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Massachusetts Institute of Technology - Physics Department Physics-8.01 Fall 1999
Solutions for Assignment # 8
by Dru Renner [email protected]
November 14, 1999 6:20 pm
Problem 8.
The subsequent motion will be elliptical with the CENTER of the planet at one focus. (Kepler’s First Law) Angular momentum of the satellite is conservedif we measure it relative to the center of the planet. (This is what is called“orbital angular momentum.”) Angular momentum is NOT conservedrelative to ANY OTHER point! Initially the magnitude of angular momentum relative to the center of the planet is
L = mRv 0 sin(20◦)
where m is the mass of the satellite. At the point of maximum distance, the apogee, the magnitude of angular momentum is
L = m 5 Rv sin(90◦)
where v is the corresponding speed. (At the apogee and perigee, the angle between the velocity vector andthe radius vector from either foci will always be 90 ◦.) Conservation of angular momentum andhence conservation of the magnitude of angular momentum gives
mRv 0 sin(20◦) = m 5 Rv sin(90◦) =⇒ v =
sin(20◦) 5
· v 0
The gravitational force is conservative; therefore mechanical energy is conserved. Initially the mechanical energy is
E = −
GM m R
mv^20
At the point of maximum distance the mechanical energy is
E = −
GM m 5 R
mv^2 = −
GM m 5 R
m
( sin(20◦) 5
· v 0
) 2
Conservation of mechanical energy gives
GM m R
mv 02 = −
GM m 5 R
m
( sin(20◦) 5
· v 0
) 2
Solving the above for v 0 gives
v 0 =
√√ √√ 8 GM 5 R
( 1 − sin
(^2) (20◦) 25
√ GM R
Notice that this satellite will slam into the planet before it has made one rotation about the planet (why?).
Problem 8.
(a) The distance apart is f 2 πR ≈ 1710 m.
(b) nsTs = T (na − f ) (Please see handout.); thus 2Ts = T (2 − 0 .04). The period Ts is
smaller than T. The sandwich should be thrown backward.
(c) Mary’s (andPeter’s) orbital period T = 2π(R^3 /M G)^0.^5 = 5580 s ≈ 93 min. Mary
must wait T (na − f ) = (2 − 0 .04) × 93 min ≈ 3 .04 hrs before she will make the catch.
(d) nsTs = T (na − f ), thus ns(4π^2 a^3 /M G)^0.^5 = (4π^2 R^3 /M G)^0.^5 (na − f ) which lead s to
a = R[(na − f )/ns]^2 /^3 ; thus a ≈ 6800(0.987) ≈ 6710 km.
(e)
(NeedPicture)
(f ) The velocity of the spacecraft, va, is 2πR/T. va ≈ 7 .66 km/s. The velocity, vs, of
the sandwich at the point X can be obtained from the relation
ENERGY = −GM m/ 2 a =
mv s^2 − GM m/R
R = 7000 km
Since
va =
√ GM/R
vs = v
√ 2 − R/a vs ≈ 7 .61 km/s
The sandwich must be thrown relative to Peter’s motion with a velocity vs − va ≈ 7. 61 −
- 66 ≈ − 52 .1 m/s ≈ − 116 .5 mph. Too high for any human being! The “-” sign indicates that the sandwich must be thrown backwards. To reduce the speed with which the sandwich has to be thrown you could increase the number of orbits for both Mary and the sandwich before the catch is made. Using Dave Pooley’s program (as shown in lectures) we findthat for ns = na = 3 the speedis about 77 .1 mph (Roger Clemens couldjust do it!); for ns = na = 4 the speedwill be about 57.7 mph, andfor ns = na = 5 the speedis only about 46.0 mph (that is certainly doable!).
(e) The impulse is given by the change in momentum. The final momentum is zero;
therefore the impulse is given by
I 2 = ∆p = mv 2 = I 0 ·
√√ √√ 2 1 + Rr
R
r
(f ) The sum of each impulse is
I = I 1 + I 2 = ∆p = mv 1 − mv 0 = I 0
√√ √√ 2 1 + Rr
(^) + I 0 ·
√√ √√ 2 1 + Rr
R
r
= I 0
√√ √√ 2 1 + Rr
( 1 +
R
r
) − 1
(^) = I 0
√ 2
( 1 +
R
r
) − 1
Therefore the difference is
I 0 − I = I 0 ·
2 −
√ 2
( 1 +
R
r
) (^) ≥ 0
since r ≥ R.
(g) For r = 20 · R the above equation gives
I 0 − I ≈ 0. 55 · I 0
The quantity I 0 is defined above as
I 0 = mv 0 ≈ 2. 98 × 104 · m
Therefore, in terms of the mass of the spacecraft, m, the difference in the impulses is
I 0 − I ≈ 1. 6 × 104 · m
Problem 8.
(a)
Gm 1 m 2 (r 1 + r 2 )^2
m 1 v^21 r 1
m 2 v^22 r 2
v 1 =
2 πr 1 T
v 2 =
2 πr 2 T
Gm 2 (r 1 + r 2 )^2
4 π^2 r 1 T
Gm 1 (r 1 + r 2 )^2
4 π^2 r 2 T
Adding and solving for T yields:
T 2 =
4 π^2 (r 1 + r 2 )^3 G(m 1 + m 2 )
(b)
r 2 =
v 2 T 2 π
v 2 = 148 × 103 m/s T = 5.6 d ays× 86 , 400 sec/day
=⇒ r 2 = 1. 14 × 1010 m
(c)
m 1 r 1 = m 2 r 2
If x = r r^12 , then m 1 = m x^2. The period is given by
T 2 =
4 π^2 (r 1 + r 2 )^3 G(m 1 + m 2 )
4 π^2 r^32
( r 1 r 2 + 1
) 3
Gm 2
( (^) m 1 m 2 + 1
4 π^2 r 23 (x + 1)^3 Gm 2
( (^1) x + 1
)
This means, upon substituting for T and r 2
15 .9 = x^3 + 2x^2 + x
Either by plotting the right handside as a function of x or by noting that x is nearly equal to 2 andusing trial anderror, you will find x ≈ 1 .90; thus r 1 = xr 2 ≈ 2. 17 × 1010 m.
(d)
m 1 = m 2 /x ≈ 15 .8 M
Problem 8.5 (Ohanian, page 352, problem 45)
We will calculate the requiredstatic friction for rolling at a given angle of inclination θ. The force equation along the incline of the ramp is
ma = mg sin θ − Ffric
where Ffric is the static friction force. The torque equation, about the center of the hoop, is Iα = RFfric
M = ρLπR^2
where ρ is the density of brass, which we are not given. For the thinner cylinder: d 1 = L 1 2 = 0.^45 m,^ R^1 = 0.005 m and^ L^1 = 0.90 m, andthe moment of inertia is
I 1 = ρL 1 πR^21
( d^21 +
R^21 +
L^21
) = 1. 91 × 10 −^5 ρ
where ρ must be in metric units. For the thicker cylinder: d 2 = 1.0 m, R 2 = 0.03 m and L 2 = 0.20 m, andthe moment of inertia is
I 2 = ρL 2 πR^22
( d^22 +
R^22 +
L^22
) = 5. 67 × 10 −^4 ρ
where ρ must be in metric units. The total moment of inertia is
I = I 1 + I 2 = 5. 87 × 10 −^4 ρ
The total mass is the sum of each mass andis given by
M = ρ(L 1 πR^21 + L 2 πR^22 ) = 6. 36 × 10 −^4 ρ
where ρ must be in metric units.
The quantity l in the formula for the periodis the distance from the point of oscillation to the center of mass of the physical pendulum. This is given by
M l = Mthin
( L
1 2
)
( L 1 +
L 2
)
where ρ must be in metric units. Therefore the quantity l is
l =
- 39 × 10 −^3 ρ
- 54 × 10 −^3 ρ
≈ 0 .941 m
andthe period T is
T = 2π
√
- 87 × 10 −^4 ρ
- 36 × 10 −^4 ρ · g · 0. 941
≈ 1 .988 s
Problem 8.
We assume that the mass m 2 is sufficiently larger than the mass m 1 so that m 2 accelerates down. This determines which direction the kinetic friction acts for each block. The force equation for m 2 in the direction down the slope is
m 2 a =
∑ F = m 2 g sin θ 2 − μm ︸ 2 g︷︷ cos θ (^2) ︸ fric
−T 2 =⇒
T 2 = m 2 g sin θ 2 − μm 2 g cos θ 2 − m 2 a (1)
where a is the acceleration of m 2 directed down the slope and T 2 is the tension for that portion of the string. The force equation for m 1 in the direction up the slope is
m 1 a =
∑ F = T 1 − m 1 g sin θ 1 − μm ︸ 1 g︷︷ cos θ (^1) ︸ fric
T 1 = m 1 a + m 1 g sin θ 1 + μm 1 g cos θ 1 (2)
where a is also the acceleration of m 1 directed up the slope (assuming the string does not stretch) and T 1 is the tension for that portion of the string. The torque equation for the pulley is
Iα =
∑ τ
M R^2 α = RT 2 − RT 1 =⇒
α =
2(T 2 − T 1 )
M R
where α is the angular acceleration clockwise in the picture and I = 12 M R^2 is the moment of inertia about the axle. (Please see Table 12.1 on page 309.) The coefficient of RT 2 is positive because T 2 acts to increase α; the coefficient of RT 1 is negative because T 1 acts to decrease α. The constraint of no slipping between the pulley andthe string requires that
a = Rα (4)
Substituting Equation (??) into Equation (??) gives
a =
2(T 2 − T 1 )
M
Problem 8.
Initially the plane is moving too fast andthe tires too slow to satisfy the rolling constraint. Friction acts on the wheels. This creates a linear acceleration that slows the plane down and an angular acceleration that spins the tires. At some moment the linear speedof the plane andangular speedof the tires are appropriate for rolling. Static friction will then maintain the rolling. Each tire supports a weight M g where the total mass is given by MT OT = nM where n is the number of tires. Now suppose the runway exerts a friction force Ffric on each tire. The force equation for the plane is
nM a = nFfric =⇒ a =
Ffric M
The friction force while sliding is constant; hence the velocity is given by
v = v 0 − at = v 0 −
Ffric M
· t
The torque equation (about the center of a tire) is
Iα = RFfric =⇒ α =
RFfric I
Again, the friction force is constant; hence the angular velocity is given by
ω =
RFfric I
· t
Rolling occurs when
v = Rω =⇒ v 0 −
Ffric M
· t = R ·
RFfric I
· t
The solution to the above equation is
t =
v 0 Ffric
( 1 M +^
R^2 I
)
The velocity at this time is
v = v 0 ·
M R^2 I 1 + M R 2 I
v 0 1 + (^) M RI 2
Problem 8.10 (Ohanian, page 353, problem 50)
The precession frequency is given by Equation (41) on page 344 as
ωP =
rM g L where the quantities above are explainedin Section 13.6 on page 343. The angular momentum is given by
L = Iω
where I is the moment of inertia given by Table 12.1 on page 309 as
I =
M R^2
Combining the above gives
ωP =
2 rg R^2 ω For this child’s toy top the values are r = 6.0 cm = 6. 0 × 10 −^2 m, M = 0.15 kg, R = 5 .0 cm = 5. 0 × 10 −^2 m, and ω = 200 rev/s = 200 · 2 π radian/s. These values give
ωP ≈ 0 .37 radian/s
Problem 8.11 (Ohanian, page 353, problem 51)
(a) The ship will rotate clockwise in the water when viewedfrom above. The wave
exerts an upwardforce on the right axle anda downwardforce on the left axle; both forces create a torque that points to the back of the boat. The angular momentum will attempt to align with the torque; thus causing the ship to rotate.
(b) The ship will roll left, or capsize. The wave exerts a backwardforce on the left
axle anda forwardforce on the right axle; both forces create a torque that point upward. The angular momentum will attempt to align with the torque; thus causing the ship to capsize.
Problem 8.
(a) Let I = the moment of inertia of the rodthrough an axis perpendicular to the
plane of the page. An angular impulse is deliveredto the rodof magnitude F d∆t. This gives the rodan angular momentum Iω, therefore ω = F d∆t/I. The rodtumbles with this angular velocity.
(b) L is the angular momentum of the spinning rod. The situation is now exactly
the same as a gyroscope with its angular momentum horizontal actedon by gravity. ( F replaces mg). The torque is F d which causes a precessional angular velocity F d/L. (See Ohanian page 344). The precession is in the plane perpendicular to F". In a time ∆t, the axis moves through an angle F d∆t/L insteadof tumbling as in part a). The larger L is the smaller the angle will be.
