 86400 s. Angular speed = _
86400
7.3 × 10–5 rad s–1
–11
24
3
GME m ____
mv2 = _
c) _
= 6.7 × 10 × 6 × 10 7 2× 1.8 × 10 = 560 N
2
r
r
(3.6 × 10 )
4
© Oxford University Press 2014: this may be reproduced for class use solely for the purchaser’s institute
2
7.
When the child is on a level surface, the normal force between his chest and the sled
is equal to the child’s weight, and thus he has no vertical acceleration. When he
goes over the hill, the normal force on him will be reduced. Since the child is
moving on a5:curved
path, there
must be Gravitation
a net centripetal force towards the center of
CHAPTER
Circular
Motion;
the path, and so the normal force does not completely support the weight. Write
Newton’s 2nd law for the radial direction, with inward as positive.
Answers toF Questions
mg FN m v 2 r
FN mg m v 2 r
R
FN
mg
Topic 6.1a Circular Motion Problems
We see
that the
normal
force is reduced
fromismg
by thetocentripetal
force. force, and so the water
The
problem
with
the statement
is that there
nothing
cause an outward
removed from the clothes is not thrown outward. Rather, the spinning drum pushes INWARD on the
8. clothes
When aand
bicycle
rider
leans
inward,
down
thecan’t
ground
at an
water.
But
where
there the
are bike
holestire
in pushes
the drum,
the on
drum
push
on the water, and so
angle.
The
road
surfacein.
then
pushesthe
back
on the
tire tangentially
both vertically
(to rotation,
provide the
the
water
is
not
pushed
Instead,
water
moves
to
the
out the holes, in a
Conceptual
Questions
normal force
which
counteracts
gravity)
and
horizontally
toward the center of the
straight
line,
and
so
the
water
is
separated
from
the
clothes.
(These questions
are
not in an
style but instead
designed
to enabling
check yourthem
understanding
should
N
curve (to
provide
theIBcentripetal
frictional
force,
to turn). of the concept of this topic.FYou
mg
try
your
best
to
appropriately
communicate
your
answer
using
prose)
CHAPTER
5: Circular Motion; Gravitation
2. The centripetal acceleration for an object moving in circular motion is inversely proportional
Ffr to the
2
a v from
r . clothes
radius of thepeople
curve, given
a constant
So for ain
gentle
curve (which
means a large
1. Sometimes
say that
water speed
is removed
a spin-dryer
by centrifugal
force
Answers
to
Questions
9. throwing
Airplanes
bank
whenoutward.
they
because
order
to
turn,
there
mustmeans
be a a small radius), the
the
water
What
isinwrong
with
this
statement?
radius),
the
acceleration
is turn
smaller,
while
for
a sharp
curve
(which
force that will
be
exerted towards the center of a circle. By tilting the
acceleration
is
larger.
1. The problem with the statement is that there is nothing to cause an outward force, and so the
Flift water
wings, the lift force on the wings has a non-vertical component which points
removed
from
the clothes
is notproviding
thrown outward.
Rather,force.
the spinning
drum pushes INWARD on Rthe
toward
the
center
the
curve,
The
banking
3. clothes
The
force
that
the of
car
exerts
onthere
the road
isthe
thecentripetal
Newton’s
3rdthelaw
reaction
to the on the water, and so
and
water.
But
where
are
holes
in
the
drum,
drum
can’t
angle can
be computed
from
thecar,
free-body
diagram.
The sum
of
verticalinpush
normal
force
of
the
road
on
the
and
so
we
can
answer
this
question
terms out the holes, in a
the
water
is not
pushed
in.
Instead,
watera moves
tangentially
the rotation,
mg
forces
must
beforce.
zero
for
thecar
plane
to the
execute
level
turn,
and
so to
FN
of
the
normal
The
exerts
the
greatest
force
on
the
road
at
the
dip
mg
straight line, and so the water is separated from the clothes.
Flift cos twomg
. The
horizontal
component
of the
force
between
hills.
There
the normal
force from
thelifting
road has
tomust
both provide
support the
weight
AND
provide
a
centripetal
upward
force
to
make
the
car
move
in an
the centripetal
to movefor
theanairplane
in a circle.
2. The
centripetalforce
acceleration
object moving
in circular motion is inversely
proportional to the
upward
curved
path.
The
car
exerts
the
least
force
on
the
road
at
the
top
of a She wants to let go at
2
2. A girl is whirling a ball
on
a
string
around
her
head
in
a
horizontal
plane.
2
v
mg
a2 rv r as
radius
the
a constant
speedm vsensation
.tan
So
a gentle
hill. We
havecurve,
allmfelt
the
upward”
wefor
have
drivencurve
over (which means a large should
Fof
sin
v 2given
r “floating
sin
lift the right time so that the ball will hit a target on the other side of the yard. When
precisely
mg
F
cos
Rg
the crestthe
of acceleration
a hill. In thatiscase,
therewhile
mustfor
be aa sharp
net downward
centripetal
radius),
smaller,
curve (which
means force
a small radius),N the
she
let
go
of
the
string?
to cause the circular
acceleration
is larger.motion, and so the normal force from the road does not
support
10. completely
She should let
go ofthe
theweight.
string when the ball is at a position where
the
tangent
line
to
the
circle
at the
when extended,
3. The force that the car exerts on
theball’s
road islocation,
the Newton’s
3rd law reaction to the
4. normal
There
are
at
least
three
distinct
major
forces
on
the
child.
The
force of target’s
gravity
passes through
That
indicates
force ofthe
thetarget’s
road onposition.
the car, and
sotangent
we can line
answer
this question
in termsis acting downward
location
mg
on
the
child. of
There
isThe
a normal
force
the seat
thethe
horse
on the child.
thethe
direction
the velocity
that
instant,
and
ifforce
theofcentripetal
FN There
of
normal
force.
caratexerts
thefrom
greatest
on
roadacting
at theupward
dip
must
friction
between
seat
offollow
the
horse
and
child
the child
force be
is removed,
then
thethe
ball
will
that
line
horizontally.
between
two hills.
There
the
normal
force
from
thethe
road
hasastowell,
bothorsupport
thecould not be
accelerated
byprovide
thediagram.
horse.
It is that upward
friction force
that provides
There may be
See theAND
top-view
weight
a centripetal
to makethe
thecentripetal
car move inacceleration.
an
smaller
forces
as
well,
such
as
a
reaction
force
on
the
child’s
hands
if
the
child
is
holding
on to part
upward curved path. The car exerts the least force on the road at the top of a
rd
of
horse.
Any
force
that
has
a
radially
inward
component
will
contribute
to
the
centripetal
11. hill.
Thetheapple
does
exert
a
gravitational
force
on
the
Earth.
By
Newton’s
3
law,
the
force
on
the Earth
We have all felt the “floating upward” sensation as we have driven over
3. Aacceleration.
bucket
of
water
can
be
whirled
in
a
vertical
circle
without
the
water
spilling
out,
even
at the top
mg
F
N
duecrest
to theofapple
thethat
same
magnitude
as the
on the apple
due to the
Earth – the weight
of the
the
a hill.is In
case,
there must
be aforce
net downward
centripetal
force
of
the
circle
when
bucket
apple.
The
is the
also
independent
of normal
the stateforce
ofExplain.
motion
of the
So for both a hanging apple
to
cause
the force
circular
motion,
andissoupside-down.
the
from the
roadapple.
does not
5. completely
For
water
to remain
in theon
bucket,
theredue
must
be aapple
centripetal
forcing
to move in a
and the
a falling
apple,
force
the Earth
to the
is equalforce
to the
weightthe
ofwater
the apple.
support
the weight.
circle along with the bucket. That centripetal force gets larger with the tangential velocity of the
water, are
since
centripetal
at the
ofMoon
the
motion
comesisfrom
a combination
FR three
m v 2 distinct
r . The major
M topMThe
4. There
at least
forcesforce
on the
child.
force
of gravity
acting
downward
12. on
Thethe
gravitational
force
on
the Moon
isfrom
given
by
, where R is the radius of the Moon’s
G of Earth
2horse acting upward on the child. There
child.
There
is
a
normal
force
the
seat
the
of the downward force of gravity and the downward normal
R force of the bucket on the water. If the
must
beisfriction
between
the seat
ofminimum
the horsespeed,
and thethe
child
as well,
or the
child
could not
bebucket is
2
bucket
moving
faster
than
some
stay
the
bucket.
If the
orbit. This is a radial force, and so can be expressed as water
can
be changed
using
M Moon will
vMoon
R . inThis
accelerated
by
the
horse.
It
is
that
friction
that
provides
the
centripetal
acceleration.
There
may
be
moving too slow, there is insufficient force to keep the water moving in the circular
path, and
it spills
2
2
smaller
forces
as
well,
such
as
a
reaction
force
on
the
child’s
hands
if
the
child
is
holding
on
to
part
the
relationship
v
2
R
T
,
where
T
is
the
orbital
period
of
the
Moon,
to
.
If
we
4
M
R
T
out.
Moon
Moon
of
the horse.
Anyexpressions
force that has
radially
will contribute to the centripetal
equate
these two
forathe
force,inward
we get component
the following:
acceleration.
6. The three major “accelerators” are the accelerator pedal, the brake pedal, and the steering wheel.
accelerator
pedal
(orSaddle
gas River,
pedal)
used
to increase
speed
(by depressing
the pedal)
or to
© 2005The
Pearson
Education, Inc.,
Upper
NJ.can
All be
rights
reserved.
This material
is protected
under all copyright
laws as they
5.
For
theNo
water
toofinremain
in the
there
must
beor abycentripetal
forcepermission
forcing
water
tothebe
move
currently
exist.
portion
this
material
may bucket,
bewith
reproduced,
in any
any means,
without
inthe
writing
from
decrease
speed
combination
friction
(byform
releasing
the pedal).
The
brake
pedal
can
usedintoa
publisher.
circle
along
with
the
bucket.
That
centripetal
force
gets
larger
with
the
tangential
velocity
of
the
decrease speed by depressing it. The steering wheel
is used to change direction, which also is an
101
2
water,
since FThere
.
The
centripetal
force
at
the
topalso
of the
comes
from a combination
m
v
r
acceleration.
are
some
other
controls
which
could
be motion
considered
accelerators.
The
R
parking
brake
can
be
used
to
decrease
speed
by
depressing
it.
The
gear
shift
lever
can
be
used
of the downward force of gravity and the downward normal force of the bucket on the water. Iftothe
decrease
byfaster
downshifting.
the car has
a manual
transmission,
clutch Ifcan
used to
bucket
is speed
moving
than someIfminimum
speed,
the water
will stay then
in thethe
bucket.
thebebucket
is
© 2005moving
Pearson Education,
Inc.,there
Upperis
Saddle
River, NJ. force
All rights
Thiswater
materialmoving
is protected
all copyright
laws as
theyit spills
too slow,
insufficient
toreserved.
keep the
inunder
the circular
path,
and
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
out.
publisher.
1.
6.
100
The three major “accelerators” are the accelerator pedal, the brake pedal, and the steering wheel.
The accelerator pedal (or gas pedal) can be used to increase speed (by depressing the pedal) or to
decrease speed in combination with friction (by releasing the pedal). The brake pedal can be used to
decrease speed by depressing it. The steering wheel is used to change direction, which also is an
acceleration. There are some other controls which could also be considered accelerators. The
parking brake can be used to decrease speed by depressing it. The gear shift lever can be used to
decrease speed by downshifting. If the car has a manual transmission, then the clutch can be used to
5
© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
F m vR R . Equate
these
R
T two expressions
GTfor the force on the moon, and substitute the relationship
Thus
a
value
for
the
mass
of
Pluto
can
be
calculated
knowing the period and radius of the moon’s
for a circular orbit that v 2 R T .
orbit.
GmM mv 2 4 2 mR
4 2 R3
M
.
R2
R
T2
GT 2
Thus a value
for the mass of Pluto can be calculated knowing the period and radius of the moon’s
Solutions
to Problems
orbit.
1. (a) Find the centripetal acceleration from Eq. 5-1.
Topic 6.1a Circular Motion Problems
v2 r
aR
1.25 m s
2
Solutions to Problems
1.42 m s 2
1.10 m
(b) The net horizontal force is causing the centripetal motion, and so will be the centripetal force.
1. (a) Find the
acceleration
FR centripetal
maQuestions
25.0
kg 1.42from
m s 2Eq. 5-1.
35.5 N
Calculation-based
R
2
aR v 2 r 1.25 m s 1.10 m 1.42 m s 2
2. Calculate
Find The
the centripetal
acceleration
from Eq.
1.
thehorizontal
centripetal
acceleration
of5-1.
the
Earth in
its orbit
thethe
Sun,
and theforce.
net force
(b)
net
force
is causing
the
centripetal
motion,
andaround
so will be
centripetal
2.
2
exerted onFthe
Earth. What
exerts
this sforce
on1the
2
s kg
gN Earth? Assume that the Earth’s orbit is a circle
2
2
ma 525 m25.0
1.42 m
35.5
R r 11 R
a
v
45.94
m
s
4.69
R
3
of radius 1.5x10 6.00
m. You
upg'son the Internet or data booklet.
10 may
m need to look other
9.80 mconstants
s2
[3 marks]
Find the centripetal acceleration from Eq. 5-1.
525 m s
2
3.
2
2
2
aR v r
45.94 m s
6.00 10 3 m
2
The centripetal acceleration is aR v REarth
orbit
1g
2 REarth2 T 4.694g's2 REarth
9.80 morbits
orbit
. The force (from
2
REarth
T
orbit
2
2
nd
3.
REarth
T converted
4 REarth
Newton’s 2 law) is FR mEarth aR . The period is2 one
year,
into seconds.
orbit
orbit
2
The centripetal
is aR 11v REarth
. The force (from
4 2 Racceleration
2
2
Earth
R
T
orbit
4
1.50
10
m
Earth
orbit
aR
5.97 10 3orbit
m s2
2
2
7
T
Newton’s 2nd law) is FR 3.15
mEarth10aR sec
. The period is one year, converted into seconds.
4 2 REarth
FR ma
5.97 102 24 kg 5.9711 10 3 m s 2
3.56 10 22 N
4
1.50
10
m
orbit
aR exerts 2this force on the Earth.2 It is 5.97
10 3 m s 2force.
The Sun
a gravitational
7
T
3.15 10 sec
4.
The speed can be found from
the centripetal
and centripetal
acceleration.
24
3 force
2
22
4.
The speed can be found from the centripetal force and centripetal acceleration.
FR maforce5.97
10 kg
5.97 10
s
3.56 10
2. A horizontal
of 210N
is exerted
on am2.0kg
discus
as itNrotates uniformly in a horizontal circle
210
N
0.90
m
F
r
2
R
(at
arm’s
ofvforce
radius
Calculate
the speed of
the discus.
FR length)
maR this
m
r on0.90m.
9.7 m s
The
Sun
exerts
thev Earth.
It is a gravitational
force.
m
2.0 kg
[2 marks]
210 N 0.90 m
FR r
2
FREducation,
maR Inc.,mUpper
v r Saddle River,
v NJ. All rights reserved. This material is protected
9.7 munder
s all copyright laws as they
© 2005 Pearson
m in any form or2.0
kg means, without permission in writing from the
currently exist. No portion of this material may be reproduced,
by any
publisher.
104
3. Suppose the space shuttle in orbit 400km from the Earth’s surface, and circles the Earth about once
© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
every
the
centripetal
acceleration
ofmeans,
the space
shuttleininwriting
its orbit.
currently
exist.90
Nominutes.
portion of thisFind
material
may
be reproduced, in
any form or by any
without permission
from theExpress your
publisher.
answer in terms of g, the gravitational acceleration at the Earth’s surface. You may need
to look up
104
Giancoli
Physics: Principles with Applications, 6th Edition
some other constants on the Internet.
[3 marks]
5.
The orbit radius will be the sum of the Earth’s radius plus the 400 km orbit height. The orbital
period is about 90 minutes. Find the centripetal acceleration from these data.
60 sec
r 6380 km 400 km 6780 km 6.78 106 m
T 90 min
5400 sec
1 min
aR
4
2
r
4
2
6.78 106 m
9.18 m s 2
1g
0.937 0.9 g's
9.80 m s 2
5400 sec
Notice how close this is to g, because the shuttle is not very far above the surface of the Earth,
relative to the radius of the Earth.
6.
T
2
2
To find the period, the rotational speed (in rev/min) is reciprocated to have min/rev, and then
converted to sec/rev. Use the period to find the speed, and then the centripetal acceleration.
1 min
60 sec
sec
2 r 2 0.16 m
T
1.333
r 0.16 m
v
0.754 m s
45 rev 1 min
rev
T
1.333 sec
aR
2
v r
0.754 m s
0.16 m
2
3.6 m s 2
6
period is about
90 minutes.
Find of
thethe
centripetal
fromNewton’s
these data.
frictional
force between
the back
rider andacceleration
the wall. Write
2nd
law for the vertical forces, noting that there is no vertical
acceleration. 60 sec
r 6380 km 400 km 6780 km 6.78 106 m
T 90 min
Fy Ffr mg 0
Ffr mg
1 min
Ffr
5400 sec
mg
FN
2
If we assume that
the static
then
6.78friction
106 mforce is a maximum,
4 2r 4
1g
2
afrR
9.18
m
s
0.937 0.9 g's
F
F
mg
F
m
g
.
s 2N
N
s
2
T
9.80 m s 2
5400 sec
But the normal force must be the force causing the centripetal motion – it is the
Notice
howpointing
close this
is tocenter
g, because
the shuttle
very
above
surface
to is
the
of rotation.
Thusis Fnot
FN far
mair-hockey
v 2 r . the
Using
v 2of rthe
T Earth,
, we
havein its orbit
R
4. Aonly
flat force
puck
(mass
M)
rotated
tabletop,
and
is held
relative
to the
radius
of the
Earth.in a circle on a frictionless
by a light
4 2cord
mr connected to a dangling block (mass m) through a central hole as shown below.
FN
. Equate the two expressions for the normal force and solve for the coefficient of
2 the
that
speed
the puck
is given
by
6. Show
To find
the
theof
rotational
speed
(in rev/min)
is reciprocated to have min/rev, and then
T period,
converted
to
sec/rev.
Use
the
period
to
find
the
speed,
and then the centripetal acceleration.
friction. Note that since there are 0.5 rev per sec, the period
𝑚𝑔𝑅2 is 2.0 sec. 2 0.16 m
1 min
60 sec
sec2
=s 2#m2 s v 2 r
9.8
4 2 mr mg
T
1.333gT
r 𝑣m0.16
0.754 m s
𝑀
FN 45 rev
0.22
1 min
rev
T . 1.333 sec
s
2
2
2
T
4 r
4
4.6 m
s
[2 marks]
2
0.754
m
s
2
Any larger
of m
friction
aR vvalue
r of the coefficient3.6
s 2 would mean that the normal force could be smaller to
achieve the same frictional
0.16 mforce, and so the period could be longer or the cylinder smaller.
There is no force pushing outward on the riders. Rather, the wall pushes against the riders, so by
See the free-body
diagram
the against
textbook.
moving
in a of
circle
with
a constant
Newton’s
3rd law the
ridersin
push
the Since
wall. the
Thisobject
givesisthe
sensation
being
pressed
into the
speed,
the
net
force
on
the
object
at
any
point
must
point
to
the
center
of
the
circle.
wall.
(a) Take positive to be downward. Write Newton’s 2nd law in the downward direction.
2
mg FT1
maR m
19. Since mass mFisR dangling,
the tension
invthercord must be equal to the weight of mass m, and so
FT mg . That same tension is in the other end of the cord,
maintaining the circular motion of mass
2
4.00
m
s
2
2
m
r g M v0.300
kg
9.80 m for
s 2 the 3.73
N and solve for the
M, and so FF
the two expressions
tension
FR v Ma
r . Equate
T1T
R
0.720 m
velocity.
This2 is a downward force, as expected.
M v positive
r mg to be vupward.
mgRWrite
M .Newton’s 2nd law in the upward direction.
(b) Take
FR FT2 mg ma m v 2 r
20. A free-body diagram for the ball is shown. The tension2 in the
5.
A 0.45kg ball, attached
to the end of a horizontal
is rotated
in a circle of radius 1.3m on a
4.00
scord,
2 not only hold the ball up,
2
suspendingFcordmmust
butmalso
provide
the
FT
v
r
g
0.300
kg
9.80
m
s
9.61
T1
frictionless
horizontal
cord
will
the tension
inNit exceeds
75N, what is the
centripetal force
neededsurface.
to make If
thethe
ball
move
inbreak
a circle.
Write
0.720
m when
nd
mg
maximum
the
ballvertical
can have?
Newton’s
2speed
for the
direction, noting that the ball is not
This is
anlaw
upward
force, as expected.
[2 marks]
accelerating vertically.
7.
© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected
under all copyright laws as they
8. The
centripetal
force
that the
provides
is given
bymeans,
. Solveinthat
forfrom
thethespeed.
FR without
mv 2 rpermission
currently
exist.
No portion of
this material
maytension
be reproduced,
in any form
or by any
writing
publisher.
v
9.
FR r
75 N 1.3 m
m
0.45 kg
15 m s
108
A free-body diagram for the car at one instant of time is shown. In the diagram, the
car is coming out of the paper at the reader, and the center of the circular path is to
the right of the car, in the plane of the paper. If the car has its maximum speed, it
would be on the verge of slipping, and the force of static friction would be at its
maximum value. The vertical forces (gravity and normal force) are of the same
magnitude, because the car is not accelerating vertically. We assume that the force
of friction is the force causing the circular motion.
FR
Ffr
m v2 r
s
FN
s
mg
v
s
rg
0.80 77 m 9.8 m s 2
FN
mg
25 m s
© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.
105
7
Ffr
radius
might
stay
the same,
andbecause
the period
decrease
by athere
factormust
of be
2 ,awhich means the speed
Airplanes
bank
when
they turn
in order
to turn,
increased
factor
of towards
2 . Or the
stay the
force that by
willa be
exerted
theperiod
centermight
of a circle.
Bysame,
tiltingand
thethe radius increase by a factor
1/ 3
1/ 3
Flift
wings,
the
lift
force
on
the
wings
has
a
non-vertical
component
which
points
of 2 , which means the speed increased by the same factor of 2 . Or
if both R and T were
to
R
toward the
centerthe
of speed
the curve,
providing
force. The
banking
R 3 centripetal
T 2 would double.
double,
keeping
constant,
then the
There
are an infinite number of
angle can be computed from the free-body diagram. The sum3 of 2vertical
other
would
also
doubling
forcescombinations
must be zero that
for the
plane
to satisfy
executethe
a level
turn,of
andR soT .
mg
Flift cos
mg . The horizontal component of the lifting force must provide
13. The
gravitational
pulltoismove
the same
in each in
case,
by Newton’s 3rd law. The magnitude
of that Gravitation
pull is
Chapter
Circular Motion;
the5centripetal
force
the airplane
a circle.
M Earth M Moon
2
v
given by F G
. mg
To find the acceleration
of each body,
the gravitational pulling force is
22 r
Flift sin
m rvEarth-Moon
sin
m v2 r
tan
The
total
acceleration
is
given
by
the
Pythagorean
combination
of
the
tangential
and
centripetal
cos
Rg
Chapter
5
Circular
Motion;
Gravitation
divided
by the mass of the2 body.
Since
the
Moon
has
the
smaller
mass,
it
will
have
the
larger
2
accelerations.
a
a
a
.
If
static
friction
is
to
provide
the
total
acceleration,
then
acceleration.
Conceptual
R
tan
10. She shouldQuestions
let gototalof the string
when the ball is at a position where
2 IB style
2
(These questions
are
not
in
an
but
instead
designed
to is
check
your
ofand
thecentripetal
of this topic. You should
the
tangent
line
to
the
circle
at
theassume
ball’s
location,
when
extended,
F
ma
m
a
a
.
We
car
on
the
verge
slipping,
isconcept
on
level
The
total
acceleration
is
given
by
the
Pythagorean
combination
ofunderstanding
theoftangential
fr difference
total
R
tan the two
14. The
in force
on
sides
ofthat
thethe
Earth
from
the
gravitational
pull
ofand
either
thea Sun
or
try
your
best
to
appropriately
communicate
your
answer
using
prose)
target’s
passes through the target’s
position.
That tangent line indicates
2
2
the
Moon
is
the
primary
cause
of
the
tides.
That
difference
in
force
comes
about
from
the
fact
that
accelerations.
atanthat
. Ifinstant,
statichas
friction
is to
provide
acceleration,
then
surface,
and so
staticaRfrictional
force
itsifmaximum
valuethe
oftotal
Ffrlocation
F
mg
. If we equate
total velocity
s N
s
the
direction
of athe
the
at
and
the
centripetal
the two sides of the Earth are a different distance away from the pulling body. Relative to the Sun,
these
two
expressions
for
the
frictional
force,
we
can
solve
for
the
coefficient
of
static
friction.
force
is
removed,
then
the
ball
will
follow
that
line
horizontally.
2 gravitational force on the Earth? If so, how large a force? Consider an apple
1. Does
antotal
apple
exert
Ffr difference
ma
m distance
aR2 aatan
. We diameter)
assume that
on the
verge
slipping,
is onaverage
a level
the
in
(Earth
of the car
twoissides
from
the of
Sun,
relativeand
to the
See
the top-view
diagram.
2
2
5
F
ma
m
a
a
mg
attached
to
the
tree
and
also
falling.
fr
total
R
tan
s
Rhas
8.5 10 . value
distance
to the
is given
by 2 Rforce
The corresponding
surface, and
soSun,
the static
frictional
of Ffr
F relationship
mg . If between
we equate
Earth
Earth its maximum
s N
s
to Sun
rd
2
2
11. The
apple
does
exert
a
gravitational
force
on
the
Earth.
By
Newton’s
3
law,
the
force
on
the
2
2
these two expressions
for the
frictional
we
can
for the coefficient of static friction.Earth
18.0
m sR2 force,
5.72
m10
s 2 2 .solve
aRthe
2
R
3.3
the
is
Since
the due
relative
change
distance
much
tan
dueEarth
to theand
apple
isaMoon
the same
magnitude
as
the
force
on
the
apple
to the
Earth in
– the
weightis of
the
Earth
2 Earth
1.92 1.9
to Moon
aR2independent
atan
mg
total
s ofm
apple.FfrsThema
force
thes 2state of motion of the apple. So for both a hanging apple
g ismalso
9.80
greater
for theapple,
Earth-Moon
combination,
we
see
the Moon
is the
cause
of the
Earth’s
and aisfalling
the force
on the2 Earth
duefriction,
to that
the 2apple
is equal
to primary
the weight
of the
apple.
2
This
an exceptionally
large
of
2
18.0coefficient
m s
5.72
m s 2 and so the curve had better be banked.
tides.
aR2 atan
1.92 poles?
1.9 Explain.
2. Will ans object weigh more at the equator
orMat the
M Moon
9.80
s 2 bymotion,
27.
We
show
aweighs
top gview
of on
thethe
particle
in
circular
traveling
12. An
Theobject
gravitational
force
the
Moon
ism
given
, where R is(not
theaoppose)
radius
ofeach
the Moon’s
G Earth
15.
more
at
poles,
due
to two
effects
which
complement
other.
2
tan
clockwise.
Because
particle
is in circular
motion,
there
must
beatahad
Rso
This of
is an
large
coefficient
ofthe
friction,
the
curve
better berelative
banked.
First
all,exceptionally
the
Earth the
is slightly
flattened
at
poles and
and
expanded
the
equator,
to
a a
2
radially-inward
component
of
acceleration.
orbit. This
is a radial
force,
andthe
so
can
be expressed
as
M Moon
vthe
R . This can be changed using
perfect
sphere.
Thus
the
mass
at
the
poles
is
slightly
closer
to
Mooncenter, and so experiences a
27. slightly
We show
aa top
viewv 2ofr the particle
in circularobjects
motion, traveling
(a)
aR larger
singravitational
Secondly,
the equator
a centripetal
acceleration
the relationship
vMoon 2 Rforce.
T , where
T is the orbitalatperiod
of thehave
Moon,
to a4tan 2 M Moon
R T 2 . If we
clockwise.
Because
is2 in
circular
motion,
must
be aTo provide
due
to the rotation
ofthe
theparticle
Earth that
objects
at the
polesothere
do not
have.
that
centripetal
aR a
equate
expressionsofm
for
force,
get32.0
the following:
v these
artwo
sincomponent
2.90
mwesin
1.27 m
s of the Earth on
radially-inward
thesthe
acceleration.
acceleration,
the
apparent1.05
weight
(the
radially
outward
normal
force
an object) is
slightly
than the
pull inward.
the two effects
both make the weight of an object
(a) The
aR less
a sin
v 2gravitational
r change comes
(b)
particle’s
speed
from theSotangential
acceleration,
© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
at thewhich
equator
less
than
that
at athe
poles.
is given
atanmay
. If in
theanytangential
acceleration
currently exist.
No portion
of this by
material
becos
reproduced,
form or by any
means, withoutis
permission in writing
a R from the
v
ar sin
1.05 m s 2 2.90 m sin 32.0 o 1.27 m s
publisher.
constant, then using Eq. 2-11a,
101tangential
16. The
Moon
is not pulled
away
from
the Earth
both the
Moon and the Earth are experiencing
(b) The
particle’s
speed
change
comes
from because
the
acceleration,
v
v
a
t
tan
0
tan
tan
Calculation
Based
the same
radial
acceleration
due
to
the
Sun.
They
both
have
the
same
which is given by atan a cos . If the tangential acceleration
is period around the Sun because
2the Sun, and
o so they travel around the Sun
they are both,
onvforce
average,
the1.27
same
distance
from
v
a
t
m
s
1.05
m
s
cos
32.0
2.00
s
3.05
m s radii) above the Earth’s
3. Calculate
the
of
Earth’s
gravity
on
a
spacecraft
12,800km
(2
Earth
tan then
0 tanusing
tanEq. 2-11a,
constant,
together.
surface ifvits
mass
1350kg.
[2 marks]
v0 tan isatan
t
tan
28.
The
spacecraft
is
three
times
as
far
from
the Earth’s
center
aso when
at acceleration
the surface of
the Earth.
17. The centripetal
acceleration
Mars
smaller
that
of
Earth.
The
planet
vsince
vthe
atan t asof
1.27
m sis
1.05 mthan
s 2the
cos
32.0
2.00
s
3.05
mforce
s of each
tan
0 tan force
Therefore,
gravity
decreases
as
square
of
the
distance,
the
of gravity
on
can be found by dividing the gravitational force on each planet by the planet’s mass. The
resulting
the
spacecraft
will
be
one-ninth
of
its
weight
at
the
Earth’s
surface.
acceleration is inversely proportional to the square of the distance of the planet from the Sun. Since
28. Mars
The spacecraft
three
as farthe
from
Earth’s
center as when
at the
surface
of the Earth.
is further is
from
thetimes
Sun than
Earth
the acceleration
of Mars
will
be smaller.
Also see the
1350
kg
9.80
mthe
s 2is,
3
1
F
mg
1.47
10
N
Therefore,
since
the
force
as
gravity
decreases
as
the
square
of
the
distance,
the
force
of gravity on
equation
G below.
Earth's
9
9 its weight at the Earth’s surface.
surfacebe one-ninth of
the spacecraft will
This
could
also
have
been
found
using
law of Universal
Gravitation.
4.
At
the
surface
of
a
certain
planet,
the
acceleration
a magnitude
of 12m/s2. A
mNewton’s
s 2 gravitational
© 2005 Pearson Education, Inc., Upper1350
Saddlekg
River,9.80
NJ. All
rights
reserved. This material
is protected under g
allhas
copyright
laws as they
3
1
currently
exist.FNo
portion
this material may be reproduced, in any form
or by10
any N
means, without permission in writing from the
mgof
1.47
21.0kg
ball
G brass
Earth's is transported to this planet. What is:
9
publisher.
9 and so the mass of the ball is 21.0 kg on both the Earth and the
surface
29.
(a) Mass is independent
of location
a.could
The
mass of the brass ball on the102
Earth
and on the planet. [1 mark]
This planet.
also have been found using Newton’s
law of Universal Gravitation.
b.
The
weigh
of
the
brass
ball
on
the
Earth
and on the planet. [1 mark]
(b) The weight is found by W mg .
29. (a) Mass is independent of location and so the mass
of the ball is 21.0 kg on both the Earth and the
WEarth mg Earth
21.0 kg 9.80 m s 2
206 N
planet.
(b) The weight
ismg
found by W
W
21.0mg
kg . 12.0 m s 2
252 N .
9.
Topic 6.2 Gravitation Problems
Planet
Planet
Principles with Applications, 6th Edition
WEarth mg Earth
21.0 kg 9.80 m s 2
206Physics:
N
30. The force of gravity on an object at the surface of
a planet is given by Newton’s law of Universal
W
mg
21.0 kg 12.0 m s 2
252 N .
Gravitation,Planet
using thePlanet
mass and radius of the planet. If that
is the only force on an object, then the
22
7.35
10
kg
acceleration
of M
a freely-falling
object
is
acceleration
due
to
gravity.
5. A
hypothetical
planet
has
a
radius
1.5
times
that
of
the Earth,
but has the same mass. What is the
11
2
2
Moon
g Moon of G
6.67
10at the
N m
kg of a planet is given
1.62
m s2
2
2
30. The force
gravity
on
an
object
surface
by
Newton’s
law of Universal
6
M
m
rMoon to gravity
Moon
acceleration
due
near the surface?
(Hint:
use the Internet or your data book to find the constants).
1.74If
10 m
FG G using
mg Moon
Gravitation,
the mass
and radius of the planet.
that
is the only force on an object, then the
2
[2
marks] ofrMoon
acceleration
a freely-falling object is acceleration due to gravity.
Giancoli
31. The acceleration
M due
m to gravity at any location on or above the surface of a planet is given by
FG G Moon
mg Moon
2
2
G M Planet
, where
is theNJ.distance
theThis
center
of isthe
planetunder
to the
locationlaws
in asquestion.
© 2005 g
Pearson
SaddlerRiver,
All rightsfrom
reserved.
material
protected
all copyright
they
planet Education,
rInc.,rUpper
Moon
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission
in writing from the
M Planet
M Earth
M Earth
1
1
9.8 m s 2
publisher.
2
g planet
G
2
G
2
2
G 1122
2
g Earth
2
4.4 m s
r Upper Saddle
REarth This1.5
1.5under all copyright laws as they
1.5River,
REarthNJ. All1.5
© 2005 Pearson Education, Inc.,
rights reserved.
material is protected
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.
32. The acceleration due to gravity at any location 112
at or above the surface of a planet is given by
2
g planet G M Planet r , where r is the distance from the center of the planet to the location in question.
g planet
G
M Planet
2
G
1.66 M Earth
2
1.66 G
M Earth
2
8
1.66 g Earth
1.66 9.80 m s 2
16.3 m s 2
567
6. In the diagram shown, two point particles are fixed on an x-axis separated by a distance d. Particle
A has mass mA and particle B has mass 3.00mA. A third particle C, of mass 75.0mA is to be placed on
the axis and near
and
d, at10what
should C be placed
2 B.2In terms of distance
3
6.67particles
10 11 NA m
/kg 2.50 10 3 kg 3.00
kgx-coordinate
1.00 10 3 kg
so that Fthe
net gravitational force on particle A from particles
B and C is zero?
net
2
2 10 1 m
572
1.67 10
14
N.
CHAPTER 13
The force is directed along the diagonal between m2 and m3, toward m2. In unit-vector
have
rnotation,
from m1we
. The
gravitational force between dm and m1 is
14
ˆi
Fnet (cos 45 ˆi sin Gm
45 ˆj)dm (1.18
(1.18 10 14 N) ˆj .
Gm1 (10
M / LN)
)dr
1
dF
,
r2
r2
7. We require the magnitude of force (given by Eq. 13-1) exerted by particle C on A be
equal to
exerted
by B on
where
wethat
have
substituted
dmA. Thus,
( M / L)dr since mass is uniformly distributed. The
Fnet
direction of dF is to the right (see Gm
figure).
TheGm
total
force can be found by integrating
A mC
A mB
=
.
2
2
over the entire length of the rod:
r
d
M L d dr
Gm1M (after
1 canceling
1
Gm
M solve for r. We
We substitute Fin mBdF
= 3mGm
“mA1”)
A 1 and mB = 3mA, and
.
2
d
L is placed
r on the xL axis,Lto dthe left
d ofdparticle
( L d ) A (so it is at a
find r = 5d. Thus, particle C
negative value of x), at x = –5.00d.
Substituting the values given in the problem statement, we obtain
2
8. What
Usingwill
F =
, we find
thatMoon’s
the topmost
mass
pulls upward
onthe
theEarth’s
one atsurface?
the
7. (a)
anGmM/r
object8 weigh
on the
surface
if it weights
100N on
(b)
11
3
2
N,
and
the
rightmost
mass
pulls
rightward
on
the
one
at
the
origin
origin
with
1.9
10
Gm
M
(6.67
10
m
/kg
s
)(0.67
kg)(5.0
kg)
10
1
How manyFEarth
radii
must this same object be from the centre of the
it is to weigh the
3.0Earth
10 if N.
Thus,
the (x, y)
components
with as
1.0it does
10 d8 on
L the
d )Moon?
(N.
(0.23
m)(3.0 mof the
0.23net
m)force, which can be converted to
same
polar components (here we use magnitude-angle notation), are
17. (a) The gravitational acceleration
at the surface
of the Moon
is gmoon = 1.67 m/s2 (see
8
8
8
10 (for
, 1.85a given
10 mass)
2.13
10ratio of
60.6
.
net
Appendix C). The Fratio
of1.04
weights
is the
g-values,
so
8
= (10010
N)(1.67/9.8)
= 17 N.
Wmoon
N.
(a) The magnitude of the force
is 2.13
(b)
force on
thatforce
objectrelative
causedtobytheEarth’s
(b) For
The the
direction
of the
+x axisgravity
is 60.6to .equal 17 N, then the free-fall
acceleration at its location must be ag = 1.67 m/s2. Thus,
9. Both the Sun and the Earth exert a gravitational pull on the space probe. The net force
can be calculated by using superposition
principle.
At the point where the two forces
GmE
GmE
ag2
r2
1.5 107 m
2
balance, we have GM e m / r1 GM
r s m / r2 , where
ag Me is the mass of Earth, Ms is the mass
of the Sun, m is the mass of the space probe, r1 is the distance from the center of Earth to
the distance
from theof
center
of2.4
the“radii”
Sun tofrom
the probe.
substitute r2
thethe
probe,
and
r2 is need
Earth’sWe
center.
so
object
would
to be a distance
r/RE =
= d r1, where d is the distance from the center of Earth to the center of the Sun, to find
8. Two
concentric
spherical
shells
with
uniformly
distributed
Macting
M
18. The
free-body
diagram
of the
force
on
the plumbmasses M1 and M2 are situated as
e
=of thes net
.
2
2 gravitational force on a particle of mass m,
shown
in
the
diagram.
Find
the
magnitude
line is shown to the right. The mass of rthe
sphere
d r1is
1
due to the shells, when the particle is located at radial distance (a) a, (b) b and (c) c.
4
4
M
V
R3
(2.6 103 kg/m3 )(2.00 103 m)3
3
3
8.71 1013 kg.
The force between the “spherical” mountain and the plumb line is F GMm / r 2 .
Suppose at equilibrium the line makes an angle with the vertical and the net force
acting on the line is zero. Therefore,
9
575
where dr 1.70 m as in Sample Problem – “Difference in acceleration at head and feet.”
This yields (in absolute value) an acceleration difference of 7.30 10 15 m/s2.
(e) The miniscule result of the previous part implies that, in this case, any effects due to
the differences of gravitational forces on the body are negligible.
23. (a) The gravitational acceleration is ag =
GM
= 7.6 m/s 2 .
2
R
(b) Note that the total mass is 5M. Thus, ag =
G 5M
3R
2
= 4.2 m/s 2 .
24. (a) What contributes to the GmM/r2 force on m is the (spherically distributed) mass M
contained within r (where r is measured from the center of M). At point A we see that M1
+ M2 is at a smaller radius than r = a and thus contributes to the force:
Fon m
G M1 M 2 m
.
a2
(b) In the case r = b, only M1 is contained within that radius, so the force on m becomes
GM1m/b2.
(c) If the particle is at C, then no other mass is at smaller radius and the gravitational
force on it is zero.
25. Using the fact that the volume of a sphere is 4 R3/3, we find the density of the sphere:
1.0 104 kg
M total
4
R3
3
4
3
1.0 m
3
2.4 103 kg/m3 .
When the particle of mass m (upon which the sphere, or parts of it, are exerting a
gravitational force) is at radius r (measured from the center of the sphere), then whatever
mass M is at a radius less than r must contribute to the magnitude of that force (GMm/r2).
(a) At r = 1.5 m, all of Mtotal is at a smaller radius and thus all contributes to the force:
Fon m
GmM total
r2
m 3.0 10 7 N/kg .
(b) At r = 0.50 m, the portion of the sphere at radius smaller than that is
M=
4 3
r = 1.3 103 kg.
3
10
Answers to exam-style questions
Topic 6
Where appropriate, 1 ✓ = 1 mark
1 A
2 C
3 B
4 C
5 C
6 B
7 D
8 D
9 C
10 A
11 a Velocity arrow. ✓
Acceleration arrow. ✓
velocity
acceleration
2π
= 4.488 ≈ 4.5 rad s −1. ✓
1.40
The linear speed is v = ω r = 4.488 × 0.22 = 0.987 ≈ 0.99 m s −1. ✓
c At maximum distance the frictional force will be the largest possible, i.e. f max = µs N = µsmg( = 0.434 N) . ✓
µg
v2
ω 2r 2
, hence r = s 2 ✓
µsmg = m = m
r
r
ω
0.82 × 9.8
r=
= 0.399 ≈ 0.40 m ✓
4.488 2
µg
µs g
d i Using r = s 2 we find ω =
✓
r
ω
b The angular speed is ω =
0.82 × 9.8
= 6.0 rad s −1 ✓
0.22
ii The static frictional force can no longer supply the larger centripetal force required. ✓
The body will then slide and the static frictional force is now replaced by the even smaller sliding frictional
force; hence the disc will slide off the rotating platform. ✓
ω=
PHYSICS FOR THE IB DIPLOMA © CAMBRIDGE UNIVERSITY PRESS 2015
11
ANSWERS TO EXAM-STYLE QUESTIONS – TOPIC 6
1
1
12 a From energy conservation: mv 2 = mgL so v = 2 gL , ✓
2
v = 2 × 9.8 × 2.0 = 6.26 ≈ 6.3 m s −1. ✓
v 2 6.26 2
=
= 19.6 ≈ 20 m s −2. ✓
L
2.0
c Weight vertically downwards. ✓
Larger arrow for tension upwards. ✓
d i A particle is in equilibrium if it moves with constant velocity. ✓
This particle moves on a circle and so cannot be in equilibrium. ✓
mv 2
✓
ii T − mg =
L
mv 2
5.0 × 6.26 2
T =
+ mg =
+ 5.0 × 9.8 = 147 ≈ 150 N ✓
L
2.0
mv 2
m × 2 gL
(or better: T =
+ mg =
+ mg = 3mg = 3 × 5.0 × 9.8 = 147 ≈ 150 N)
L
L
b a=
13 a Correct arrows for tension. ✓
Correct arrow for weight. ✓
tension
mg
b A particle is in equilibrium if it moves with constant velocity. ✓
This particle moves on a circle and so cannot be in equilibrium. ✓
c i The vertical component of the tension equals the weight and so T cos θ = mg , i.e. T =
The horizontal component of the tension is T sin θ and T sin θ = m
Combining gives the answer v =
mg
.✓
cos θ
v2
v2
=m
✓
r
L sin θ
gL sin 2 θ
.
cos θ
ii The angular and linear speeds are related by v = ω r = ω L sin θ . ✓
So ω =
gL sin 2 θ
cos θ . ✓
L sin θ
Which is the answer ω =
d i v=
g
.
L cos θ
9.8 × 0.45 × sin 2 60°
= 2.57 ≈ 2.6 m s −1 ✓
cos 60°
9.8
= 6.5997 ≈ 6.6 rad s −1 ✓
0.45 × cos 60°
e i The air resistance force will reduce the speed of the ball. ✓
sin 2 θ
ii A graph of
shows that because the speed decreases, the angle will also decrease. ✓
cos θ
ii θ =
2
ANSWERS TO EXAM-STYLE QUESTIONS – TOPIC 6
PHYSICS FOR THE IB DIPLOMA © CAMBRIDGE UNIVERSITY PRESS 2015
12
iii The cosine of the angle will increase and hence the angular speed will decrease. ✓
(Note: These questions are best answered by considering the total energy of the ball:
E=
 sin 2 θ + 2 cos θ − 2 cos 2 θ 
1 2
1 gL sin 2 θ
1
mv + mgh = m
+ mgL (1 − cos θ ) = mgL 

2
2
cos θ
2
cos θ

The air resistance will reduce the total energy; graphing the total energy as a function of angle θ shows that
for the energy to decrease the angle must decrease.)
14 a Measuring distances from the top of the sphere and using energy conservation shows that:
1
0 = mv 2 − mgh where h is the vertical distance the marble falls. ✓
2
From trigonometry: h = R(1 − cos θ ). ✓ (see diagram that follows in b)
1
And so 0 = mv 2 − mgR(1 − cos θ ). ✓
2
Manipulating gives v = 2 gR(1 − cos θ ).
b The forces on the marble are the weight mg and the normal reaction force N:
N
R
Rcos θ
θ
mg
Taking components of the weight gives mg cos θ − N =
Hence N = mg cos θ −
mv 2
.✓
R
mv 2
.✓
R
Substituting the expression for the speed from above gives N = mg cos θ − 2mgR(1 − cos θ ) . ✓
And the result N = mg(3cos θ − 2) follows.
2
c The marble will lose contact when N → 0 , i.e. when cos θ = or θ ≈ 48°. ✓
3
15 a Calling this distance x we have that:
G16M
GM
✓
=
2
x
(d − x )2
16(d − x )2 = x 2 or 4(d − x ) = ± x ✓
Only the plus sign gives a positive distance and so x =
b Correct sign. ✓
Correct intersection. ✓
(The negative of this graph is also acceptable)
4d
.✓
5
PHYSICS FOR THE IB DIPLOMA © CAMBRIDGE UNIVERSITY PRESS 2015
13
ANSWERS TO EXAM-STYLE QUESTIONS – TOPIC 6
3
g
x/d
0.2
0.4
0.6
0.8
1
c i The force is zero. ✓
ii The force from the larger mass will be larger because the particle will be closer to it. ✓
Hence the net force will be directed towards the large mass. ✓
d It will move to the left. ✓
With increasing speed and increasing acceleration. ✓
16 a i Velocity arrow. ✓
Acceleration arrow. ✓
velocity
acceleration
ii Acceleration is the rate of change of the velocity vector. ✓
Here the velocity vector is changing because its direction is so we have acceleration. ✓
b The force on the satellite is
Using v = ω r , ✓
GMm
v2
GM
i.e.
=m
= v 2. ✓
2
r
r
r
GM
= ω 2r 2. ✓
r
From which the result ω 2r 3 = GM follows.
c i Since r decreases, from ω 2r 3 = GM the angular speed will increase. ✓
GM
= v 2 , as r decrease v increases. ✓
ii From
r
ω 2r 3
d i Using ω 2r 3 = GM we find M =
✓
G
(5.31 × 10 −5 )2 × (2.38 × 108 )3
And so M =
= 5.70 × 10 26 kg . ✓
−11
6.67 × 10
gives
ii Again using ω 2r 3 = GM we find ω T2rT3 = ω E2rE3 . ✓
3
Hence ω T = ω E
Hence T =
4
 2.38 × 108 
rE3
−5
−6
−1
=
×
×
5.31
10
 1.22 × 109  = 4.58 × 10 rad s ✓
rT3
2π
2π
1.37 × 106
6
=
=
×
=
1.37
10
s
d = 15.856 ≈ 15.9 d ✓
ω T 4.58 × 10 −6
24 × 3600
ANSWERS TO EXAM-STYLE QUESTIONS – TOPIC 6
PHYSICS FOR THE IB DIPLOMA © CAMBRIDGE UNIVERSITY PRESS 2015
14
Answers to test yourself questions
Topic 6
6.1 Circular motion
1 a The angular speed is just ω =
b The frequency is f =
2π
2π
=
= 5.07 rad s −1. The linear speed is v = ω R = 5.07 × 3.50 = 17.7 m s −1.
1.24
T
1
1
=
= 0.806 s −1.
T 1.24
2 a = 4π 2 rf 2 = 4π 2 × 2.45 × ( 3.5)2 = 1.2 × 10 3 m s −2.
∆v
. The velocity vectors at A and B and the change in the velocity ∆v
3 a The average acceleration is defined as a =
t
∆
are shown below.
2π × 2.0
The magnitude of the velocity vector is 4.0 m s −1 and it takes a time of
= 3.14 s to complete a full
4.0
3.14
revolution. Hence a time of
= 0.785 s to complete a quarter of revolution from A to B. The magnitude of
4
5.66
= 7.2 m s −2. This is
∆v is 4.0 2 + 4.0 2 = 5.66 m s −1 and so the magnitude of the average acceleration is
0.785
directed towards north-west and if this vector is made to start at the midpoint of the arc AB it is then directed
towards the center of the circle.
v 2 16.0
b The centripetal acceleration has magnitude
=
= 8.0 m s −2 directed towards the center of the circle.
r
2.0
 2π r 


4π 2 r
v
= T
= 2 = 4π 2 rf 2 . Hence
4 The centripetal acceleration is a =
r
r
T
2
2
a
=
4π 2 r
50
= 0.356 s −1 ≈ 21 min −1.
4π × 10
v2
4.00
5 a The centripetal acceleration is
=
= 10.0 m s −2. The tension is the force that provides the centripetal
r
0.400
acceleration and so T = ma = 1.00 × 10.0 = 10.0 N.
f =
2
b From T = ma = 20.0 N we have a =
v2
= 20.0 m s −2 and so v = 20 × 0.40 = 2.83 m s −1.
r
4.00 2
16.0
⇒r =
= 0.800 m
c 20.0 = 1.00 ×
r
20.0
6 With a = 9.8 m s −2 we have that a =
4π 2 r
⇒T =
T2
4π 2 × 6.4 × 106
= 5.08 × 10 3 s ≈ 85 min.
9.8
PHYSICS FOR THE IB DIPLOMA © CAMBRIDGE UNIVERSITY PRESS 2015
15
ANSWERS TO TEST YOURSELF QUESTIONS 6
1
 2π r 


4π 2 r 4π 2 × 50.0 × 10 3
v
= 3.2 × 109 m s −2
= T
= 2 =
7 a a=
( 25.0 × 10 −3 )2
r
r
T
2
2
b The forces on the probe are (i) its weight, mg, and (ii) the normal reaction force N from the surface. Assuming
the probe to stay on the surface the net force would be
mg − N =
mv 2
mv 2
v2 

⇒ N = mg −
= m  g −  = m(8.0 × 1010 − 3.2 × 109 ) > 0.

r
r
r 
This is positive so the probe can stay on the surface.
2π R
2π × 1.5 × 1011
m s −1
= 2.99 × 104 ≈ 30 km
=
T
365 × 24 × 60 × 60
2
( 2.99 × 104 )2
v
=
=
a
= 5.95 × 10 −3 ≈ 6.0 × 10 −3 m s −2
b
11
r
1.5 × 10
8 a v=
c F = ma =
mv 2
= 6.0 × 10 24 × 5.95 × 10 −3 ≈ 3.6 × 10 22 N
r
9 The components of L are:
L x = L sin θ , L y = L cos θ
We have that
v2
L sin θ = m
R
L cos θ = mg
Dividing side by side:
v2
m
L sin θ
= R
L cos θ
mg
tan θ =
v2
gR
This gives ⇒ R =
10 a
180 2
v2
=
= 4.7 km
g tan θ 9.8 × tan 35°
friction
reaction
weight
b Let the normal reaction force from the wall be N. Then
v2
N =m
r
mg = f s
For the minimum rotation speed the frictional force must be a maximum i.e. f s = µs N . I.e.
v2
N =m
r
mg = µs N
2
ANSWERS TO TEST YOURSELF QUESTIONS 6
PHYSICS FOR THE IB DIPLOMA © CAMBRIDGE UNIVERSITY PRESS 2015
16
Combining gives mg = µsm
f =
v2
i.e. v =
r
gr
=
µs
9.8 × 5.0
= 9.04 m s −1. From v = 2π rf we find
0.60
v
9.04
=
= 0.288 rev s −1 ≈ 17 rev min −1.
2π r 2π × 5.0
11 a Let v be the speed on the flat part of the road before the loop is entered. At the top the net force on the
cart is its weight and the normal reaction force from the road, both directed vertically downwards. Then,
mu 2
mu 2
⇒N =
− mg where u is the speed at the top. For the cart not to fall off the road, we must
N + mg =
R
R
1
1
2
have N > 0 i.e. u > gR . From conservation of energy, mv 2 = mg( 2R ) + mu 2 and so u 2 = v 2 − 4 gR .
2
2
Hence v 2 − 4 gR > gR , i.e. v > 5 gR = 29.7 ≈ 30 m s −1.
b For just about equal to 5 gR we get u =
gR = 13.3 ≈ 13 m s −1.
12 The tension in the string must equal the weight of the hanging mass i.e. T = Mg . The tension serves as the
Mgr
v2
v2
.
centripetal force on the smaller mass and so T = m . Hence m = Mg ⇒ v =
m
r
r
13 Let the tension in the upper string be TU and TL in the lower string. Both strings make an angle θ with the
horizontal. We have that:
TU sin θ = mg + TL sin θ
TU cos θ + TL cos θ = m
v2
r
We may rewrite these as:
TU sin θ − TL sin θ = mg
TU cos θ + TL cos θ = m
v2
r
0.50
= 0.50 ⇒ θ = 30°. Further, r = 1.0 2 − 0.50 2 = 0.866 m. Therefore the
From trigonometry, sin θ =
1.0
equations simplify to
TU − TL = 4.90
0.50 × (TU − TL ) = 2.45
or
.
TU + TL = 21.33
0.866 × (TU + TL ) = 18.48
Finally, TU = 13.1 N, TL = 8.22 N.
1 2
mv and so v = 2 gh = 2 × 9.81 × 120 = 48.9 ≈ 49 m s −1 (with this speed,
2
this amusement park should not have a licence to operate!).
14 a By conservation of energy, mgh =
b The forces on a passenger are the weight and the reaction force R both in the vertically down direction. Thus
v2
v2
R + mg = m ⇒ R = m − mg . The speed at the top is found from energy conservation as
r
r
1 2
mgH = mv + mg( 2r ) ⇒ v 2 = 9.81 × 240 − 2 × 9.81 × 60 = 1177 . Hence
2
1177
R = 60 ×
− 60 × 9.81 = 1765 ≈ 1800 N.
30
50 2
= 30 m s −2 (some passengers will be fainting
c Using v 2 = u 2 − 2as we get 0 = 49 2 − 2a × 40 and so a =
2 × 40
now, assuming they are still alive!).
PHYSICS FOR THE IB DIPLOMA © CAMBRIDGE UNIVERSITY PRESS 2015
17
ANSWERS TO TEST YOURSELF QUESTIONS 6
3
6.2 The law of gravitation
15 a F = G
Mm
5.98 × 10 24 × 7.35 × 10 22
−11
=
.
×
×
= 1.99 × 10 20 N
6
67
10
( 3.84 × 108 )2
R2
b F =G
Mm
1.99 × 10 30 × 1.90 × 10 27
−11
=
.
×
×
= 4.17 × 10 23 N
6
67
10
(7.78 × 1011 )2
R2
c F =G
Mm
1.67 × 10 −27 × 9.11 × 10 −31
−11
=
.
×
×
= 1.0 × 10 −47 N
6
67
10
(1.00 × 10 −10 )2
R2
16 a Zero since it is being pulled equally from all directions.
b Zero, by Newton’s third law.
m(m + M )
Mm
m2
m2
=
+G
=G
F
G
c F =G
,
(d)
2
2
2
4R
4R
4R 2
4R
 GM 
 (9R )2 
g
1
17 A =
=
gB
81
 GM 
 R2 
 G 2M 
 ( 2R )2  1
g
18 A =
=
gB
2
 GM 
 R2 
19 Since star A is 27 times as massive and the density is the same the volume of A must be 27 times as large. Its radius
 G 27M 
g A  ( 3R )2 
must therefore be 3 times as large. Hence
=
= 3.
gB
 GM 
 R2 
 GM / 2 
 ( R / 2)2 
g
20 new =
=2
g old
 GM 
 R2 
21 Let this point be a distance x from the center of the Earth and let d be the center to center distance between the
earth and the moon. Then
G 81M
GM
=
2
x
(d − x )2
81(d − x )2 = x 2
9(d − x ) = x
x
9
=
= 0.9
d 10
22 a At point P the gravitational field strength is obviously zero.
b The gravitational field strength at Q from each of the masses is
g=
GM
3.0 × 10 22
−11
= 1.0 × 106 N kg −1. The net field, taking components, is directed from Q
=
.
×
×
6
67
10
2
9 2
( 2 × 10 )
R
to P and has magnitude 2 g cos 45° = 2 × 1 × 106 cos 45° = 1.4 × 106 N kg −1.
4
ANSWERS TO TEST YOURSELF QUESTIONS 6
PHYSICS FOR THE IB DIPLOMA © CAMBRIDGE UNIVERSITY PRESS 2015
18
GM
2π r
v2
GMm
4π 2 r 3
m
=
⇒ v2 =
23 We know that 2
and so we deduce that T 2 =
. Therefore
. But v =
r
r
T
r
GM
r=
3
GMT 2
=
4π 2
24 a From v 2 =
3
6.67 × 10 −11 × 6.0 × 10 24 × ( 24 × 60 × 60)2
= 4.2 × 107 m.
4π 2
GM
we calculate v =
r
6.67 × 10 −11 × 6.0 × 10 24
= 7.5828754 × 10 3 ≈ 7.6 × 10 3 m s −1.
(6.4 + 0.560) × 106
6.67 × 10 −11 × 6.0 × 10 24
b The shuttle speed is v =
= 7.5831478 × 10 3 m s −1. The relative speed of the shuttle
6
6.9595 × 10
104
−1
= 36711 s ≈ 10 hrs.
and Hubble is 0.2724 m s and so the distance of 10 km will be covered in
0.2724
25 a
Gm
2π r
v2
Gm1m2
m
=
⇒ v 2 = n − 11 . But v =
and so
2
n
r
r
T
r
4π 2 r 2 Gm1
= n −1
T2
r
2 n +1
4π r
T2 =
Gm1
2
 2π r  = Gm1 giving


T
r n −1
b For this to be consistent with Kepler’s third law we need n + 1 = 3 ⇒ n = 2
PHYSICS FOR THE IB DIPLOMA © CAMBRIDGE UNIVERSITY PRESS 2015
19
ANSWERS TO TEST YOURSELF QUESTIONS 6
5
Topic 6 (New) [54 marks]
An electron moves in circular motion in a uniform magnetic field.
The velocity of the electron at point P is 6.8 × 10 5 m s –1 in the direction shown.
The magnitude of the magnetic field is 8.5 T.
1a. State the direction of the magnetic field.
[1 mark]
Markscheme
out of the page plane / ⊙
Do not accept just “up” or “outwards”.
[1 mark]
1b. Calculate, in N, the magnitude of the magnetic force acting on the electron.
[1 mark]
Markscheme
1.60 × 10–19 × 6.8 × 105 × 8.5 = 9.2 × 10 –13 «N»
[1 mark]
1c. Explain why the electron moves at constant speed.
[1 mark]
Markscheme
the magnetic force does not do work on the electron hence does not change the electron’s
kinetic energy
OR
the magnetic force/acceleration is at right angles to velocity
[1 mark]
20
1d. Explain why the electron moves on a circular path.
[2 marks]
Markscheme
the velocity of the electron is at right angles to the magnetic field
(therefore) there is a centripetal acceleration / force acting on the charge
OWTTE
[2 marks]
A small ball of mass m is moving in a horizontal circle on the inside surface of a frictionless
hemispherical bowl.
The normal reaction force N makes an angle θ to the horizontal.
2a. State the direction of the resultant force on the ball.
Markscheme
towards the centre «of the circle» / horizontally to the right
Do not accept towards the centre of the bowl
[1 mark]
On the diagram, construct an arrow of the correct
21 length to represent the weight of the
[1 mark]