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Fowles & Cassidy 8.1 -

Recall

We've applied Newton's laws to systems of particles and found that the motion can be broken down in a top down fashion into the motion of the CM, collective motion about the CM (rotation), and finally internal motion of the particles relative to the CM.

Variable Mass Motion: Rockets

Consider a rocket lifting off from earth. The only external force on the rocket is gravity, which tends to hold the rocket on the ground. How does a rocket lift off and move hundreds or thousands of kilometers from earth?

Traditional rockets move by ejecting a part of their mass at high speed opposite the desired direction of motion (really force, but motion sounds better).

We want to come up with a form of Newton's law that takes into account a changing mass. Consider two objects initially moving separately, at different velocities, but then combining into one a short time later (this is the time reversed picture of a rocket, as done in the text). Initially the momentum is P_t = mv + (v + V)Dm where V is the relative velocity of the small mass relative to the primary object. After the two objects combine they are one object moving at a new velocity P_t+Dt = (m+Dm) (v+Dv). The change in momentum is

DP = P_t+Dt - P_t = (m+Dm) (v+Dv) - mv - (v+V)Dm = -VDm + (m+Dm)Dv = F_extDt

Dividing through by Dt and taking the limit as Dt goes to zero yields:

F_ext = m dv/dt - V dm/dt

This is the desired equation that tells us how to evaluate the motion of rockets or train cars being filled with sand.

Example: Train Cars and Fog

First, let's apply this result to an object moving through fog, collecting mass as it proceeds. Since the fog is at rest relative to the object, the relative velocity of the fog is V = -v where v is the velocity of the object. Our equation of motion becomes:

F_ext = m dv/dt + v dm/dt = d/dt (mv)

Typical questions are then to find the force required to keep a filling train car moving with constant velocity, or what is the speed as a function of time for an object moving through fog with no external forces. I'll leave these for homework problems.

Example: Rockets

Of course, the "coolest" application is the motion of rockets. To keep this example simple, let's look at the motion of a rocket with no external forces, such as gravity. In this case, V can be considered constant, independent of v. The equation of motion is:

0 = m dv/dt - V dm/dt

This equation can be solved by separation of variables. The variables are m and v.

dv/dt = (V/m)dm/dt

Integrating both sides gives us:

v - v₀ = V ln(m₀/m)

where v₀ is the initial velocity and m₀ is the initial mass. This result means that the velocity increases as the logarithm of the initial to the final mass. For instance, if the rocket exhaust is ejected at V = 10³ m/s and 80% of the mass is expelled as exhaust, then the velocity changes by only 10³ ln(5) = 2.5×10³ m/s. That's 80% of the mass to increase the velocity by 2.5 times the velocity of the rocket exhaust. One can of course try to increase the velocity of the exhaust, but there are limits. The ultimate velocity difference is the speed of light. Ion propulsion accelerates charged ions to nearly the speed of light then expels them. It takes V to its ultimate limit, but the rate for expelling ions is rather low. More familiar solutions include rocket stages or boosters that are discarded once the fuel inside is expended.

Rigid Bodies: Planar Motion

A rigid body is a system of particles where the relative locations don't change. More commonly, we describe such bodies as solid, stiff, and inflexible. Firstly, note that a rigid body is a system of particles, so all the results from the last chapter apply. However, it is often impracticle to calculate the infinite sums in the results. Let's look again at how to transform the infinite sums from the last chapter into integrals.

r_CM = (1/M) S_i m_ir_i ==> (1/M)ò dm r

where we now need to relate m and r in order to perform the integral. The most common situation is that the object has a uniform mass density. The mass density is commonly expressed either as a volume density, surface density, or linear density.

Example: Hemispherical Shell

Let's calculate the CM of a hemispherical shell with a uniform surface mass density s and radius a. First, what's the total mass of the shell?

m = s dA = 2psa²

Next we set up the coordinates such that z is along the axis of the shell and x and y lie in the plane that contains the edge of the shell. With these coordinates, we can immediately say that x_cm and y_cm are zero because of the symmetry of the shell. We still need to calculate z_cm -- there's a lesson here, don't calculate if you can arrive at the answer by a simple route. Getting back to the calculation of z_cm, I note that instead of integrating small surface elements over the surface that we can take rings at an angle q from the z axis and sum up the rings to form a hemisphere. The mass in a ring is 2psradq where r is the radius of the ring and dq is the width. The radius of a ring at height z is r = \sqrt{a² - z²} and dq = dz/r

z_cm = 2psa/m ò₀^azdz = psa³/m = psa³/2psa² = a/2

So, the CM of a hemispherical shell is at a point along the axis of symmetry and half a radius from the center.

Example: Hemisphere

Deviating from the text, I'll find the CM of a solid hemisphere of uniform mass density by summing the solutions for the hemispherical shell. If we have an object composed of a number of components, we can find the CM by summing over the CM's of the components. Note that this technique can be used to handle an object with a "hole" in it; simply treat the hole as a component with negative mass!

Now back to the hemisphere. We can find the CM by adding together the weighted sums of the CM's of an infinite number of hemispherical shells of increasing radius from 0 to a. Since x_cm and y_cm are zero for all the shells, they will also be zero for the hemisphere. For z_cm we will need to sum over an infinite number of shells. In other words, we'll be integrating over all the shells. The hemisphere has a uniform volume mass density r. This is related to the surface mass density s of the shells by s=rdr. But first, the mass of a hemisphere or radius a can be written down immediately remembering the formula for the volume of a sphere and taking half and multiplying by r:

m = 2/3 pra³

z_cm = (1/m)S_im_iz_cmi = (1/m)S_i2psr_i² r_i/2 = pr/m ò₀^ar³dr = pra⁴/4m= pra⁴/(8/3) pra³ = 3a/8

© 28 Feb 1999 R. Harr