Physics 1116 Fall 2011 Homework 5 Due: Friday, Sep 30

Reading: KK Ch. 3 & Note 3.1

1. Calculate center-of-mass.

a. A catenary, y(x) = cosh x, describes a massive cable strung between x = ±x0. Thecatenary has uniform linear mass density, λ(s) = λ0. Where is the center of mass?

b. A disk of radius R is centered on the origin of the xy plane. The disk has a uniformmass density, σ(x, y) = σ0, except that a circular region of radius R/2, centered at(x, y) = (−R/2, 0) has been completely removed. Where is the center of mass?

Solution:

a. The center of mass is given by rcm = 1M

∫rdm, where M =

∫dm. Since we know

the configuration of the chain as a function of x, these integrals will have to bein x. Thus we’ll need to convert the dm, as follows:

dm =dm

dxdx

=dm

ds

ds

dxdx

= λ0ds

dxdx

= λ0

√1 + y′(x)2 dx

Note that ds2 = dx2 + dy2 so ds/dx =√

1 + y′(x)2 =√

1 + sinh2(x) = coshx.

The last step follows from the hyperbolic trig identity, cosh2 x − sinh2 x = 1.(http://en.wikipedia.org/wiki/Hyperbolic function)

Therefore, the x and y centers of mass are given as follows.

xcm = λ01

M

∫ x0

−x0

x coshxdx = 0

This integral is zero because the integrand is an odd function (x) times an evenfunction (coshx), over a symmetric interval.

ycm = λ01

M

∫ x0

−x0

y coshx dx = λ01

M

∫ x0

−x0

cosh2 x dx =λ0

M(x0 + sinhx0 coshx0)

The integral may be done by hand relatively easily if you first convert coshx =(ex + e−x)/2. Otherwise, in Mathematica (or at wolframalpha.com) you couldrequest: Integrate[Cosh[x]∧2, {x,−x0, x0}].

Now we just need to get the total mass:

M =

∫ x0

−x0

dm

=

∫ x0

−x0

λ0

√1 + sinh2(x) dx

= 2λ0 sinhx0.

Putting this in for M , we find

ycm =1

2

[y0 +

x0

sinhx0

].

where y0 ≡ coshx0.

It’s interesting to check to limiting cases: x0 → very small, and x0 → very large.The results should make sense. (They do.)

b. The peculiar shape described here – a disk with a disk removed from within it –can be easily handled by treating the missing disk as the sum of two disks, onewith mass density σ0, and one with mass density −σ0. Obviously this is just amathematical trick. It works because the formula for the center of mass is linearin the density, so we can treat zero density as σ = σ0−σ0 = 0, and then divide theproblem into two pieces, with with mass density σ0, and one with mass density−σ0.

Here goes. Let region C denote the large crescent region, which is basically abig disk with the smaller disk cut out, and let region d denote the region of themissing smaller disk. Let’s also let D denote the full region of the large disk(with the hole filled in). D is the union of regions C and d. If we define massescorresponding to each of these, we would have MD = MC +md.

MC rcm =

∫C

r dm

=

∫C

rσ0 dA+

∫d

r 0 dA (note the zero density in the 2nd integral)

=

∫C

rσ0 dA+

∫d

r(σ0 − σ0) dA

=

[∫C

rσ0 dA+

∫d

rσ0 dA

]−∫d

rσ0 dA

But the term in the square brackets is just MDXcmD = (0, 0). And the second

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term is md rcmd = md(−R2, 0). Therefore

xcmC =MDxcmD −mdxcmd

MC

=−md(−R

2)

MD −md

=R

2

[σ0πR

2/4

σ0πR2 − σ0πR2/4

]=

R

6.

2. Exploding projectile. KK 3.4

Solution:

Let the mass of the projectile be M+m, where the large mass M = 3m will fly forward,and the smaller mass, m, will fly backward and wind up at the launch site.

When the thing was launched, it had an initial velocity; let the x-component of thisbe called v0x. For convenience, let’s assume v0x > 0. The initial horizontal momentumcomponent is therefore p0x = (M + m)v0x, and since there are no forces acting in thehorizontal direction, this momentum must be conserved.

After the explosion, m travels backward with some velocity v1 < 0, and M travelsforward with some other velocity v > 0. Conservation of horizontal momentum tells us

(m+M)v0x = mv1 +Mv.

Now we also know the m lands right on the original launch point, so it has reversedits trajectory exactly. This implies v1 = −v0x. Plugging this information into themomentum equation gives us

v =

(1 +

2m

M

)v0x.

But how far does M go? Well, the time it takes to return to the ground will be equalto the time it took to rise to the peak of the trajectory, which had to be the time totravel from x = 0 to x = L, which is T = L/v0x. So in addition to having travelled adistance L horizontally before the explosion, the large mass now travels an additionaldistance vT , meaning it will strike the ground a distance L′ = L+ vT from the launchpoint. Combining above equations and now using M = 3m, we get

L′ = L+

(1 +

2m

M

)L =

8

3L.

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3. Springy blocks. KK 3.7

Solution:

Let x be measured from the equilibrium point of the spring. The spring is compresseduntil x = −`/2, and then released. m2 accelerates rightward. Up until it reaches x = 0,it is being pushed by the force of the compressed spring; and meanwhile m2 is stillmotionless, pressed against the wall at the left.

At the point when m2 is at x = 0, however, the spring is neither stretched nor com-pressed, so there is no force (on either mass). At this instant, the total momentum ofthe system is just m2v, where v is a velocity we’ll have to calculate. Just after that, thespring goes into extension, and m1 is pulled away from the wall; now there are no moreexternal forces acting on the m1,m2 system, and the total momentum is now fixed at(m1 + m2)V = m2v. Since this is a conserved quantity, we know that V is a constanthenceforth, and V = (m2/(m1 + m2))v. Naturally there will now be a lot of “inter-nal” motion as the two masses bounce towards and away from each other, alternatelyattracting and repelling each other; but the C.M. moves along at a constant, statelypace.

Now we only need to find v. The simplest way is to use conservation of kinetic +potential energy, in which case 1

2m2v

2 = 12k(`/2)2. The other way is to recall the

general solution for motion of a harmonic oscillator, x(t) = A sinωt + B cosωt, whereω =

√k/m2. To satisfy the initial condition that x(0) = −`/2 and x(0) = 0, we

must have A = 0 and B = −`/2. Therefore the motion of m2 after release is x(t) =−(`/2) cosωt. The moment at which it just reaches x = 0 must correspond thereforeto ωt = π/2. Then v ≡ x(t = π/2ω) = ω`/2, which is the same as found above by theenergy method.

So putting it all together:

V =

(m2

m1 +m2

)v

=

(m2

m1 +m2

)`

2

√k

m2

4. Continuous flow. KK 3.16

Solution:

F = v0dmdt

, and dmdt

= π4D2ρv0 so F = π

4D2ρv2

0.

5. Rocket Science. KK 3.20

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Solution:

There are two confusing issues with this problem. The first is that while the statementof the problem would lead you to write the drag force as f = −bv, in this form it is notsolvable. You need to set f = −Mbv to get a differential equation that can be solved(easily, at least).

The second issue concerns mass. One might think that the rocket mass at launchtime consists of two parts, M0, the structural mass of the rocket itself, and m0, themass of the fuel payload. Then after a time t, when a total mass m(t) has beenexpelled by the rocket engines, the total mass would be M(t) = M0 +m0 −m(t). Thisreasonable, but in order to have a solvable problem, and to get the “answer clue” theyprovide, you have to impose M0=0. This means that M(t) = m0−m(t), or equivalentlyM+m = m0 = const. In this form the problem is solvable, but a bit strange: the rocketis basically just a big bag of flying fuel and nothing more. Oh well. We only have tosolve the equations, we don’t have to go for a ride in this thing.

From these statements about mass, we see that dM/dt = −dm/dt. The problem alsosays “dm/dt = γm, where m is the instantenous mass of the rocket...”. This is aslightly confusing statement, but it appears to make sense if “m” on the left side refersto the expelled mass, and “m” on the right refers to the total remaining mass. Thusdm/dt = γM . Then dM/dt = −γM , which says that the rocket (ie fuel) mass willdecrease exponentially, M(t) = M(0)e−γt = m0e

−γt.

Now we can apply the rocket equation, KK 3.18, p137, setting the left side to the knownphysical agents of force, gravity and drag:

−Mgr−Mbvr = M rdv

dt− ur

dm

dt.

Notice that here, by writing u = −ur I am keeping the natural idea that |~u| ≡ ushould be a positive number, but explicitly acknowledging that the direction of themass emission by the rocket engines is opposite (−r) to the direction of the rocket(+r). This slightly confusing matter of the direction associated to ~u is addressed at thebottom of p137.

Substituting dm/dt = γM and gathering terms, we get

−Mg −Mbv = Mdv

dt− uγM. (1)

At this point the time-dependent factors of M cancel out, and we have

(γu− g)− bv =dv

dt.

This is a differential equation of the general form dv/dt = A − Bv which is solved byseparation of variables to give v = (A/B)(1 − e−Bt) + v0e

−Bt. In this case, where theinitial velocity is zero, the second term vanishes and we have:

v =γu− gb

(1− e−γt)

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6. A Classic Problem. A dog of mass m = 10kg is standing at one end of a raft whichis floating in a lake; the dog is 10m from the shore. The raft can glide in the waterwithout any friction. The raft has a mass M = 40kg and length L = 7m long; themass is uniformly distributed. The dog walks 5m on the raft towards shore and thensits down. How far is he from shore?

Solution:

The central idea is that the dog and raft form an isolated system so their Center of Masscannot move, even though they may internally rearrange their positions with respectto each other.

Set up a shore-based coordinate system. See Fig. 1.

The overall center of mass is given below in the “before” coordinates (x0, X0), and the“after” coordinates (x1, X1). Lower case variables refer to the dog, upper case to theraft. All x and X coordinates are measured from the shore. The goal of the problemis to find x1.

We are given m, M , and x0, and one more quantity ∆x = 5m which is the distancethe dog walks on the raft. Note that ∆x is measured in coordinates fixed to the raft.

The fact that the CM is unchanged in going from the initial to the final state gives usthis equation:

mx0 +MX0

m+M=mx1 +MX1

m+M. (2)

In coordinates measured on the raft, the dog has moved from x0 −X0 to x1 −X1, so

∆x = (x0 −X0)− (x1 −X1)

= (x0 − x1) + (X1 −X0) (3)

We can use Eqn (2) to eliminate X1 −X0 from Eq (1), and this gives the answer,

x1 = x0 −(

M

m+M

)∆x = 6m.

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Figure 1: Dog (m) on raft (M). Above: initial configuration; below: final configuration.Note the vertical dotted line indicating the fixed position of the center of mass of the wholesystem.

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