Download Problem Set 9 Angular Momentum Solution

MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
Physics 8.01
Fall 2012
Problem Set 9 Angular Momentum Solution
Problem 1 Sedna
90377 Sedna is a large trans-Neptunian object, which as of 2012 was about three times as
far from the Sun as Neptune. Sedna has the longest orbital period of any known large
object in the Solar System, calculated at approximately 11,400 years. Its maximum speed
is 4.64 km ! s "1 . Its orbit is extremely eccentric, with an aphelion (furthest distance form
the Sun) estimated at 937 AU and a perihelion (closest distance form the Sun) at about 76
AU, the most distant perihelion ever observed for any Solar System object. (An
astronomical unit (abbreviated as AU, is a unit of length equal to exactly
1 AU =1.49,597,870,700 m or approximately the mean Earth–Sun distance.) The mass
and its radius are not well known. Its orbit is shown in the figure below. 1
Figure: The orbit of Sedna (red) set against the orbits of Jupiter (orange), Saturn
(yellow), Uranus (green), Neptune (blue), and Pluto (purple).
a) At what points in its orbit do its minimum and maximum speeds occur?
b) What is Sedna’s minimum speed?
c) What is the ratio of Sedna’s maximum kinetic energy to its minimum kinetic
energy?
Solution:
a) At what points in its orbit do its minimum and maximum speeds occur?
The potential energy function is given by
1
http://upload.wikimedia.org/wikipedia/commons/thumb/e/e3/Sedna_orbit.svg/220pxSedna_orbit.svg.png
1
U (r) = !
GmmSun
, U (") = 0 .
r
(1)
The energy is therefore
GmmSun
1
mv(r)2 !
.
(2)
2
r
Because the energy is constant, K(r) is maximum when U (r) is at a minimum. Because
U (r) < 0 , the smallest value for the potential energy occurs when r has its minimum
value. This occurs when Sedna is closet to the Sun. This point in the orbit is called the
perihelion, which we denote by rp . Hence the maximum speed of Sedna occurs at
E = K(r) + U (r) =
perihelion. Similarly, K(r) is minimum when U (r) takes on its largest value when r has
its maximum value. This occurs when Sedna is furthest from the Sun. This point in the
orbit is called the aphelion, which we denote by ra . Hence the minimum speed of Sedna
occurs at aphelion.
b) What is Sedna’s minimum speed?
Let’s assume that the only significant interaction is the gravitational force between the
Sun and Sedna and points towards the center of sun, (call that the point S ). The torque
!
!
!
!
about the center of the Sun due to this gravitational force is zero, ! S , F = rS ,G " FG = 0 ,
!
!
because rS ,G and FG are anti-parallel. Therefore the angular momentum about the center
!
of the Sun is constant. When Sedna is at its furthest approach (aphelion), rS ,a = rar̂ , and
!
!
the velocity v a ! rS ,a , and so the magnitude of the angular momentum about is S is
LS ,a = ra mva .
(3)
!
!
!
When Sedna is at its nearest approach (perihelion), rS , p = rpr̂ , and the velocity v p ! rS , p ,
and so the magnitude of the angular momentum about is S is
LS , p = rp mv p .
(4)
The angular momentum about the center of the Sun is constant and therefore
ra mva = rp mv p .
(5)
So the minimum speed of Sedna is given by
2
va =
rp
ra
vp .
(6)
Substituting in the given values we determine that
va =
(76 AU)
(4.64 km ! s "1 ) = 0.38 km ! s "1 .
(937 AU)
(7)
c) What is the ratio of Sedna’s maximum kinetic energy to its minimum kinetic
energy?
The ratio of kinetic energies is
Kp
Ka
=
(1 / 2)mv 2p
(1 / 2)mva2
=
v 2p
va2
=
v 2p
(rp / ra )2 v 2p
=
ra2
rp2
(8)
Substituting in the given values we determine that
Kp
(937 AU)2
=
= 1.5 ! 102 .
2
Ka
(76 AU)
(9)
3
Problem 2 Tetherball
A tetherball of mass m is attached to a post of radius R by a string. Initially it is a
distance r0 from the center of the post and it is moving tangentially with a speed v0 .
a) The string passes through a hole in the center of the post at the top (figure above on
left). The string is gradually shortened by drawing it through the hole. Ignore gravity.
Until the ball hits the post, is angular momentum about the center of the post constant?
Explain your reasoning.
b) The string is attached to the outside of the post. As the ball wraps around the post, the
string is gradually shortened (figure above on right). Ignore gravity. Until the ball hits the
post, is angular momentum about the center of the post constant? Explain your reasoning.
Solutions
Part a) The tension force points in the radially inward direction.
hence torque about central point is zero
! !
! !
! S = rS ,T " T = 0
Therefore angular momentum about central point is constant.
!
! system
Lsystem
=
L
S, f
S ,0
For all radial forces, (example gravitation), the angular momentum about the central point
is a constant of the motion. A small displacement of the ball has an inward radial
component
4
so the work done by tension is not zero,
! ! !
dW ext = T ! dr " 0 .
Hence the mechanical energy is not constant.
Part b): The tension force points towards contact point.
Hence torque about central point is not zero.
! !
! !
! S = rS ,T " T # 0
Therefore the angular momentum about central point is not constant.
!
! system
Lsystem
!
L
S, f
S ,0
Small displacement is always perpendicular to string since at each instant in time ball
undergoes instantaneous circular motion about string contact point with pole.
Therefore the tension force is perpendicular to the displacement, and the work done is
zero
! ! !
dW ext = T ! dr = 0
Hence mechanical energy is constant.
5
Problem 3 Bug Walking on Pivoted Ring
A ring of radius R and mass m1 lies on its side on a frictionless table. It is pivoted to the
table at its rim. A bug of mass m2 walks on the ring with constant speed v relative to the
ring, starting at the pivot, when the ring is initially at rest.
What is the direction and magnitude of the rotational velocity of the ring when the bug is
(a) halfway around and (b) back at the pivot.
Solution: We begin by choosing our system to consist of the bug and the ring. Choose
the positive z -direction to point into the figure above. Because there are no external
torques about the pivot point (call it S ), the angular momentum of the system consisting
of the bug and the ring is constant about point S . The initial angular momentum of the
system is zero because the bug starts at the pivot point S and the ring is initially at rest.
Let ! 1 denote the angular ring with respect to a reference frame fixed to the ground
when the bug is halfway around the ring. The angular momentum of the ring at that
instant is given by
!
L S ,ring ,1 = I S! 1k̂
(1)
We use the parallel axis theorem to calculate the angular momentum of the ring about the
pivot point S ,
I S = m1 R 2 + I cm = m1 R 2 + m1 R 2 = 2m1 R 2
(2)
Substituting Eq. (2) into Eq. (1) yields
!
L S ,ring ,1 = I S! 1k̂ = 2m1 R 2! 1k̂
(3)
The bug is moving tangentially with respect to the ground so we can choose polar
coordinates. Denote the velocity of the bug with respect to the ground when the bug is
halfway around the ring by
6
!
v b,1 = vb,1!ˆ .
(4)
where vb,1 < 0 . At that instant, a point on the rim of the ring has velocity
!
v r ,1 = vr ,1!ˆ = 2 R" 1!ˆ .
(5)
!
Therefore the relative velocity, v rel = vrel !ˆ = v!ˆ , of the bug to the ring is given by the
difference of the velocity of the bug and the point on the rim of the track,
!
!
!
v rel = v b,1 ! v r ,1 = (vb,1 ! 2Rw1 )"ˆ = !v"ˆ .
(6)
Therefore the ! -component of the velocity of the bug with respect to a reference frame
fixed to the table is
vb,1 = !v + 2R" 1 .
(7)
When the bug is half around the track, the angular momentum of the bug with respect to
pivot point S is
!
!
!
L S ,bug ,1 = rS ,b,1 ! m2 v b,1 = 2Rr̂ ! m2 vb,1"ˆ = 2 Rm2 vb,1k̂ .
(8)
Substituting Eq. (7) into Eq. (8) yields
!
L S ,bug ,1 = 2Rm2 (!v + 2R" 1 )k̂
(9)
Because the angular momentum of the system is constant about the point S ,
! !
!
0 = L S ,bug ,1 + L S ,ring ,1 = (2Rm2 (!v + 2R" 1 ) + 2m1 R 2" 1 )k̂ .
(10)
Therefore the z -component of the angular momentum about the point S of the bug and
ring satisfies
(11)
0 = 2Rm2 (!v + 2R" 1 ) + 2m1 R 2" 1 .
We now solve the Eq. (11) for the angular speed of the ring
!1 =
m2 v
.
(2m2 + m1 )R
(12)
b) When the bug returns to the pivot point, the angular momentum of the bug about S is
zero, so the angular momentum of the ring about the pivot point is zero also. Hence the
ring has zero angular speed.
7
Problem 4 Inelastic Collision
A rigid hoop of radius R and mass m is lying on a horizontal frictionless table and
pivoted at the point P (shown in the figure). A point-like object of the same mass m is
moving to the right (see figure) with speed v0 . It collides and sticks to the hoop at the
midpoint of the hoop. The moment of inertia of a hoop for axis through the center of
mass and perpendicular to the plane of the hoop is I cm = mR 2 . After the collision, the
hoop rotates counterclockwise about its pivot point with angular speed ! f .
What is the change in mechanical energy of the hoop and object due to this completely
inelastic collision? Express your answer in terms of R , m , and v0 as needed but do not
use ! f in your answer.
8
9
10
Problem 5 Satellite Target Practice
A satellite is a distance ri from the center of the earth which has mass M e . While
hanging motionless in space, the satellite fires an instrument package with an unknown
speed vi at an angle ! i with respect to a radial line between the center of the planet and
the satellite. The package has mass m which is much smaller than the mass of the
spacecraft and the earth. When the package reaches the opposite side of the earth, a
distance r f from the center of the earth, it makes an angle ! f with respect to a radial line
between the center of the planet and the spacecraft. The universal constant of gravitation
is G .
What is the speed v f of the package at that instant? Express your answer in terms of
M e , m , G , ri , ! i , r f , and ! f as needed. Do not use the unknown speed vi in your
answer.
Solution:
Mechanical energy is constant because there is no non-conservative work and we assume
the package and planet are isolated form the rest of the universe. Therefore
mM e 1 2
mM e
1 2
mvi ! G
= mv f ! G
.
2
ri
2
rf
The only force acting on the package is the gravitational force which points towards the
center of the planet, (call that the point S ). The torque about the center of the planet due
!
!
!
!
!
!
to this gravitational force is zero, ! S , F = rS ,G " FG = 0 , because rS ,G and FG are
antiparallel. Therefore the angular momentum about the center of the planet is constant.
With initial and final states shown in the figure above, the condition that angular
momentum is constant is
ri mvi sin ! i = r f mv f sin ! f .
We can solve this equation for the initial speed
11
vi =
r f sin ! f
ri sin ! i
vf .
We now substitute this into the energy equation and find that
2
mM e 1 2
mM e
1 " r f sin ! f %
m$
vf ' ( G
= mv f ( G
.
2 # ri sin ! i &
ri
2
rf
We now solve for the final speed
vf =
" 1 1%
2GM e $ ! '
# r f ri &
" " r sin ( % 2 %
f
$1 ! $ f
' '
$# # ri sin ( i & '&
.
12
Problem 6 Satellite Motion
A spherical non-rotating planet (with no atmosphere) has mass m1 and radius r1 . A
projectile of mass m2 << m1 is fired from the surface of the planet at a point A with a
speed vA at an angle ! = 30! with respect to the radial direction. In its subsequent
trajectory the projectile reaches a maximum altitude at point B on the sketch. The
distance from the center of the planet to the point B is r2 = (5 / 2)r1 .
a) Is there a point about which the angular momentum of the projectile is constant?
If so, use this point to determine a relation between the speed vB of the projectile
at the point B in terms of vA and the angle ! .
b) Determine the initial speed , vA , in terms of G , m1 , and r1 .
Solution:
a) The only force of interest is the gravitational force, which is always directed toward
the center of the planet; hence angular momentum about the center of the planet is a
constant.
At point A, the component p! of the satellite’s linear momentum perpendicular to the
radius vector is
m v
p! = m2 vA sin " = 2 A ,
(1)
2
Using sin 30! = 1 / 2 . The magnitude of the angular momentum about the center of the
planet is then
rm v
LA = r1 p! = 1 2 A .
(2)
2
13
At point B (the apogee), the velocity vector is perpendicular to the radius vector and the
magnitude of the angular momentum is the product of the distance from the center of the
planet and the speed,
5
LB = r2 vB = r1 vB .
(3)
2
There is no torque on the satellite, so LB = LA ; so equating the expressions in Equations
(3) and (2) yields
rm v
v
5
r1 vB = 1 2 A ! vB = A .
(4)
2
2
5
b) The mechanical energy E of the satellite as a function of speed v and radius (distance
from the center of the planet) r is
mm
1
E = m2 v 2 ! G 1 2 .
(5)
2
r
Equating the energies at points A and B, and using rB = r2 = (5 / 2)r1 , rA = r1 and
vB = vA / 5 from part a) (Equation (4) above), we have that
mm
mm
1
1
m2 vA2 ! G 1 2 = m2 vB2 ! G 1 2
2
rA
2
rB
2
m
m
1 2
1" v %
vA ! G 1 = $ A ' ! G 1
2
r1
2# 5 &
5r1 / 2
m
1 2"
1%
vA $ 1 ! ' = G 1
2 #
25 &
r1
"
2%
$# 1 ! 5 '&
5 m1
G
( vA =
4 r1
5 G m1
4 r1
vA2 =
(6)
It’s worth noting that vA is less than the escape velocity
vescape =
2G m1
r1
(7)
5
! 0.79 .
8
(8)
of the planet, but not by much;
vA
vescape
=
14
Problem 7 Elastic Collision
A rigid rod of length d and mass m is lying on a horizontal frictionless table and pivoted
at the point P on the one end (shown in the figure). A point-like object of the same mass
m is moving to the right (see figure) with speed v0 . It collides elastically with the rod at
the midpoint of the rod and rebounds backwards with speed v f . The moment of inertia of
1
md 2 . After the collision, the rod rotates
12
clockwise about its pivot point P with angular speed ! f . Find the angular speed ! f .
a rod about the center of mass is I cm =
Express your answer in terms of d , m , v0 , and v f as needed.
Solution:
Comment: The statement of the problem asks to find ! f in terms of d , m , v0 , and v f
as needed. We can use either energy or angular momentum to determine a relationship
for ! f . When we use both equations we can eliminate v f and find ! f in terms of d ,
and v0 .
The collision is elastic therefore the mechanical energy is of the rod and object are
constant. We can equate the initial and final mechanical energies and determine that
1
1
1
mv02 = mv 2f + I P! 2f .
2
2
2
(1)
We use the parallel axis theorem the moment of inertia about the pivot point P to find
that
2
2
! d$
! d$
1
1
I P = m # & + I cm = m # & + md 2 = md 2
12
3
" 2%
" 2%
(2)
15
Substitute Eq. (2) into Eq. (1) and after rearranging terms, we have that
(
)
1
1
m v02 ! v 2f = md 2" 2f .
2
6
(3)
We can rewrite Eq.(3) as
! 2f =
3(v0 " v f )(v0 + v f )
d2
.
(4)
This implies that
!f =
3(v0 " v f )(v0 + v f )
d2
.
(5)
The pivots forces exert no torque about the pivot point. During the collision, the collision
forces are internal and the torques about P on the rod and object due to these collision
forces add to zero). Therefore the angular momentum about the pivot point P is
constant. Choose unit vectors as shown in the figure below.
Before the collision, the angular momentum about the pivot point P is entirely due to the
!
motion of the object, rP, f
!
!
!
L P,0 = rP,0 ! p0 = ((d / 2) ĵ + x0 î) ! mv0 î = "(d / 2)mv0k̂ ,
(6)
After the collision, the angular momentum about the pivot point P is the sum of the
contributions from the object and the rotating rod
!
!
!
L P, f = rP, f ! p f " I P# f k̂ = ((d / 2) ĵ + x f î) ! ("mv f î) = ((d / 2)mv f " I P# f )k̂ ,
(7)
where I P is the moment of inertia of the rod about the pivot point about P . Inserting Eq.
!
(2) into Eq. (7), and then equating the expressions for the z -components of L P,0 and
!
L P, f yields
16
!
d
d
1
mv0 = mv f ! md 2" f ,
2
2
3
(8)
We can rearrange Eq. (8) as
v0 + v f =
2d
! .
3 f
(9)
We now substitute Eq. (9) into Eq. (4) and solve for the angular speed ! f ,
!f =
2(v0 " v f )
d
.
(10)
We can now use both of our results, Eqs. (4) and (10), to find ! f in terms of d , and
v0 . We first solve Eq. (9) for v f ,
vf =
2d
! " v0 .
3 f
(11)
We substitute Eq. (11) into Eq. (10) and solve for ! f just in terms of v0 and d ,
!f =
2v0 2 # 2d
v
&
7
" % ! f " v0 ( ) ! f = 4 0 .
d
d$ 3
3
d
'
(12)
12 v0
.
7 d
(13)
Therefore
!f =
We now substitute ! f into Eq. (11) and solve for v f ,
vf =
2d ! 12 v0 $
1
' v0 = v0 .
#
&
3 " 7 d%
7
(14)
As a check, we substitute Eq. (14) into Eq. (4) and find that
! =
2
f
3(v0 " (1 / 7)v0 )(v0 + (1 / 7)v0 )
d2
2
144 v0
12 v0
=
#
!
=
,
f
49 d 2
7 d
(15)
in agreement with Eq. (13).
17
Problem 8 Angular Momentum Experiment
In the angular momentum experiment,
shown to the right, a washer is dropped
smooth side down onto the spinning
rotor.
The graph below shows a plot of the
rotor angular speed as a function of time.
Assume the following:
• The rotor and washer have the
same
moment
of
inertia
"5
2
I = 5.0 ! 10 kg # m .
•
!
The friction torque ! f on the
rotor is constant during the
measurement.
In the figure below, let t 0 = 0 s , t1 = 1.5 s , t 2 = 2.0 s , and t 3 = 4.0 s ,
! 0 " ! (t 0 ) = 280 rad # s $1 , ! 1 " ! (t1 ) = 220 rad # s $1 , ! 2 " ! (t 2 ) , and
! 3 " ! (t 3 ) = 60 rad # s $1 .
In the following questions first determine an analytic expression prior to substituting the
values obtained from the plot of ! vs. t . Express your analytic answers in terms of I ,
t1 , t 2 , t 3 , ! 0 , ! 1 , and ! 3 as appropriate.
18
a) Determine an expression for the magnitude of the frictional torque in the rotor,
!
! f , and determine its value from the measured quantities.
b) Determine an expression for the magnitude of the angular impulse during the
collision, and determine its value from the measured quantities.
c) Determine an expression for the magnitude of the angular velocity ! 2 of the two
washers immediately after the collision is finished (at time t2 ), and determine its
value from the measured quantities.
d) How much mechanical energy is lost to the work done due to frictional torque
during the collision?
e) How much mechanical energy is lost to the work done due to friction between the
rotor and the washer during the collision?
Solution:
a) Determine an expression for the magnitude of the frictional torque in the rotor,
!
! f , and determine its value from the measured quantities.
Choose the positive z -direction so that z -component of the angular velocity is positive
throughout the motion.
The z -component of the angular acceleration of the motor and washer from the instant
when the power is shut off until the second washer was dropped is given by
!1 =
"1 # " 0
.
t1
(1)
The z -component of the angular acceleration is due to the frictional torque in the motor,
!" friction = I#1 .
(2)
So the magnitude of the frictional torque is given by
19
# $ #1
!
.
! f = I "1 = I 0
t1
(3)
Substituting the given values, the magnitude of the frictional torque is
!
((280 rad $ s #1 ) # (220 rad $ s #1 ))
! f = (5.0 " 10#5 kg $ m 2 )
= 2.0 " 10#3 N $ m .
(1.5 s)
(4)
b) Determine an expression for the magnitude of the angular impulse during the
collision, and determine its value from the measured quantities.
During the collision with the second washer, the frictional torque exerts an angular
impulse (pointing along the z -axis in the figure),
Jz =
"
t2
t1
! f ,z dt = #! f (t2 # t1 ) = I($ 1 # $ 0 )
(t2 # t1 )
.
t1
(5)
Substituting the given values, the magnitude of the angular impulse is
Jz =
(5.0 ! 10"5 kg # m 2 )((280 rad # s "1 ) " (220 rad # s "1 ))(2.0 s " 1.5 s)
(1.5 s)
.
(6)
"3
J z = 1.0 ! 10 N # m # s
c) Determine an expression for the magnitude of the angular velocity ! 2 of the two
washers immediately after the collision is finished (at time t2 ), and determine its
value from the measured quantities.
The z -component of the angular momentum about the rotation axis of the motor changes
during the collision,
!Lz = L2,z " L2,z = (I r + I w )# 2 " (I r )# 1 = 2I# 2 " I# 1 .
(7)
The change in the z-component of the angular momentum is equal to the z-component of
the angular impulse
J z = !Lz .
(8)
Thus, equating the expressions in Equations (5) and (7),
I(! 1 " ! 0 )
(t2 " t1 )
= I(2! 2 " ! 1 ) .
t1
(9)
20
We can solve Equation (9) for the angular velocity immediately after the collision,
!2 =
! 1 (! 1 " ! 0 )(t2 " t1 )
+
2
2t1
(10)
Substituting the given values, the magnitude of the angular velocity ! 2 is
!2 =
(220 rad " s #1 ) ((220 rad " s #1 ) # 280 rad " s #1 )(2.0 s # 1.5 s))
+
2
2(1.5 s)
(11)
! 2 = 100 rad " s #1
d) How much mechanical energy is lost to the work done due to frictional torque
during the collision?
.
The mechanical energy lost due to the bearing friction is the product of the magnitude of
the frictional torque and the total angle "! through which the bearing has turned during
the collision. A quick way to calculate "! is to use
1
!" = # ave !t = (# 2 + # 1 )(t2 $ t1 )
2
(12)
Equivalently, the z -component of the angular acceleration during the collision is given
by
(" # " 1 )
.
!2 = 2
(t2 # t1 )
Integrating we find that
t
! (t) " ! 1 = $ # 2 dt =
t1
(! 2 " ! 1 )(t " t1 )
.
(t2 " t1 )
Integrating again we have that
t2
t2
%
(# " # 1 )(t " t1 ) (
! (t2 ) " ! (t1 ) = $ # (t) dt = $ ' # 1 + 2
* dt
(t2 " t1 )
)
t1
t1 &
(# " # 1 )(t " t1 )2
= # 1 (t2 " t1 ) + 2
2(t2 " t1 )
t2
t1
1
1
= # 1 (t2 " t1 ) + (# 2 " # 1 )(t2 " t1 ) = (# 2 + # 1 )(t2 " t1 ) = # ave (t2 " t1 )
2
2
21
The energy loss due to torsional friction between the rotor and the dropped washer during
the collision is
I(% 0 " % 1 )(% 2 + % 1 )(t2 " t1 )
!
!Ebearing = " # f !$ = "
2t1
="
(5.0 & 10"5 kg ' m 2 )((280 rad ' s "1 ) " (220 rad ' s "1 ))((100 rad ' s "1 ) + (220 rad ' s "1 ))(0.5 s)
2(1.5 s)
= "1.6 & 10"1 N ' m
(13)
.
e) How much mechanical energy is lost to the work done due to friction between the
rotor and the washer during the collision?
The change in kinetic energy during the collision is
!K = K 2 " K1 =
1
1
1
2I # 22 " I# 12 = I(2# 22 " # 12 )
2
2
2
( )
(14)
Substituting the given values and ! 2 = 100 rad " s #1 yields,
1
(5.0 " 10#5 kg $ m 2 )(2(100 rad $ s #1 )2 # (220 rad $ s #1 )2 )
.
2
#1
= #7.1 " 10 N $ m
!K =
The change in kinetic energy is equal to the energy lost due to the bearing friction,
!Ebearing , and the energy, !Erotor ,washer , lost to the work done due to friction between the
rotor and the washer during the collision
!K = !Ebearing + !Erotor ,washer
(15)
Therefore
!Erotor ,washer = !K " !Ebearing
!Erotor ,washer = ("7.1 # 10"1 N $ m) " ("1.6 # 10"1 N $ m) .
(16)
= " 5.5 # 10"1 N $ m
22