Physics Lecture 11 - Momentum Conservation and Center-of-mass

Momentum Conservation Continued

There are many examples of "difficult" problems we can solve with momentum conservation that are impractical or impossible to satisfy with the typical approach of F = ma. Examples of the impractical would be instances of a bullet hitting a block on a frictionless surface. The problem in that case is to find the final velocity of the bullet and block in the case where the bullet lodges inside the block. In this case we could use Newton's Second Law to calculate the acceleration of the block and bullet at each instant, but we would need a lot of detailed information on the interaction of the bullet and block, so much in fact that it is highly impractical to consider it. However, as we saw, momentum conservation makes quick work of such a problem precisely because it does not care about the details. Choosing the system so that there are no external forces in the horizontal plane means that the momentum of the system stays the same no matter how the interaction proceeds. That is the power of the momentum approach. A case where F = ma is impossible to use in solving the problem would be calculating the final velocity of a rocket that starts with a certain mass of fuel which it burns and ejects to increase the rocket velocity. Here, the mass is not constant so we do not have the mathematics in hand to handle this case. Again, as the book describes, momentum conservation can be used to give an involved, but still accessible solution.

We can now consider some problems that are tougher to handle. Usually, the toughest part of such problems is choosing a good reference frame for determining the momenta of the particles involved. Relative velocities are usually involved when objects are splitting apart so we must be careful. Consider the following:

Example 1:
A cannon, initially at rest with respect to the earth, has a mass of 1950 kg and fires a shell of mass 50 kg. If the shell is fired in the horizontal direction with a speed of 40 m/s with respect to the cannon, find the recoil velocity of the cannon assuming that it is free to move without friction in the horizontal direction.
Solution: The earth makes a good reference frame since the gravitational force and normal forces applied by the ground on the cannon operate only in the vertical direction and we are concerned only with motion in the horizontal direction. Therefore, the system of cannon and cannonball has no external forces to affect the horizontal motion and momentum in the horizontal plane is conserved. We need the velocities of both the cannon and cannonball in the horizontal. Let's call the final velocity of the cannon in this reference frame V. The problem asks us to find V. The velocity of the cannonball in this frame is v_E, the velocity relative to the earth. Since we can assume that V points to the left in the figure above, we should have v_E = v - V since v = 55 m is relative to the cannon and the cannon is moving backward.
The initial momentum of the system before the cannon fires is zero. The final momentum is mv_E - MV, so we need to solve for V as follows:

p₀ = p_f

0 = mv_E - MV

0 = m(v - V) - MV ==>

0 = mv - (m + M)V ==>

V = (mv)/(m + M)

= (50 kg)(40 m/s)/(50 kg + 1950 kg)

V = 1 m/s

Suppose in the above example, a second shell is fired after the first one. What is the final velocity of the cannon relative to the earth?

Solution: Things are complicated slightly by the second shell. We begin by noting that the initial mass of the cannon + two cannon shells is now M + 2m and the mass of what remains in the cannon plus the cannon after the first shot is fired is M + m. So, we calculate the final velocity of the cannon and one remaining cannonball after the first shot is fired:

p₀₁	= p_f1
0	= mv_E1 - (m + M)V₁
0	= m(v - V₁) - (m + M)V₁ ==>
0	= mv - (2m + M)V₁ ==>
V₁	= mv/ (2m + M)
	= (50 kg)(40 m/s)/[2(50 kg) + 1950 kg]
V₁	= 0.98 m/s

The shot of the second cannonball is still 40 m/s relative to the cannon, but now the cannon is moving at about 1 m/s. We go through the same procedure of momentum conservation, being very careful to correctly account for signs in a consistent manner.

p₀₂	= p_f2
-(m + M)V₁	= mv_E2 - MV₂
	= m(v - V₂) - MV₂ ==>
	= mv - (m + M)V₂ ==>
V₂	= [mv + (m + M)V₁]/(m + M)
	= [(50 kg)(40 m/s) + (50 kg + 1950 kg)(0.98 m/s)/ (50 kg + 1950 kg)
V₂	= 1.98 m/s

We can see the advantage of shooting one shell after another if our purpose is to increase the speed of the cannon! Although this is not desirable for artillery firing, it is the same principle behind using multi-stage rockets to lift payloads into orbit.

Example 2:
A cannonshell is fired with an initial velocity of 50 m/s at 37° to the horizontal. When the shell is at the highest point of its trajectory, it explodes into two pieces, m₁ and m₂. m₁ falls straight down from the highest point after the explosion. If m₁ = 2m₂, where does m₂ land relative to the initial firing point?

Solution: This problem can be solved quickly by taking into account information we already know about free-fall trajectories. First, the highest point on the trajectory occurs when the shell is a distance ½ R from the firing point, R being the range and equal to 2v₀²sin(theta)cos(theta)/g. We also know how to calculate the height of this highest point. Second, the velocity of the shell when it explodes is only along the horizontal since, by definition, the shell has no vertical velocity at the highest point. That means that if we consider m₁ and m₂ to be our system, then we only need worry about momentum conservation along the horizontal direction to find the speed of m₂ just after the explosion. Since we can calculate the velocity of m₂ at a position we know, we can then use the equations for motion to find the final landing point. To get the velocity of m₂ just after the explosion, we note that momentum conservation does not care about the complexity of the explosion itself. We can relate what happens just before the explosion to what happens after as long as no external forces in the horizontal direction are acting and we account for the momentum of every piece of matter just after the explosion.

p₀ = p_f

(m₁ + m₂)v_x = m₁*0 + m₂v₂ ==>

(m₁ + m₂)v₀cos(theta)/m₂ = v₂

(2m₂ + m₂)v₀cos(theta)/m₂ = v₂

3v₀cos(theta) = v₂

Thus mass m₂ flies off with 3 times the x velocity of the original intact shell. We could now go through the equations of motion to find the horizontal distance m₂ travels from the highest point, but we can do things faster by noting the symmetry of the situation. Normally, the whole shell would have traveled a horizontal distance of ½ R from the midpoint because it takes just as much time to go down to ground level as it did to go up to the highest point from ground level and the horizontal velocity would have been constant. The explosion imparts m₂ with a new x velocity that is 3 times the old x velocity, but does not change its y velocity. Hence, m₂ will travel 3 times as far from the midpoint as the intact shell would have since it will still take the same amount of time to fall to the ground from the highest point. Therefore, the horizontal distance m₂ travels from the initial firing position is ½ R + 3*½ R = 2R = 4*v₀²sin(theta)cos(theta)/g = 4(50 m/s)²sin37° cos37° /(9.8 m/s²) = 490 m.

To see an interactive demonstration of 1D momentum conservation in action, turn to the interactive textbook by clicking here.

Center-of-mass

We have treated all the above problems as if the objects involved were point particles. Strictly speaking, as defined so far, Newton's Laws apply only to objects which have no spatial extension, i.e. they are mathematical points in space. In fact, the utility of Newton's Laws for these problems is predicated on the notion that the behavior of real bodies in space is approximated, at least for translational motion (we discuss rotation later), by the motion of a hypothetical point that represents the position of the "mass" of the body. Although the mass is distributed throughout any typical object, there does always exist one point on the body for which Newton's Second Law is valid for describing and predicting the motion. This point is called the center-of-mass. For extended bodies, Newton's Second Law becomes F_{net, ext} = dP/dt where P is the momentum of the center-of-mass of the body or system. If the mass of the body or system is not changing, then F_{net, ext} = Ma_cm
To find the center-of-mass for a system of point particles that comprise an extended body, we can calculate the net force on the system by looking at the net forces acting on each particle in the system. If there are N particles in all, then F_{net, ext} = m₁a₁ + m₂a₂ + .. + m_Na_N
If we define the center-of-mass position to be the mass-weighted average of the positions of the particles r_cm = m₁r₁ + m₂r₂ + ... + mr_N/M
where M is the mass of all the particles. Now we note that, for any particle, say particle i, we have a_i = d²/dt²(r_i) and, coincidentally, v_i = d/dt(r_i). Therefore, the net momentum of the system is

P_net	= m₁v₁ + m₂v₂ + ... + m_Nv_N
	= d/dt(m₁r₁) + d/dt(m₂r₂) + ... + d/dt(m_N r_N
	= d/dt(m₁r₁ + m₂r₂ + ... + m_Nr_N)
	= M d/dt[m₁r₁ + m₂r₂ + ... + m_Nr_N)/M]
	= M d/dt[r_CM]
P_net	= d/dt (Mr_CM) = dP_cm/dt

So the momentum of the center-of-mass is identical to the momentum of the whole system. That means that Newton's Second Law (in momentum terms) can be used to determine the change in motion of an entire system of particles by considering the change in motion of just one point, as long as that point is the center-of-mass as defined above.

Our expression for the net force acting on the system can be rewritten (if the mass is not changing) as

F_{net, ext}	= d²/dt²( m₁r₁ + m₂r₂ + ... + m_Nr_N)
	= M d²/dt² ( m₁r₁ + m₂r₂ + ... + m_Nr_N)/M
F_{net, ext}	= M d²/dt² (r_cm) = Ma_cm

That complets the proof of the assertion made earlier that Newton's Second Law corresponds to the motion of the center-of-mass of a system of particles.

An immediate application of this concept is the problem we did earlier about the exploding shell. The center-of-mass motion does not change because of the explosion since all the forces involved are internal to the shell itself. In that case, the above problem becomes even easier since the center-of-mass follows the same trajectory path as it would have in the absence of the explosion, i.e. The center-of-mass winds up at position R with respect to the original firing point. We know that m₁ winds up at ½ R, so we can find the position for m₂ by demanding that its final position be such as to give a center-of-mass position at R, so

x_cm	= (m₁x₁ + m₂x₂)/ (m₁ + m₂)
R	= (2m₂½ R + m₂x₂)/ (2m₂ + m₂) ==>
3m₂R	= m₂(R + x₂) ==>
x₂	= 2R

So we get the identical result, as we must.

Calculating Center of Mass for Continuous Mass Distributions

Finding the location of the center of mass for a real 3-dimensional object can be quite tricky. In this case, we have to resort to integral calculus if we wish to calculate the position. The technique makes logical sense: we imagine breaking the object into an infinite number of infinitesimal ``mass elements''. Each mass element is a point so we can know its position. We call its infinitesimal mass dm. Then we can find the x position of the center of mass of the object, for example, by integrating the product of x dm where x is the x position of a given mass element. The integral sums up the contribution to the center of mass position of each mass element. We can try this for something relatively simple where we know what the answer should be and then extend it to something more complex.

Example 3:
Find the center of mass for a uniform rectangular slab of mass M, length b, and height h assuming the origin is at the lower left-hand corner of the slab.
Solution: It should be clear by symmetry that the CM position will turn out to be at the center of the slab, hence we should find x_CM = b/2 and y_CM = h/2. How do we prove it? Let's consider just the x position for now. We can consider breaking the slab into small strips as shown below. For each strip, we posit that the CM of the strip lies at the strip center (by symmetry and the uniformity of how mass is distributed). Clearly if we make enough strips so that the strips are infinitesimally thin, we can believe ala calculus that the center of the "line" strip is where the CM for that strip is located. For now, let's do 10 strips and then see what happens as we increase that number.

We look at each strip and see that it has width b/10, so if the CM position of each strip is at the center, then the mass of each strip can be considered as being at positions ½ b/10, 3/2 * b/10, 5/2 * b/10 , etc. The general formula would be (2i - 1)b/(2*10) as i goes from 1 to 10. The mass of each strip should be 1/10 of the total mass. If we call the total mass of the slab M, then the x position of the CM for the whole slab is defined as being located at

Looks awesomely hard, but luckily Maple is really good for handling these kinds of things. Give it a try by clicking here. There are two values for the number of strips there. You should verify that we get the same result for any number of strips you care to try.
So that's not so bad, but it still wasn't very challenging. After all, what about shapes that aren't quite as nice?

Example 4:

Find the center-of-mass of the uniform isosceles trapezoidal plate shown in the the figure below.

Solution: We can divide the plate into sections for which we can find the center-of-mass. Once we have the center-of-mass position for each piece, then we can find the center-of-mass for the whole plate. The most sensible division of the plate would be that shown below.

Now we have to find the centers-of-mass of two similar triangles and a rectangular slab. The rectangular slab we did previously, so we know its center-of-mass position will be X_CM = ½ a and Y_CM = ½ h. To find the center-of-mass of an isosceles triangular slab with width w and height h, we can consider breaking it into strips of width x and height dy as shown in the figure below.

If we make the origin of the coordinate system the bottom left corner, then the vertical position of the strip from the bottom of the triangle is given by y. The mass of strip is dm. Since the material is uniform, we assume its mass is evenly distributed and therefore the mass, dm, is related to the mass M of the whole triangle as the ratio of areas. Approximately,

dm/M = (x dy)/(½ wh)

where the approximation arises from the determination of the area of the strip as a rectangle. The shape is only approximately rectangular, but we can show that the approximation to a rectangle is better and better in the limit where dy goes to zero, a limit we will assume in forming the definite integral for the calculation of the center-of-mass position for x and y. In fact, the x position of the center-of-mass is

where we have taken the Riemannian sum over to an integral (it will be a definite integral when we have correctly parameterized dm). Since we know that dm = M(x dy)/(½ wh) we need to find x in terms of y. The easiest connection between x and y is through the tangent of the angle theta since

tan(theta) = x/(h - y) = w/h ==> x = (w/h)(h - y)
We can now replace every instance of x with one of y and integrate over the whole y range, 0 to h. So, the x position of the CM is

To get the y position of the CM, we don't have to do much more work. We note that we can relate the y position of any of the strips we've used before relative to the top point of the triangle through the angle theta. Thus, we can find the y position of the CM relative to that point by noting that

tan(theta) = b/h = x/(h - y) ==> y = h - h/b x Now we can find the y position of the CM

We have to remember that we did this relative to the top of the triangle, so with respect to our origin, we find Y_CM = 2/3 h. Applying these results to our original trapezoidal triangle piece, we find the CM position to be 2/3 (b - a)/2 = 1/3 (b - a) from the leftmost corner and 2/3 h above the bottom of the trapezoid. Since the triangle on the right side of our trapezoidal slab is identical to the triangle we just handled, the CM is located a distance 1/3 (b - a) from the rightmost corner and 2/3 h above the bottom of the trapezoid. The fraction of mass of the whole contained in each triangular piece is the same as the ratio of the triangle's area to the area of the whole trapezoid. The area of the trapezoid is 2[½ (b - a)h/2] + a*h = ½ (b + a)h, therefore the fraction of mass contained in each triangle is ¾ (b - a)h/[½ (b + a)h] = ½ (b - a)/(b + a). The fraction of mass contained in the central square piece is a*h/[½ (b + a)h] = 2a/(b + a). We combine the CM position and masses for each of the three pieces to get the final result for the CM of the trapezoidal slab relative to the leftmost bottom corner position.

X_CM	= [1/3 (b - a)][(b - a)/2(b + a)] + [½ (b - a) + ½ a][2a/(b + a)] + [b - 1/3 (b - a)][(b - a)/2(b + a)]
	= (b - a)²/[6(b + a)] + b/2 * 2a/(b + a) + [2b/3 + a/3][½ (b - a)/(b + a)]
	= (b - a)²/[6(b + a)] + ba/(b + a) + (2b + a)/6 * [(b - a)/(b + a)]
	= 1/[6(b + a)][b² - 2ba + a² + 6ba + 2b² - ba - a²]
	= 1/[6(b + a)][3b² + 3ba]
	= 3b/[6(b + a)][b + a]
X_CM	= b/2

Y_CM	= (2h/3)[½ (b - a)/(b + a)] + [½ h][2a/(b + a)] + (2h/3)[½ (b - a)/(b + a)]
	= (2h/3)[(b - a)/(b + a)] + h[a/(b + a)]
	= [h/(b + a)][2b/3 - 2a/3 + a]

Y_CM	= (2h/3)[½ (b - a)/(b + a)] + [½ h][2a/(b + a)] + (2h/3)[½ (b - a)/(b + a)]
	= (2h/3)[(b - a)/(b + a)] + h[a/(b + a)]
	= [h/(b + a)][2b/3 - 2a/3 + a]
	= [h/(b + a)][2b/3 + a/3]
Y_CM	= h(2b + a)/[3(b + a)]

Now how would you do something more complex, like finding the center of mass of a uniform sphere of radius R and mass M for example? The simplest thing is to resort to symmetry - the center of mass must be at the geometric center of the sphere. This is much faster than calculation and you can easily convince yourself that it must be correct! To prove it explicitly though, you have to break the sphere into infinitesimal pieces. In this case it be easiest to consider strips for a spherical shell. You consider a hollow, very thin-walled sphere and break it into strips which each go all the way around the sphere. If you can't picture that, try this animation. The strips are integrated over the whole sphere from pole to pole. This finds the center of mass of each spherical shell. You can then integrate over spherical shells of smaller and smaller radius from the maximum, R, to the minimum, 0. Each of these shells will have its center of mass in the same place (the center), so you can verify the result of our symmetry argument. Although this is a lot of work for something we can easily ``see'', it is a useful technique for finding the center of mass of things like hemispheres (which do not have enough symmetry for you to trivially calculate the center of mass position). We also use this same technique later when talking about rotational inertia.

Hits since 10/26/99:

p₀	= p_f
0	= mv_E - MV
0	= m(v - V) - MV ==>
0	= mv - (m + M)V ==>
V	= (mv)/(m + M)
	= (50 kg)(40 m/s)/(50 kg + 1950 kg)
V	= 1 m/s

p₀	= p_f
(m₁ + m₂)v_x	= m₁*0 + m₂v₂ ==>
(m₁ + m₂)v₀cos(theta)/m₂	= v₂
(2m₂ + m₂)v₀cos(theta)/m₂	= v₂
3v₀cos(theta)	= v₂