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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/220px-Sedna_orbit.svg.png

2

U (r) = !

GmmSun

r, U (") = 0 . (1)

The energy is therefore

E = K(r) +U (r) =

12

mv(r)2 !GmmSun

r. (2)

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 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 = rar , and

the velocity !va !

!rS ,a , and so the magnitude of the angular momentum about is S is

LS ,a = ramva . (3) When Sedna is at its nearest approach (perihelion),

!rS , p = rpr , and the velocity

!v p !

!rS , p , and so the magnitude of the angular momentum about is S is

LS , p = rpmvp . (4) The angular momentum about the center of the Sun is constant and therefore

ramva = rpmvp . (5) So the minimum speed of Sedna is given by

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va =

rp

ra

vp . (6)

Substituting in the given values we determine that

va =

(76 AU)(937 AU)

(4.64 km ! s"1) = 0.38 km ! s"1 . (7)

c) What is the ratio of Sedna’s maximum kinetic energy to its minimum kinetic

energy? The ratio of kinetic energies is

K p

Ka

=(1 / 2)mvp

2

(1 / 2)mva2 =

vp2

va2 =

vp2

(rp / ra )2 vp2 =

ra2

rp2 (8)

Substituting in the given values we determine that

K p

Ka

=(937 AU)2

(76 AU)2 = 1.5!102 . (9)

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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 0v .

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 S T! = " =r T 0!!!!

Therefore angular momentum about central point is constant.

, ,0system systemS f S=L L! !

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

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so the work done by tension is not zero,

extdW d= ! "T r 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 S T! = " #r T 0!!!!

Therefore the angular momentum about central point is not constant.

, ,0system systemS f S!L L! !

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

extdW d= ! =T r 0!! !

Hence mechanical energy is constant.

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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

!LS ,ring ,1 = IS!1k (1)

We use the parallel axis theorem to calculate the angular momentum of the ring about the pivot point S , IS = m1R2 + Icm = m1R22 + m1R2 = 2m1R2 (2) Substituting Eq. (2) into Eq. (1) yields

!LS ,ring ,1 = IS!1k = 2m1R

2!1k (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

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!vb,1 = vb,1! . (4)

where

vb,1 < 0 . At that instant, a point on the rim of the ring has velocity

!v r ,1 = vr ,1! = 2R"1! . (5)

Therefore the relative velocity,

!v rel = vrel! = v! , 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 =

!vb,1 !!v r ,1 = (vb,1 ! 2Rw1)" = !v" . (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

!LS ,bug ,1 =

!rS ,b,1 ! m2!vb,1 = 2Rr ! m2vb,1" = 2Rm2vb,1k . (8)

Substituting Eq. (7) into Eq. (8) yields

!LS ,bug ,1 = 2Rm2 (!v + 2R"1)k (9)

Because the angular momentum of the system is constant about the point S ,

!0 =!LS ,bug ,1 +

!LS ,ring ,1 = (2Rm2 (!v + 2R"1) + 2m1R

2"1)k . (10) Therefore the z -component of the angular momentum about the point S of the bug and ring satisfies 0 = 2Rm2 (!v + 2R"1) + 2m1R

2"1 . (11) We now solve the Eq. (11) for the angular speed of the ring

!1 =

m2v(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.

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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 Icm = mR2 . 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.

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10

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Problem 5 Satellite Target Practice A satellite is a distance ri from the center of the earth which has mass Me . 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

rf 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

Me , m , G , ri , !i , rf , 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

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mvi2 ! G

mMe

ri

=12

mv f2 ! G

mMe

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

rimvi sin!i = rf mv f sin! f .

We can solve this equation for the initial speed

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vi =

rf sin! f

ri sin!i

v f .

We now substitute this into the energy equation and find that

12

mrf sin! f

ri sin!i

v f

"

#$

%

&'

2

( GmMe

ri

=12

mv f2 ( G

mMe

rf

.

We now solve for the final speed

v f =

2GMe

1rf

! 1ri

"

#$

%

&'

1!rf sin( f

ri sin(i

"

#$

%

&'

2"

#$$

%

&''

.

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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

p! = m2 vA sin" =

m2 vA

2, (1)

Using sin30! = 1 / 2 . The magnitude of the angular momentum about the center of the planet is then

LA = r1 p! =

r1 m2 vA

2. (2)

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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,

LB = r2 vB =

52

r1 vB . (3)

There is no torque on the satellite, so LB = LA ; so equating the expressions in Equations (3) and (2) yields

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r1 vB =r1 m2 vA

2! vB =

vA

5. (4)

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

E =

12

m2 v2 ! Gm1 m2

r. (5)

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

12

m2 vA2 ! G

m1 m2

rA

=12

m2 vB2 ! G

m1 m2

rB

12

vA2 ! G

m1

r1

=12

vA

5"

#$%

&'

2

! Gm1

5r1 / 2

12

vA2 1!

125

"#$

%&'= G

m1

r1

1!25

"#$

%&'

vA2 =

54

Gm1

r1

( vA =54

G m1

r1

(6)

It’s worth noting that vA is less than the escape velocity

vescape =

2G m1

r1

(7)

of the planet, but not by much;

vA

vescape

=58! 0.79 . (8)

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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

a rod about the center of mass is Icm =

112

md 2 . After the collision, the rod rotates

clockwise about its pivot point P with angular speed ! f . Find the angular speed

! f .

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

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mv02 =

12

mv f2 +

12

IP! f2 . (1)

We use the parallel axis theorem the moment of inertia about the pivot point P to find that

IP = m d

2!"#

$%&

2

+ Icm = m d2

!"#

$%&

2

+1

12md 2 =

13

md 2 (2)

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Substitute Eq. (2) into Eq. (1) and after rearranging terms, we have that

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m v02 ! v f

2( ) = 16

md 2" f2 . (3)

We can rewrite Eq.(3) as

! f

2 =3(v0 " v f )(v0 + v f )

d 2 . (4)

This implies that

! f =

3(v0 " v f )(v0 + v f )

d 2 . (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

!LP,0 =

!rP,0 !!p0 = ((d / 2) j+ x0 i) ! mv0 i= "(d / 2)mv0k , (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

!LP, f =

!rP, f !!p f " IP# f k = ((d / 2) j+ x f i) ! ("mv f i) = ((d / 2)mv f " IP# f )k , (7)

where IP 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

!LP,0 and

!LP, f yields

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!

d2

mv0 =d2

mv f !13

md 2" f , (8)

We can rearrange Eq. (8) as

v0 + v f =

2d3! 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 ,

v f =

2d3! f " v0 . (11)

We substitute Eq. (11) into Eq. (10) and solve for

! f just in terms of v0 and d ,

! f =

2v0

d"

2d

2d3! f " v0

#$%

&'()

73! f = 4

v0

d. (12)

Therefore

! f =

127

v0

d. (13)

We now substitute

! f into Eq. (11) and solve for v f ,

v f =

2d3

127

v0

d!

"#$

%&' v0 =

17

v0 . (14)

As a check, we substitute Eq. (14) into Eq. (4) and find that

! f

2 =3(v0 " (1 / 7)v0 )(v0 + (1 / 7)v0 )

d 2 =14449

v02

d 2 #! f =127

v0

d, (15)

in agreement with Eq. (13).

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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 I = 5.0 !10"5 kg #m2 .

• The friction torque

!! f on the

rotor is constant during the measurement.

In the figure below, let t0 = 0 s , t1 = 1.5 s , t2 = 2.0 s , and t3 = 4.0 s , !0 " ! (t0 ) = 280 rad # s

$1 , !1 " ! (t1) = 220 rad # s$1 , !2 " ! (t2 ) , and

! 3 " ! (t3) = 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 , t2 , t3 , !0 , !1 , and ! 3 as appropriate.

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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

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!! f = I "1 = I

#0 $#1

t1. (3)

Substituting the given values, the magnitude of the frictional torque is

!! f = (5.0 "10#5 kg $m2 )

((280 rad $ s#1) # (220 rad $ s#1))(1.5 s)

= 2.0 "10#3 N $m . (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 = ! f ,z dt

t1

t2" = #! 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 #m2 )((280 rad # s"1) " (220 rad # s"1))(2.0 s "1.5 s)

(1.5 s)Jz = 1.0 !10"3 N #m # s

. (6)

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 = (Ir + Iw )#2 " (Ir )#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 Jz = !Lz . (8) Thus, equating the expressions in Equations (5) and (7),

I(!1 "!0 )

(t2 " t1)t1

= I(2!2 "!1) . (9)

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We can solve Equation (9) for the angular velocity immediately after the collision,

!2 =

!1

2+

(!1 "!0 )(t2 " t1)2t1

(10)

Substituting the given values, the magnitude of the angular velocity !2 is

!2 =(220 rad " s#1)

2+

((220 rad " s#1) # 280 rad " s#1)(2.0 s #1.5 s))2(1.5 s)

!2 = 100 rad " s#1

(11)

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

!" =#ave!t = 1

2(#2 +#1)(t2 $ t1) (12)

Equivalently, the z -component of the angular acceleration during the collision is given by

!2 =

("2 #"1)(t2 # t1)

.

Integrating we find that

! (t) "!1 = #2 dt

t1

t

$ =(!2 "!1)(t " t1)

(t2 " t1).

Integrating again we have that

!(t2 ) "!(t1) = # (t) dtt1

t2

$ = #1 +(#2 "#1)(t " t1)

(t2 " t1)%

&'(

)*dt

t1

t2

$

=#1(t2 " t1) +(#2 "#1)(t " t1)2

2(t2 " t1)t1

t2

=#1(t2 " t1) +12

(#2 "#1)(t2 " t1) =12

(#2 +#1)(t2 " t1) =# ave(t2 " t1)

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The energy loss due to torsional friction between the rotor and the dropped washer during the collision is

!Ebearing = "!# f !$ = "

I(%0 "%1)(%2 +%1)(t2 " t1)2t1

= "(5.0 &10"5 kg 'm2 )((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"1N '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 = K2 " K1 =

12

2I( )#22 "

12

I#12 =

12

I(2#22 "#1

2 ) (14)

Substituting the given values and !2 = 100 rad " s#1 yields,

!K =12

(5.0 "10#5 kg $m2 )(2(100 rad $ s#1)2 # (220 rad $ s#1)2 )

= #7.1"10#1 N $m.

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"1N $m)

= " 5.5#10"1 N $m

. (16)