## D-5 BALLISTIC PENDULUM

Instructor's notes.

The collision of the ball with the catcher is inelastic, so kinetic energy is not conserved in the impact. Momentum is always conserved, so:

 [1]

mv = (m+M)V

M is the mass of the pendulum and m is the mass of the projectile. V is the velocity of the catcher and ball after impact, v is the velocity of the ball before impact.

If one is tempted to consider conservation of linear and angular momentum as a means for analyzing the data, consider this: The momentum of the swinging catcher decreases smoothly to zero at the extreme of the swing. It loses its horizontal component of momentum to the apparatus as a whole through forces at the pendulum support point. However these constraint forces, which restrict the motion to the arc of a circle, also give the catcher a vertical component of momentum, which is finally lost due to the opposing force of gravity. (The vertical component of catcher momentum, initially zero, increases after the impact, then decreases to zero.) The angular momentum of the swinging catcher also decreases smoothly to zero, this being primarily to the earth through the gravitational interaction (the gravitational torque about the pivot point opposes the motion). If the pivot point is frictionless it can not affect the catcher's angular momentum about that point.

Two things are in our favor: (1) The impact is so brief that it is over before the catcher has a chance to move much. (2) The nearly frictionless nature of the pivot is the key to proceeding with the solution, for there's only small energy loss there. Therefore, conservation of kinetic and potential energy may be used.

 [2]

(1/2)(m+M)V2 = (m+M)gh , or V2 = 2gh

These may be combined:

 [3]

v = [(m+M)/m]√(2gh) , or better, v = [1+(M/m)]√(2gh)

A straightforward (but tedious) error propagation calculation gives the determinate-error equation:

 [4]

Δv/v = [M/(M+m)][ΔM/M - Δm/m] + (1/2)[Δh/h]

The quantity M/(M+m) is about 0.8. The fractional error in M is 0.2% and that in m is 0.07%, and in h is about 1.4%. The error due to measurement of h is dominant, and the errors due to mass measurements are negligible.

From my notes,

```Quantity         uncertainty   % uncertainty

m = 67.91 gm     0.05 gm       0.07
M = 264.7 gm     0.05 gm       0.2
R = 71.1 cm      0.2 cm        0.3
h = 74.1 cm      0.1 cm        0.13 (reading uncertainty)
Δh = 7.25 cm     0.1 cm        1.4  (pendulum c.m. rise)
v = 584 cm/s, typical
```

However, there are other sources of uncertainty in the firing apparatus, which cause the ball to have different velocity each time the gun is fired. While the determination of the velocity for a given trial is quite good, the reproducibility of the velocity from trial to trial may be poor, and shows up as a large variation in the values of h, larger than the simple uncertainty in reading Δh.

Defects of the above analysis.

Several things were ignored:

(1) Energy loss in the swing, due to the ratchet and the pivot. This makes Eq. 2 wrong. This alone would make the calculated velocities nearly 7% too small. (Beck apparatus, Wall, Ref. 2.)

(2) The pendulum's effective length is 0.93 L, causing the calculated velocity to be 7% too large. (Beck apparatus, Wall, Ref 2.)

(3) The center of percussion of the pendulum is above the point at which the collision occurs. This makes Eq. 2 wrong. Sandin (Ref. 3) says this gives a forward impulse to the pendulum (through the reaction force at the pivot bearing), resulting in more linear momentum after the collision than before, about 1% more for the Cenco apparatus, and 3% more for the Beck apparatus (due to its sturdier arm construction). Sandin notes that this could be eliminated by adding weights below the collision point until the center of percussion and center of mass coincide.

Important design considerations not usually mentioned.

(1) The resting angle of the pendulum should be the same whether or not it has the ball captured in it. This should be checked by the student.

(2) The ball should not 'rattle around' in the catcher during collision and subsequent swing. This would change the center of mass of the ball-pendulum combination during the swing. The ball should not end up lodged in the catcher lower or higher after the collision than before it.

Things which can be done which many manuals don't mention.

(1) The radius of gyration can be found by measuring the small-angle period of the pendulum in order to find the pendulum's center of mass.

(2) The energy loss in the swing can be measured from the decrease in amplitude of pendulum swing, starting from the amplitude of maximum deflection.

Critique of some instruction manuals:

The instruction sheets for the Cenco Precision Ballistic Pendulum, Catalog #31379. No date, probably around 1988. The apparatus has a counterweight above the pendulum rotation axis.

To simplify calculation, the support rod K has a counterweight W whose purpose is to indicate the position of the center of gravity of the rod at the axis. Thus, only the mass of the bob and the projectile need to be taken into account. This allows different projectile masses to be used, without the need for calculating a new center of gravity and radius of gyration for each mass.

Radius of gyration isn't mentioned elsewhere in Cenco's instructions! Center of percussion isn't mentioned either.

The location of the pendulum arm's center of mass at its rotation axis does make it unnecessary to correct for the center of mass when using different mass projectiles. When the catcher is taken off the arm, the arm and its counterweight have neutral equilibrium. Therefore when measuring the swing of the pendulum one needs only take into account the mass of the catcher in the equation for the pendulum swing.

Later in the Cenco document:

ERRORS: Part of the momentum of the impact will be transmitted to the pendulum frame through imperceptible flexing and energy will be lost through windage and bearing friction. To correct for these factors, remove the spring gun by removing 2 wing screws on the bottom. Position the stop bar to allow the pendulum to be consistently and accurately located at a selected angle to the right. Pull the pendulum back to this position, measure with the indicator and release the pendulum. Ideally, the pendulum will swing as far left as it was released right. The observed difference is due to the above-mentioned error.

This is misleading in several ways. Inelastic flexing of the apparatus represents a way energy is lost to the frame; exchange of momentum doesn't require consideration of inelastic flexing of the frame, but motion of the frame, giving the frame a velocity. In the worst case, if the apparatus were on a frictionless table, the fraction of the momentum lost to the frame would be M/Mf. In reality it is no worse than M/(Mf + Mt) where Mf is the mass of the frame of the apparatus and Mt is the mass of the table on which it sits. This is hardly a concern.

Windage and bearing friction do represent losses of energy worth investigating (but difficult to do).

It may be significant that Cenco no longer sells this version of the apparatus.

The instruction manual for the Beck Ball Pendulum, Model M-965, includes this 'IMPORTANT NOTE':

In performing this classic lab exercise, the fact that there is a small amount of pendulum rotation after the collision is disregarded and it is incorrectly assumed that the equation for conservation of linear momentum is all that is needed for the analysis. This assumption results in a 4% to 5% error when determining the initial speed of the ball. Other errors, such as those due to friction, are also disregarded. Since these errors tend to cancel each other, the net results obtained by using this classic technique are sometimes deceptively 'accurate'.

This paragraph implies that the conservation of linear momentum does not apply in this apparatus! No wonder students get strange ideas. Conservation of linear momentum always applies. So does conservation of angular momentum. In the collision, we are entirely justified in using Eq. 1 (with the understanding that some negligible tiny fraction of the momentum is given to the frame of the apparatus, the table, etc.)

Data from the Beck instructions:

```Mass of ball (m)                 57.5 g
Mass of pendulum (M)            143.5 g
Radius of gyration of ball       30.7 cm
Radius of gyration, ball+pend    27.3 cm
Period of ball+pendulum           1.08 s
Rise of center of mass           10.2 cm
Calculated ball velocity        636.0 cm/s
% discrepancy between two methods   4.4%
```

RESULTS

```For either apparatus, v ≅ 600 cm/s = 6 m/s
Projectile mass ≅ 65 g = 0.065 kg
Projectile kinetic energy ≅ 1.2 J
Discrepancy between two methods ≅ 5%
Work done in cocking the gun ≅ (11 kg)(9.8 m/s2)(0.043 m) = 4.6 J
Energy lost in the firing of gun ≅ 3.4 J
Efficiency of gun ≅ 26%
Fraction of energy lost in the collision of ball and pendulum
≅ 2/3 (This is strictly determined by the ratio of
m/M of ball and pendulum. See question 5.
```

BIBLIOGRAPHY:

(1) C. N. Wall and R. B. Levine, Physics laboratory Manual 2nd ed. )rentice-Hall, 1962, Experiment 5, pp. 38-40.
(2) C. N. Wall. "Apparatus Review: The Beck Ball Pendulum." AJP, 36, p. 1161 (Dec. 1968).
(3) T. R. Sandin. "Nonconservation of Linear Momentum in Ballistic Pendulums." AJP, 41, p. 426 (March 1973).
(4) A. Sachs. "Blackwood Pendulum Experiment Revisited." AJP, 44, p. 182 (Feb. 1976).
(5) P. D. Gupta. "Blackwood Pendulum Experiment and the Conservation of Linear Momentum." AJP, 53, p. 267 (March 1985).

THE CENTER OF MASS AND RADIUS OF GYRATION CORRECTIONS

Cenco provides a small pointer on the pendulum from which to measure the change in height. The Cenco Selective Experiments in Physics No 71990-M75b, Momentum: Ballistics merely refers to this as "the index point for the center of gravity," without further explanation! This source tells the student to measure the change of height of this pointer. The implication here is that you are merely correcting the height measurement for the fact that the potential energy change is due to the rise in the center of mass of the pendulum-ball combination. But what about the radius of gyration correction? Why no mention of that; isn't it important?

In the above analysis [Eq. 1-2] we assumed that the kinetic energy of the pendulum just after impact was translational, mv2/2. This ignored the fact that the pendulum is rotating, and its kinetic energy should be written Iw2/2. So, the analysis should go something like this:

 [5]

(1/2) I w2 = (m+M)gh , or (1/2) I [V2/R2] = (m+M)gh

Now what actually is the moment of inertia, I? If we simply put I = (m+M)R2 we'd get the same result as before [Eq. 1]. But, using the parallel axis theorem, I = Io + Mpl2, and letting Mp = (M+m) we get:

 [6]

I = MpR2 + kMpr2 = MpRg2 , where Rg2 = R2 + kr2

Io is the moment of inertia about the symmetry axis and k is the factor appropriate to the shape of the pendulum catcher. [See the table of moments of inertia for simple geometric shapes, found in any textbook, or see the Wikipedia.] For a sphere, k = 0.4. For a solid cylinder of the length used in this catcher, k = 0.6 approximately, and for a cylindrical shell k is nearly 1. Our composite mass, a spherical ball in a cylindrical catcher, may be taken to have k = 0.7 approximately without serious loss of precision. [We are ignoring the small correction for the rotational inertia of the vertical arm of the pendulum.] Rg is the radius of gyration.

So Eq. 2 should have read,

 [7]

MpRg2V2 = Mpgh , or Rg2V2 = 2gh

In this apparatus, the radius of the catcher cylinder, r, is about 0.1 R. Therefore

 [8]

Rg2 = R2[1 + 0.7(.1)2] = 1.007 R2

This factor 1.007 transfers to the measurement of vertical displacement, or horizontal displacement, of the catcher. This quantity is square rooted. So, neglect of the radius of gyration correction causes only about 0.7% determinate error in the height calculation and 0.35% in the velocity calculations. [In fact, the correction for the inertia of the aluminum rod is a determinate error of the opposite sign, and the two corrections tend to offset each other.] Compared to other error sources, these are negligible.

```Data:		Aluminum rod: 30gm, length 28 cm.
Support to center of ball, 31 cm.
Catcher 230 gm, ball 60 gm.
```

8. QUESTIONS:

(1) Did this experiment provide verification of the law of conservation of momentum? If not, why not, and what role did that law play in this experiment? If you say that the experiment did verify the law, specify which part of the experiment did this, and also state the limits of error within which the law was verified.

No, it did not verify conservation of momentum, it assumed momentum was conserved in order to find the velocity of the ball. However, if the ball's velocity was independently measured (by the trajectory method) then the ballistic and trajectory methods can independently be used to determine the initial mometum of the ball, and these two results compared.

(2) Did this experiment provide verification of the law of conservation of energy? Justify your answer in the same manner as indicated in item (5).

No, it did not verify conservation of energy. In the swing of the pendulum the conservation of energy was assumed, and probably not independently checked.

(3*) Suppose you had not bothered to restrain the apparatus to prevent recoil. The mass of the Cenco apparatus is about 7 kilograms. The coefficient of kinetic friction between the rubber feet and varnished wood is probably about 1. Now if you had performed the calculations of this experiment without regard for the recoil, how much systematic error would this cause in the experimental value of vo? In each case indicate whether the error would make the result too high or too low in value.

This is a tricky question. Recoil has no appreciable effect on the velocity of the ball. The compressed spring has potential energy, and when released, the ends of the spring exert forces (a) on the projectile ball, and (2) an equal and oppositely directed force on the gun and its heavy frame. This is true whether or not the gun and frame move. In either case the momentum given to the ball is equal and opposite to that given the gun. So in either case the ball is released with the same velocity. In the restrained case the energy given to the gun and frame goes, not into much motion, but into dissipative processes in its materials and in the restraint.

See Effect of gun recoil on bullet velocity.

But motion of the frame also moves the target, and initiates a small swing of the pendulum. So by the time the ball reaches the catcher, the catcher may be in motion. This will not cause large error in the results, but is difficult to analyze, for it depends on the time delay from launch to catch and on the period of the catcher pendulum.

One could also argue that during the recoil the gun moves backward, and therefore the velocity of the ball is slightly lower. How much lower? Approximately 1% reduction in the velocity, since m/M of the momentum goes to the apparatus, where m/M = 70/7000 = 1/100. However, now the pendulum is moving toward the oncoming ball, and the relative velocity of ball and pendulum is about the same as before. This could be checked by suspending the entire apparatus as a pendulum. This is too involved for the usual lab period length.

(4) How much systematic error would be caused if the gun were misaligned and fired slightly upward, an angle α above the horizontal? Consider separately the systematic error this would cause in your calculated vo for each of the two methods, if the calculations had been done without knowledge of the misalignment.

(5) Kinetic energy is lost in the impact between the ball and the pendulum. Suggest what could have happened to the that energy. Did you observe anything to indicate into what form some of it might have been converted? What fraction of the ball's energy is lost in the impact with the pendulum if the ball has mass 75g, speed 60 cm/s and the pendulum has mass 150 g?

Actually most of the kinetic energy of the ball is lost here, to thermal energy and sound. There's also some energy imparted to the rest of the apparatus through the forward component of force the pendulum exerts on its upper support. The fraction lost for this totally inelastic collision is strictly dependent on the mass ratio of ball to pendulum.

mv = (m+M)V

½mv2 = ½(m+M)V2 + Heat = ½(m+M)m2v2/(m+M)2 + Heat

= m2v2/[2(m+M)] + heat

Heat = (mv2/2) - [m2/(m+M)](v2/2) = [M/(m+M)](mv2/2)

= [1/(1+m/M)](mv2/2)

or Ek = (1+M/m)Epot

When m/M = 70/150 = 1/2, then Heat = (2/3)Ek

Therefore 2/3 of the kinetic energy is lost to heat.

Alternate method, use v = (1+M/m)(2gh)1/2

½mv2 = (1+M/m)2(mgh) = [(m+M)2/m)]gh = [(m+M)/m](m+M)gh

Ek = (1+M/m)Epot

(6*) Suppose the ball hit the catcher a bit offside, so that it bounced off without being captured. This may have happened when you did the experiment. In this case, will the pendulum swing higher, less high, or the same height as when the ball is properly captured? Of course you must justify your answer by appeal to the physical laws and the details of your apparatus and procedure.

The pendulum will swing much higher, for this is now a nearly elastic collision of metal on metal. The center of mass moves with the same speed after collision whether the collision is elastic or inelastic. No kinetic energy is lost in an elastic collsion. So the velocities of both the ball and pendulum will be greater in the elastic case.

(7*) The speed of the pendulum after impact decreases smoothly to zero during its swing. Is the linear momentum of the loaded pendulum conserved during its swing? Is its angular momentum conserved during its swing? Discuss this, considering the changes of momentum of the pendulum during its swing, and the forces and torques which cause those changes.

Note the key words 'of the loaded pendulum'. The x-component of momentum is large at first, decreasing smoothly to zero at the extreme of the swing. Obviously the pendulum momentum x-component changes, some of its momentum being given to the apparatus through the pivot bearing.

The y-component of momentum is zero initially and finally. But during the swing the y-component of momentum begins at zero, increases as the catcher moves upward, then decreases to zero. So, one may ask, "What is giving y-momentum to the pendulum during the early part of the swing?" It is the reaction force at the pivot, exerting an upward component of force on the pendulum. This force is initially (M+m)g upward, balanced by the gravitational force downward on the pendulum, for zero net upward force. But as the pendulum moves to angle q, the reaction force increases in size, and having rotated to angle q, now has a horizontal component, effective in decreasing the horizontal component of momentum.

The angular momentum also decreases to zero during the swing. We can't simply say it was given to the apparatus, for the pivot is sufficiently frictionless that it provides insufficient torque. The angular momentum is decreased by the torque of gravity on the pendulum, and the earth gains the angular momentum the pendulum loses.

The kinetic energy of the pendulum rises from zero during the collision, then decreases smoothly to zero. The kinetic energy it loses goes mostly to the earth, through the gravitational coupling (gravity exerts a retarding torque on the pendulum). The kinetic energy Iω2/2 is the kinetic energy mv2/2, so these need not be discussed separately.

© 1997, 2004, 2012 by Donald E. Simanek.