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Massachusetts Institute of Technology - Physics Department Physics-8.01 Fall 1999
Solutions for Assignment # 7
by Dru Renner [email protected]
November 7, 1999 9:50 pm
Problem 7.1 (Ohanian, page 271, problem 55)
Throughout this problem we will use the rocket equation:
vf − vi = u ln
( Mi Mf
)
Consider two rockets: (1) a multiple-stage rocket with initial mass M 0 andwith a final mass of the last stage M ; (2) a single-stage rocket with initial mass M 0 andfinal mass M. Suppose all stages of (1) andthe single stage of (2) have the same exhaust speed u. We know that the terminal speedof the single-stage rocket is
v 2 = u ln
( M
0 M
)
Rocket (1) has less fuel than rocket (2) because each jettisonedstage has some mass; we shouldexpect that rocket (1) will have a smaller terminal speedthan rocket (2). To see this, consider a two stage rocket. Suppose the mass of the fuel for the first stage is MF 1 , the mass of the first jettisonedstage is M 1 , andthe mass of the fuel for the secondstage is MF 2. Then the final mass will be M = M 0 − MF 1 − M 1 − MF 2. After burning M F 1, the speedof the rocket is
v′^ = u ln
( M
0 M 0 − MF 1
)
Then the first stage is jettisoned, so the mass is reduced to M 0 − MF 1 − M 1. If we assume the first stage moves away at zero velocity relative to the remainder of the rocket then the remainder of the rocket will not speed up due to the jettisoned stage. Then MF 2 is burned, andthe speedof the rocket is
v′′^ = u ln
( M
0 −^ MF 1 −^ M 1
M 0 − MF 1 − M 1 − M 2
) = u ln
( M
0 −^ MF 1 −^ M 1
M
)
Thus the terminal speedfor rocket (1) is
v 1 = v′^ + v′′^ = u ln
( M
0 M 0 − MF 1
)
( M
0 −^ MF 1 −^ M 1
M
)
= u ln
( M
0 M 0 − MF 1
M 0 − MF 1 − M 1
M
)
= u ln
( M
0 M
) − u ln
( M
0 −^ MF 1
M 0 − MF 1 − M 1
)
= v 2 − u ln
( M
0 −^ MF 1
M 0 − MF 1 − M 1
)
Since M 1 > 0 the secondpiece in the above expression is negative, so
v 1 < v 2
For each stage that is jettisonedthere will be similar terms which further reduce v 1.
Problem 7.
Consider velocities positive if along the initial direction of the spacecraft. Let the final velocity of the spacecraft be v′^ andthe final velocity of the planet be V ′. Conservation of momentum gives
mv + M V = mv′^ + M V ′^ =⇒ V ′^ − V = m M · (v − v′) (1)
Mechanical energy is conserved. If we consider the initial and final moments of the system to occur when there is sufficient separation between the planets, then the gravitational potential energy can be ignored. Therefore kinetic energy is conserved.
1 2 mv^2 +
M V 2 =
mv′^2 +
M V ′^2 =⇒
m M
· (v − v′) · (v + v′) = (V ′^ − V ) · (V ′^ + V )
Using Equation (??) we can rewrite the above equation as
m M
· (v − v′) · (v + v′) = m M
· (v − v′) · (V ′^ + V ) =⇒
V ′^ = v + v′^ − V
Substituting this expression back into Equation (??) we obtain
v + v′^ − 2 V = m M
· (v − v′) =⇒ v′^ =
m M −^1 1 + (^) Mm
· v +
1 + (^) Mm
· V
Problem 7.4 (Ohanian, page 320, problem 26)
For a sphere of uniform density, the center of mass is actually the geometric center; thus the moment of inertia about a diameter is actually the moment of inertia about an axis through the center of mass. Therefore in the language of the parallel axis theorem
ICM =
M R^2
Now we note that any axis tangent to the surface is parallel to some diameter, thus we can use the parallel axis theorem to findthe moment of inertia, I, about such an axis
I = ICM + M R^2 =
M R^2 + M R^2 =
M R^2
Problem 7.5 (Ohanian, page 322, problem 41)
Let M = 1. 5 × 1030 kg, R = 20 km = 20 × 103 m, ω 0 = 2.1 rev/s = 2. 1 · 2 π radian/s, α 0 = − 1. 0 × 10 −^15 rev/s^2 = − 1. 0 × 10 −^15 · 2 π radian/s. (Remember it is necessary to convert all angular quantities from “rev” to “radian.”) The rotational kinetic energy is
K =
Iω^2
where the moment of inertia is given in Table 12.1 on page 309 as
I =
M R^2 = 2. 4 × 1038 kg m^2
Therefore the time rate of change of K is
dK dt
I
dω^2 dt = Iωα
Initially ω = ω 0 and α = α 0 , so the initial rate of change of K is
r 0 = Iω 0 α 0 ≈ − 1. 99 × 1025 J/s
If this rate is to remain constant then
dK dt
I
dω^2 dt = r 0 =⇒ ω^2 = 2 r 0 I · t + ω^20
The time T for the rotation to stop is given by
2 r 0 I T + ω^20 =⇒ T = − ω^20 I 2 r 0 = 1. 05 × 1015 s ≈ 3. 3 × 107 years
Problem 7.6 (Ohanian, page 322, problem 45)
Choose an x-y-z coordinate system with origin at the center of the square, z axis perpendicular to the square, and x and y axes parallel to sides of the square. In the language of the perpendicular axis theorem, we want to calculate Iz , but it is easier to calculate Ix and Iy instead. Let M be the mass of the square and l be the length of each side. Also let h be the small depth for the square. Then the volume of the square is
V = hl^2
andthe density of each rodis
ρ =
M
V
M
hl^2
which is a constant that can be “pulledout” of the integration. Then the moment of inertia about the x axis is
Ix =
∫ ρr^2 dV
=
∫ (^) l 0
dx
∫ 2 l − 2 l
ρ hdxdy
= ρh · l ·
2 l^3 8 =
M
hl^2 · h · l^4 12 =
M l^2
By symmetry, the moment of inertia about the y axis is the same
Iy =
M l^2
Using the perpendicular axis theorem gives
Iz = Ix + Iy =
M l^2 +
M l^2 =
M l^2
Problem 7.7 (Ohanian, page 324, problem 59)
First we needthe equations of motion for projectile flight. For the angle 45 ◦^ we find cos 45◦^ = √^12 andsin 45 ◦^ = √^12 The equations of motion are given by
x =
√^1
v 0 t
v =
√^1
v 0 xˆ −
√^1
v 0 yˆ
The magnitude of the angular momentum is given by the product of the distance and the component of velocity perpendicular to r,
|L| = m| r| · |vy| = v^20 g
m √ 2
v 0 = m √ 2
v 03 g
The direction is perpendicular to the plane of motion. For each of the three instants, the magnitude of angular momentum was different; hence the angular momentum was indeed not conserved. (This indicates that at least one component of angular momentum was not conserved. The torque is due to gravity and is perpendicular to the plane of motion; hence the components of angular momentum in the other directions are conserved.)
Problem 7.8 (Ohanian, page 348, problem 11)
Please see Figure 13.34 on page 348. The only forces acting on the mass m are gravity andthe tension due to the string T. If we let a be the acceleration down then
ma = mg − T (2)
The only forces acting on the mass M are gravity, the force of the support, andthe tension due to the string. All three forces balance to give no translational motion, but only tension creates a torque about the axis of the disk. (The other two forces act through the center of the disk; hence there torque vanishes.) If the string is massless the tension throughout the string from mass m to mass M is constant. If we let α be the angular acceleration clockwise then
Iα = RT (3)
where R is the radius of the disk. If the disk has a uniform density then I is given by Table 12.1 on page 309 as
I =
M R^2 (4)
Finally if we assume the string does not slip against the disk then
a = αR (5)
Substituting Equations (??) and (??) into Equation (??) gives
M R^2 ·
a R
= RT =⇒ T =
aM 2
Substituting this result into Equation (??) gives
ma = mg − · aM 2 =⇒ a = g 1 + 2 Mm
We can check that for M m that a → g.
Problem 7.
Choose the y axis parallel to the rodandthe x axis along the direction of the hit, which is said to be perpendicular to the rod. Let M = 3 kg, l = 50 cm, and d = 15 cm.
(a) The impulse of the hit has magnitude I = 4 kg · m/s anddirection ˆx, so let I = I xˆ;
this is given by
I =
∫ F dt^ =
∫ (^) d P dt
dt = ∆ P
where P is the total momentum of the system. The object is initially at rest, so the total momentum before the hit is zero. The total momentum after the hit, which is relatedto the center of mass velocity, is then
M vCM = P af ter = ∆ P = I
The point C is the center of the rod, which is the center of mass assuming the rod has uniform mass; therefore the translational speedof C is
| vCM | =
|I |
M
I
M
m/s
(b) If we assume that the hit occurs over a short enough time that the roddoes not
move appreciably, i.e. r is constant, then ∫ r × F dt = r ×
∫ F dt^ = r × I
The change in angular momentum is then given by
∆L =
∫ (^) dL dt dt =
∫ τ dt =
∫ r × F dt = r × I
The object is initially at rest, so the total angular momentum before the hit is zero, and the total angular momentum after the hit is
T cos α − F = ma =⇒ F = T cos α − ma (6)
where T is the tension due to the pull, F is the friction, and a is the acceleration in the positive horizontal direction. The torque equation about the center of the yo-yo is
F R 2 − T R 1 = Iα (7)
where α is the angular acceleration. If the pull is “gentle” enough, then the friction force will be able to create rolling without sliding, i.e. the relationship between a and α is
a = R 2 α
We can use this relationship to eliminate α from Equation (??) giving
F R 2 − T R 1 =
I
R 2
a (8)
Now we use Equation (??) to eliminate F from the above equation giving
(T cos α − ma)R 2 − T R 1 =
I
R 2
a =⇒ a =
R 2 (cos α − R R^12 ) I R 2 +^ R^2 M^
· T
Therefore, if cos α > R R^12 then a > 0 andthe yo-yo will roll in the direction of the pull; if cos α < R R^12 then a < 0 andthe yo-yo will roll in the opposite direction of the pull. For “gentle” enough pulls, the yo-yo will not roll at the critical angle cos α^ = R R^12. (Once you pull hardenough the yo-yo will either slide or lift off the floor.)
Problem 7.
We needto be careful because: (1) Angular momentum is NOT conservedand (2) Kinetic energy of rotation is NOT conserved
(a) The frictional force will dissipate some of the kinetic energy.
(b) An external torque is needed; you will sense this in your hands as you push the
wheels together.
(c) The appliedtorque will change the angular momentum.
(d) When the two disks are no longer slipping against one another their circumferential
speeds must be the same. Thus
ω 1 R 1 = ω 2 R 2 and ω 1 ω 2
R 2
R 1
F 2 and F 1 are the frictional forces on disks 2 and 1, respectively. |F 1 | = |F 2 | = F. For a uniform disk the moment of inertia is 12 M R^2. Since the torque equals the moment of inertia times the angular acceleration:
F R 1 =
M 1 R^21 α 1 − F R 2 =
M 2 R^22 α 2
α 1 and α 2 are the angular accelerations of the disks. Disk #1 is spun down; disk #2 is spun up. Notice the vectors F and the radii R are perpendicular to each other.
Hence |α α^12 | = M M^21 RR^21. Since α = dωdt
dω 1 = −
M 2 R 2
M 1 R 1
dω 2
ω 1 − ω = −
M 2 R 2
M 1 R 1
ω 2
Using the relation in Part (d) we have two equations and two unknowns. Solving, we get
|ω 1 | = ω 1 + M M^21
|ω 2 | =
ω R R^12 1 + M M^21
Since the disks are in all respects identical except for their radii, M M^21 =
( (^) R 2 R 1
) 2 andthus
|ω 1 | =
ω 1 +
( (^) R 2 R 1
) 2 |ω 2 | =
ω R R^12 1 +
( (^) R 2 R 1
) 2
Notice that for R 2 = R 1 we obtain ω 2 = ω 1 as one wouldexpect (why?); they both equal to ω 2 (not so obvious!).
