How things work college course/Momentum transfer under elastic and inelastic collisions

Wikipedia and other pages

 * Impulse (physics)
 * Inelastic_collision
 * Momentum
 * Physics_equations/Impulse,_momentum,_and_motion_about_a_fixed_axis

Experimental results
All experiments involved a compound pendulum made from pine. The inverted pendulum was a "knock-down" experiment in which the critical path length for knockdown was obtained. We measured path length because it was easiest to measure. The potential energy associated with the path length $$\ell$$ is $$mg\ell\sin\alpha$$ where $$\alpha$$ is the angle of inclination. Most experiments involved finding the sad/happy pathlength ratio. The simplest theory says that happy is fully elastic and sad is fully inelastice, and the sad/happy pathlength ratio is 4.

Preliminary Results (inverted)
Since a happy ball delivers twice the momentum as a sad one, it stands to reason that the sad ball needs to go twice as fast to knock down the block. But recall that by energy conservation, the speed of a ball sliding down a ramp varies as the square-root of the height. At a 12 degree ramp we found that the happy ball path-length was 43 cm while the sad path length was 80.5 cm. The sad/happy pathlength ratio was 80.5/43=1.87

Steeper ramp (first run)
Here we performed experiments on the inverted pendulum (knockdown) and a stable pendulum (in which amplitude was measured).

Inverted (knockdown) mode
We changed the ramp slope 21.3 degrees. The paths were 62.5 and 139.5 cm long, respectively for happy and sad. The sad/happy pathlength ratio was 139.5/62.5=2.23 The block was pine with height 14, thickness 4, width 7 cm. The mass of the block was block 174.4 gm. The path length (continued straight) from the ramp's edge to the block was 2.6 cm. The bottom of the ramp is 7.2 cm above the bottom of the block.

Balanced block on its side so that almost 50% was over and edge and measured bounce of happy. It fell down 26 cm and came back up 8.5 cm. On the table the happy fell down 100 cm and came back up 87.5 cm.


 * Ball mass and diameter: happy 9.0 gm, sad 9.8 gm, diameter 2.44 cm.


 * The block was almost touching the ramp and warped so that the wall was perpendicular to the sloped ramp.


 * This was the last experiment that did not use the 2-meter sticks to center the ball on the ramp.--(20 November 2014)

Stable (amplitude measurement)mode
Thursday 21 November 2014:


 * Two 2-meter rulers were place on the track. This lifted the ball up and also centered it as it rolled along the ramp.  The 2-meter rulers had a thickness of .6 cm and created a groove that was 0.6 cm wide. We estimate that the ball struck the block between .5 cm and 1 cm from the bottom of the block.


 * Push pins were inserted about 0.1 cm below the top of the block and the block was supported so it's bottom was 12.3 cm above the table. The distance to between the block and the ramp was 2.2 cm, as measured by placing a ruler along the top of a 2-meter ruler that was used to align the rolling balls.  A pendulum amplitude 2.05 cm was achieved with both happy and sad.  This amplitude was measured at the bottom of the block (from equilibrium). The top of the 2-meter ruler was 11.25 cm above the table at it's lowest point.


 * The happy and sad paths as measured by the 2-meter rulers was 25 cm and 70 cm, respectively. Sad/happy pathlengh ratio = 70/25 = 2.8.

Steeper ramp (second run)
We moved the table and repeated the previous experiment, making some improvements in the design:
 * 1) Clamps were placed on the two meter sticks, permitting the ramp to be extended farther.
 * 2) In knockdown mode (inverted pendulum), we used a heavy metal plate to align the block so that it was always in the same place.
 * 3) In stable (compound pendulum) mode, we used tape to assure that the pendulum was always at the same location.

Knockdown mode
Sad/happy pathlength ratio = 169cm/83cm=2.04

Stable mode
Sad/happy pathlength ratio = 112cm/57cm=1.965

stable mode

 * Sad/happy pathlengs = (140+1)cm/(70+1)cm = 1.985
 * Amplitude = 3cm/13.5cm = .2222 radian

knockdown
The ball struck roughly 2 cm below the top.
 * Sad/happy ratio 205+2cm/80+2cm=2.52 ([the 205 was estimated because we could not exceed 200)

Linear and angular momentum transfer equations
For a non-spinning ball and a simple pendulum with an inelastic collision, the formula is a simple matter of momentum conservation, mv0 = (m+M) v1, where m, M, v0, v1 are the mass of the bullet, pendulum, speed of bullet and final speed, respectively. For a spinning ball that does not embed itself into the pendulum, the situation is somewhat different from the traditional ballistic pendulum because the ball can roll along the pendulum after the initial collision. The figure to the right shows an inelastic collision between a spinning ball and a compound pendulum. The spinning (solid) ball has mass, $$m$$, and radius $$r$$. The ball's initial angular and linear speeds are, $$\omega_0$$ and $$\vec v_0$$, and the corresponding final state of the ball is ,$$\omega_1$$ and $$\vec v_1$$. The ball strikes the block at a point $$\ell$$ from the axis or rotation, and $$I$$ is the pendulum's moment of inertia. The pendulum's initial angle is $$\theta_0=0\,$$, and it acquires and angular velocity of $$\dot\theta_1$$ immediately after the collision with the ball.

To visualize this situation it is helpful to imagine that a spring mechanism quickly brings the ball at rest with respect to the pendulum. Here, "at rest" means that the ball and pendulum have no relative motion at the point of contact between them. The ball's center of mass might be moving if the ball is rolling. If the collision is completely inelastic, the spring mechanisms will compress, perhaps like ideal springs, but they do not decompress once the springs have come to rest. This model is somewhat "uncomfortable" because there are actually two spring mechanisms, for vertical and horizontal motion. Only if both mechanisms are carefully balanced will both springs come to maximum compression at the same time. If this is a well-posed physical problem (i.e., with a unique solution), this assumption that both springs come to rest simultaneously is not necessary. We shall speak no more of this blemish in the derivation.

During the time it takes the springs to compress, both the ball and pendulum will change their velocities. But we assume that the impulse time is so short that neither moves an appreciable distance during this first phase of the collision. (The second phase consists of the ball rolling along the block as it exerts a normal force and hence further torque on the pendulum.)

Four equations in four variables
Once we place the ball on the ramp, it is not difficult to calculate the how the ball is moving (i.e. angular and linear velocity) just before it strikes the block. But the rest is rather complicated as we now have four variables to deduce, $$v_{x1}, v_{y1}, \omega_1, \dot\theta_1$$. This requires at least four equations and quite a bit of algebra is involved. Two involve the fact that the ball rolls along a moving pendulum:


 * $$v_{x1}=\ell\dot\theta_1$$  and    $$v_{y1}=-r\omega_1$$

Another constraint is conservation of total angular momentum about the axis about which the pendulum rotates.

This discussion is getting to complicated for "How things work". For a really long calculation see:
 * Topic:Advanced_Classical_Mechanics/Compound_ballistic_pendulum_with_spinning_ball

Doing the algebra
%Third run (w/Teaching Assistant) Sad stable pendulum clc, close all, clear all M=.1744;% kg (kilograms) is mass of block mSad=.0098% kg mass of sad L=.14;% m (meters) is length of block pathSad= 1.40;%m is the distance the sad ball rolled down the ramp incline=21.3*pi/180;%degrees; is the incline angle in RADIAMS ghSad=9.81*pathSad*sin(incline);%initial potential energy of sad Iapprox=sqrt(M*L^2/3) %kg m^2 is the approximate moment of inertial (thin rod model) vSad=sqrt(ghSad*10/7)*cos(incline) thetaDot = mSad*vSad*L/Iapprox thetaMax = sqrt(2*Iapprox/9.81/M/L) thetaObserved = .222 BigOverSmall=thetaMax/thetaObserved

The period of a compound pendulum
This lab could be done at home or in as simply equipped classroom. All you need is two sewing pins and a block of wood such as a 2x4 (inch) board that is about 6 inches long.

In the approximation that the block is a thin rod of length L the period is the same as for a simple pendulum of 2/3 its length :
 * $$T=2\pi\sqrt{ \frac{2L}{3g}  }$$

How fast do objects roll downhill?
It would be nice to perform this lab with an actual ramp and two different objects that roll (they need not have the same dimensions or mass). But you can use the photofinish image to do this at home without equipment. In the space below, use the animated gif on the left and the photofinish on the right to write a lab that students can do at home.

The theory is explained in various places on the internet. For example,ir we neglect wind drag and rolling resistance the speed of a rolling solid sphere is:


 * $$v_0=\sqrt{\frac{10}{7}g\ell\sin\alpha}$$, where $$\ell$$ is this distance rolled along the ramp and $$\alpha$$ is the plane's inclination.

References and footnotes

 * One dimensional elastic and inelastic collisions by Richard Fitzpatrick at the University of Texas at Austin.
 * Real-world Physics: Bouncing-ball
 * Articles by Rod Cross (See also PHYSICS OF BOUNCE)
 * The coefficient of restitution for collisions of happy balls, unhappy balls, and tennis balls by Rod Cross
 * Grip-slip behavior of a bouncing ball by Rod Cross
 * Impact of a ball on a surface with tangential compliance by Rod Cross
 * Footnotes: