PC235 Winter 2013 Classical Mechanics
Assignment #3 Solutions #1 (5 points) JRT Prob. 3.4 Two hobos, each of mass m mass mh , are standing at one end of a stationary railroad flatcar with frictionless wheels and mass m f c . Either hobo can run to the other end of the flatcar and jump off with the same speed u (relative to the car). (a) Use conserv conservation of momentum momentum to find the speed of the recoiling recoiling car if the two two men run and jump simultaneously. (b) What is it if the second man starts running only after the first has already jumped? Which procedure gives the greater speed to the car? Solution (a) Let v Let v be the speed of recoil of the flatcar, so that u − v is the speed of either hobo relative relative to the ground just after they jump. Conserv Conservation ation of momentum implies that 2m 2mh (u − v ) = m f c v, from which we find v =
2mh u. 2mh + + m mf c
(1)
(b) Let v be the recoil speed of the flatcar after the first hobo jumps and v the speed after the second hobo jumps. jumps. Conserv Conservation ation of momentum after the first jump works just as in part (a), but with only one hobo jumping (and the mass of the second hobo remaining): mh (u − v ) = (mf c + m + mh )v , or ′
′′
′
′
v =
′
mh u. 2mh + m + mf c
(2)
For the second jump, the procedure is the same, except the flatcar is already moving with speed v speed v . Here, conservation of momentum implies that m h (u − + mf c )v , or, v ) − mf c v = −(mh + m ′
′′
′′
′
(mh + m + mf c )v mh u + (m = v = mh + mf c ′
′′
3m + 2m 2m h
fc
2mh + 2m 2 mf c
2mh u. 2mh + mf c
Clearly, v > v (since the first fraction in v is greater than 1). ′′
′′
1
(3)
#2 (5 points) JRT Prob. 3.8 A rocket (initial mass m0 ) needs to use its engines to hover stationary, just above the ground. (a) If it can afford to burn no more than a mass λm 0 of its fuel, for how long can it hover? (b) If v ≈ 3000 m/s and λ ≈ 10%, for how long could the rocket hover just above the earth’s surface? ex
Solution (a) If the rocket is hovering, the thrust must balance gravity; i.e. −mv ˙ = mg. This requires that −dm/m = (g/v )dt, which integrates to give − ln(m/m0 ) = gt/v . The maximum hovering time occurs when we reach the limit m = (1 − λ)m0 (that’s the remaining amount of fuel). Therefore, t = − ln(1 − λ)v /g. ex
ex
ex
max
ex
(b) With the given values of λ = 0.1 and v = 3000 m/s, this gives t seconds. ex
max
= 32
#3 (5 points) JRT Prob. 3.12 To illustrate the use of a multistage rocket, consider the following: (a) A certain rocket carries 60% of its initial mass as fuel (that is, the mass of fuel is 0.6m0 ). What is the rocket’s final speed, accelerating from rest in free space, if it burns all of its fuel in a single stage? Express your answer as a multiple of v . ex
(b) Suppose instead it burns the fuel in two states as follows: In the first stage it burns a mass of 0.3m0 of fuel. It then jettisons the first-stage fuel tank, which has a mass of 0.1m0 , and then burns the remaining 0.3m0 of fuel. Find the final speed in this case, assuming the same value of v throughout, and compare. ex
Solution (a) If it uses all the fuel in a single burn, then v = v ln ex
m 0
0.4m0
= v ln(2.5) = 0.92v . ex
ex
(4)
(b) After the first stage, the speed is v1 = v ln ex
m 0
0.7m0 2
= v ln ex
1 0.7
(5)
and after the second stage it is v = v ln ex
0.6m 0
0.3m0
+ v1 = v ln ex
0.6 1 ·
0.3 0.7
= v ln(2.86) = 1.05v . ex
ex
(6)
#4 (5 points) JRT Prob. 3.22 Use spherical coordinates r, θ,φ to find the CM of a uniform solid hemisphere of radius R, whose flat face lies in the xy plane with its center at the origin. Solution Let the hemisphere’s mass be M and its density be ρ = M/V , where, for a hemi1 rdV sphere, the volume V = 32 πR3 . The CM position is R = M ρrdV = V 1 (because ρ is independent of position) where the integral runs over the volume of the hemisphere. By symmetry, it is clear that the x− and y −components of the CM are both zero. As for the z −component, recall from chapter 1 that in spherical coordinates, z = r cos θ. Furthermore, the differential volume element in spherical coordinates is dV = r 2 sin θ dr dθ dφ. Then,
1 Z = V
3 z dV = 2πR 3
R
π/ 2
2π
r=0
θ =0
(r cos θ)r 2 sin θ dr dθ dφ
φ=0
where the integral over θ has an upper limit of π/2 because we are only integrating over a half-sphere. Evaluating these integrals gives 3 3 R 4 1 · · · 2π = R, Z = 2πR 3 4 2 8 and therefore the CM lies at R = (X , Y , Z ) = 0, 0, 38R .
#5 (10 points) Prove that the moment of inertia of a thin spherical shell of mass M and radius R, rotating about an axis that passes through its center, is equal to I = 2M R2 /3 (hint: use spherical coordinates with R constant ). Solution This problem is deceptively tricky. In spherical coordinates with constant radius R, the differential area is dA = R 2 sin θ dθ dφ. Then, remember that “r2 ” term in the moment of inertia integral doesn’t refer to the variable r, but to the perpendicular distance of the differential area element to the rotation axis . This is a function of θ; to be precise, it is equal to ( R sin θ)2. Noting that the area of the shell is 4πR 2, we 3
can write: I =
π M MR2 2π 2 2 2 (R sin θ) R sin θ dθ dφ = sin3 θ (7) ρa r dA = dφ dθ 2 4πR 4π 0 0 π 2 2 1 2 MR M R 4 = cos3 θ − cos θ = = M R2 . (8) · 2π · · 4π 3 2 3 3 0
You can also solve this problem by considering the spherical shell as a solid sphere with a concentric and slightly smaller solid sphere removed from its center, in the limit as the size of the two spheres approach each other (note that this method is only easier if you happen to recall that I = (2/5)MR2 for a solid sphere of uniform density). Here’s how it works: the density of a solid sphere is ρ = M/V = M/( 43 πR 3) = 3M/4πR 3 . Therefore, we have I =
8 πρR5. 15
(9)
If we remove a smaller sphere of radius R − δ from its center, the new moment of inertia is 8 (10) I = πρ (R5 − (R − δ )5 ), 15 where 3M (11) ρ = 4π(R3 − (R − δ )3) ′
′
(remember, the mass M is given; if we remove a smaller sphere, the density of the remaining shell must increase accordingly). Therefore, we have for the “hollow sphere”
2 R5 − (R − δ )5 2 R5 − R5 1 − = M I = M 3 5 R − (R − δ )3 5 R3 − R3 1 −
δ R δ R
5
3
≈
2 (−5δR 4 ) 2 = MR2 , (12) M 2 5 (−3δR ) 3
where we used a Taylor series (binomial) expansion in the second-to-last step; the approximation becomes exact in the limit of infinitesimally small δ , which represents the spherical shell.
#6 (5 points) JRT Prob. 3.29 A uniform spherical asteroid of radius R0 is spinning with angular velocity ω0 . As the aeons go by, it picks up more matter until its radius is R. Assuming that its density remains the same and that the additional matter was originally at rest relative to the asteroid (on average), find the asteroid’s new angular velocity. What is the final angular velocity if the radius doubles?
4
Solution Since the matter accreted by the asteroid is initially at rest, it provides no change to the asteroid’s angular momentum. Therefore, I ω = I 0 ω0 , where I 0 and I are the initial and final moments of inertia. Now, if the density of the asteroid is ρ, we have 2 2 I = M R2 = 5 5
4 3
πρR3 R2 =
8 πρR5 . 15
(13)
Since ρ is a constant, conservation of angular momentum implies that R 05 ω0 = R 5 ω, and therefore the new angular velocity of the asteroid is ω = (R0 /R)5 ω0 . If the radius doubles, then we have ω = ω 0 /32.
#7 (5 points) JRT Prob. 3.34 A juggler is juggling a uniform rod one end of which is coated in tar and burning. He is holding the rod by the opposite end and throws it up such that, at the moment of release, it is horizontal, its CM is traveling vertically up at speed v0 and it is rotating with angular velocity ω0 . To catch it, he wants to arrange that when it returns to his hand it will have made an integer number of complete rotations. What should v0 be if the rod is to have made exactly n rotations when it returns to his hand? Solution The CM moves just like a point mass M . We know from previous classes that an object under the influence of gravity and an initial velocity v0 has a height Y = v0 t − 21 gt2 . Setting Y = 0 gives us the time taken for the object to travel up and then down to its inital position: t = 2v0 /g. Now, we must realize that after the rod is released, there can not be any torque around the center of mass, and thus the angular momentum L = Iω is constant. This means that ω = ω0 is constant, and the number of revolutions in the time t is n = t/T , where T = 2π/ω0 is the period of one rotation of the rod. Thus, n = ω 0 t/2π = ω 0 v0/πg. Therefore, he must arrange that nπg v0 = , ω0 where n is an integer.
#8 (10 points) JRT Prob. 4.4 A particle of mass m is moving on a frictionless, horizontal table and is attached to a massless string, whose other end passes through a hole in the table, where I am holding it. Initially, the particle is moving in a circle of radius r0 with angular velocity ω 0, but now I pull the string down through the hole until a length r remains between the hole and the particle. 5
(a) What is the angular velocity now? (b) Assuming that I pull the string so slowly that we can approximate the particle’s path by a circle of slowly shrinking radius, calculate the work I did by pulling the string. (c) Compare this with the particle’s gain in kinetic energy. Solution (a) Conservation of angular momentum implies that mr2 ω = mr02 ω0 ,
(14)
ω = (r0 /r)2 ω0 .
(15)
so therefore, (b) We will need to solve Newton’s second law, F = m a. From the chapter 1 notes, ˙ 2 , you will recall that the 2D polar components of acceleration are a r = r¨ − rφ ¨ 2r˙ ˙φ. As the string is pulled, the path of the mass spirals and aφ = r φ + toward the hole in the middle of the table; the radius and angular velocity both change. This is where we must assume that I pull the string slowly ...if the radius of the mass’s orbit changes very slowly, then the term ¨r in ar , and both of the terms in a φ can be neglected in comparison to the term −r ˙φ2 in a r ¨ as you can determine (very slow changes in r will cause very small values of φ, by conservation of angular momentum.) Thus, the only force on the mass is the centripetal force F r = mar , which is balanced by the string’s tension to keep the mass on a (roughly) circular path. From the last assignment, you know that the new angular velocity (calculated by conservation of angular momentum) is ω = (r0 /r)2 ω0 . Thus, the force that I exert is 2 r0 2 mω02r04 2 (16) F (r) = mar = mω r = m ω0 r = . r r3 To calculate the work done, remember that the distance I pull the string in any small displacement is −dr. Then, r r 1 1 1 dr 2 4 2 4 = − 2 . (17) W = F (r )(−dr ) = −mω0 r0 mω r 0 0 3 2 r2 r0 r r r
′
′
0
′
′
0
(c) Recalling that for circular motion, velocity v = rω, we see that the KE for a mass m undergoing circular motion is T = 21 mv2 = 21 mr2ω 2 . The change in KE is then 1 1 1 1 ∆T = m(r2 ω 2 − r02 ω02 ) = mω02 r04 2 − 2 . (18) 2 2 r r0 This is identical to the result of part (b), as we expect from the work-KE theorem.
6
#9 (10 points) JRT Prob. 4.24 An infinitely long, uniform rod of mass µ per unit length is situated on the z axis. (a) Calculate the gravitational force F on a point mass m a distance ρ from the z axis, given that the gravitational force between two point masses is F = −GMmˆ r/r2 . (b) Rewrite F in terms of the rectangular coordinates (x,y,z ) of the point and verify that ∇ × F = 0. (c) Show that ∇ × F = 0 using the appropriate expression in cylindrical coordinates from inside the back cover of the text. (d) Find the corresponding potential energy U . Solution (a) Consider first the differential force on m due to a short segment dz of the rod at a height z above m. This force has magnitude dF = Gmµdz/r2 in the direction shown in the side view (see figures below,) where r is the distance from the element dz to the mass m, and µdz = dm is the differential mass of the segment dz . To find the total force, we must integrate over all of these differential forces from z = −∞ to ∞. When we do this, the z components f z from points z and −z will cancel (every “upward” pull is balanced by a corresponding “downward” pull.) In addition, the component of force perpendicular to the page (in the side view) is clearly zero. Thus, we need only consider the component of force in the direction of ρˆ (the unit vector in the ρ direction, pointing away from the z axis): dz dz (19) dF ρ = −GMµ cos α 2 = −Gmµρ 3 , r r where the last step follows because cos α = ρ/r. The total force is then ∞
F ρ =
−Gmµρ
−∞
dz = −Gmµρ r3
∞
−∞
dz (z 2 + ρ2 )3/2
=
−Gmµ
ρ
π/ 2
cos α dα, π/ 2
−
(20) where we have made the substitution z = ρ tan α. The integral just evaluates to 2, and we conclude that F =
−
2Gmµ ρ ˆ. ρ
(21)
(b) The unit vector ρˆ lies in the xy plane. If we denote it as in the figure, then ρ ˆ =
x + sin φˆ y, cos φˆ
7
(22)
where cos φ = x/ρ and sin φ = y/ρ. Therefore, the answer to part (a) reads F =
−
2Gmµ x + yˆ y + 0ˆ z) . (xˆ ρ2
It is then fairly trivial to show that ∇×
F =
ˆ x
0 ∂ ∂
x
ˆ y
ˆ z
∂ ∂x
∂ ∂y
∂ ∂z
y
∇ × F =
∂ ˆ = x (0) − (y) ∂y ∂z = 0.
ˆ −y
∂x
(23)
0 using the determinant method: (24)
(0) −
∂ ∂ ∂ y (z ) + ˆ (x) − (y) (25) ∂z ∂y ∂z (26)
(c) In cylindrical coordinates, it’s trivial to show that ∇ × F = 0, since F only has a ρˆ component, and it is independent of φ or z , thus making all of the partial derivatives in the curl equal to zero.
(d) The potential energy U (r) is given by the integral − F · dr taken from any arbitrary reference point r0 to the point of interest r. Since the integral is independent of path, we can choose a path which is the most geometrically “convenient.” Label the reference point with the coordinates r0 = (ρ0 , φ0 , z 0 ) and the point of interest r = (ρ,φ,z ). To be completely general, we will consider this path in three stages: 1. Start at r0 and move parallel to the z axis until you reach the desired final value z . 2. Next, move in a circlular arc of constant ρ and z until you reach the desired final value φ. 3. Finally, move radially to the desired final value ρ. Now, recall from part (a) that the force only has a radial component F ρ . In the first two stages of the path, dr has only a z component and a φ component, respectively. These are both perpendicular to the force, and thus they do no work. For the third stage of the path, both F and dr point in the ρ direction, and the work integral is easily written as an integral over the single coordinate ρ: ρ
U (r) = −
2Gmµ ρ0
ρ
′
′
dρ = 2Gmµ ln(ρ/ρ0 ).
(27)
#10 (5 BONUS points) In the class notes, we solved JRT Prob. 3.33 to find I for a uniform thin 8
Fig. 1: Geometry for problem #9
square of side 2b about an axis that is perpendicular to the square and which passes through its center. Now, calculate I for the same square, for an axis that is perpendicular to the square, but which passes through the point (x, y) = (x0 , y0 ). Note that this point does not need to lie within the boundary of the square. Solution The only difference between this problem and the one covered in the notes is that here the position of each differential bit of mass relative to the axis of rotation isn’t r = x2 + y 2, it’s r = (x − x0 )2 + (y − y0 )2. Here is a rigorous solution:
M M (x x ) + (y y ) dxdy r dA = 4b 4b x + x 2xx + y + y 2yy dxdy M 4b x M + xx 2xyy dy x x + xy + xy 4b 3 2b M + 2bx + 2by + 2by 4byy dy 4b 3 2by 2by y M 2b y + 2bx y + + 2by y 4b 3 3 3 4b M 4b + 4b x + + 4b y 4b 3 3 2b b
I =
2
b b
2
2
−
0
2 0
2
−
2 0
0
−
0
(29)
b
3
0
b b
−
2
2
2 0
0
−
0
(30)
b
−
3
2
2
(28)
−
−
=
−b
b
−
=
−b
2
0
b
2
2
−
2
b
=
b
2
2 0
0
−
0
(31)
b
−
3
=
3
2
b
2
2 0
2 0
0
−
(32)
b
−
4
=
4
2
2
2 0
2
2 0
(33)
2
= M
3
+ x20 + y02
(34)
2 (35) M b2 + M (x20 + y02 ). 3 This is a bit ugly, but you can also solve it simply by invoking the parallel axis theorem (which you may or may not have learned in an earlier math course). This =
9
theorem tells us how the moment of inertia changes when the rotation axis shifts away from the center of mass...if the shift has a magnitude of r, then I = I CM + M r 2, which gives us the same answer.
10
(36)
