Energy Efficiency Lab

Purpose §

To figure out the efficiency (energy remaining) of our ball-and-ramp system as the ball travels from its initial position to when it first hits the floor.

Experiment Method §

1. Prop up one end of the metal ramp. Ensure that it is 34.0 centimeters above the table, and that one side reaches the table and leaves a little bit of room for horizontal motion.
2. Attach the plastic racecar tracks to each other as tightly as possible, to avoid unintended energy loss or gain. Align them so that one edge of the tracks leads off the end of the table, and the other leads off the top of the ramp. It is okay if there is a little overhang at the top of the ramp, but not at the edge of the table.
3. Affix the tracks to the ramp with double-sided pieces of tape, ensuring that the bottom of the tracks is as flush with the ramp as possible. Also add a piece of tape under the part of the track that lies on the table, to keep it in place. Press down to flatten any bumps that may form because of the tape.
4. Record the height of the table relative to the ground.
5. Choose a starting point for the ball. Record its height relative to the table.
6. Mark the starting point of the ball with a piece of masking tape.
7. Ensure that the front of the ball starts at the back, more elevated edge of the tape. Perform a test run to see where the ball lands when you start it at the point you’ve chosen. Make sure to hold down the part of the track that lies on the table, so that it is as flat as possible to minimize vertical motion before the ball is launched. Mark the spot you are holding it down from, where the back of the tape signifies the correct location.
8. Wherever the ball lands, put down a ‘sandwich’ of one piece of normal paper, one piece of carbon paper shiny side down, and another piece of normal paper on top. Tape down the normal paper at the bottom to make sure that it does not move. The other papers do not have to be taped down.
9. Hang a plumb bob below the edge of the table, then mark that same spot on the floor with masking tape.
10. Release the ball again. This time, it will make a mark on the taped-down white paper.
11. Using a meter stick, measure and record the distance from the back of the masking tape on the ground to the back of the center of the circular mark on the paper.
12. Repeat steps 5-11 until you have recorded the result of releasing the ball from five points.

Data Collection §

General Values §

Type	Value	Unit
Mass of Ball	66.91	g
Floor to Table	91.7	cm
Radius of Ball	1.2	cm
Weight of Ball	0.656	N

Points §

Point	Table to Point (cm)	Floor to Point (cm)	Dist Impact From Table (cm)
Point 1	34.0	125.7	91.9
Point 2	30.7	122.4	87.0
Point 3	27.6	119.3	83.0
Point 4	16.5	108.2	65.6
Point 5	5.6	97.3	36.9

Results/Analysis §

Note to self: check angular motion formula sheet and Physics 1 review notes.

Here is the math behind solving getting the values for the data collected point/trial 1.

Variables §

Variable	Value	Variable	Value
$d_x$	$0.919\text{ m}$	$m$	$0.06691\text{ kg}$
$h_0$	$1.257\text{ m}$	$g$	$9.81{\text{ m/s}}^2$
$h_{table}$	$0.917\text{ m}$	$v_0$	$0\text{ m/s}$
$R$	$0.012\text{ cm}$	$w_0$	$0\text{ m/s}$

Velocity Calculations §

Floor Vertical Velocity §

$$\begin{array}{rl}h& =\frac{g{t}^2}{2}\\ 0.917& =\frac{9.8t^2}{2}\\ t& =\sqrt{\frac{2(0.917)}{9.81}}\\ & \approx 0.43235\text{ s}\end{array}$$

Substitute $t$ in equation for vertical velocity:

$$\begin{array}{rl}v& =gt\\ & =(9.81)(0.43256)\\ & \approx 4.24164\text{ m/s}\end{array}$$

Horizontal Velocity §

Using $t$ as determined above:

$$\begin{array}{rl}d& =v_{x}t\\ 0.919& =0.43235v_x\\ v_x& =\frac{0.919}{0.43235}\\ & =2.12545\text{ m/s}\end{array}$$

Total Velocity §

$$\begin{array}{rl}{v_{tot}}^2& ={v_x}^2+{v_y}^2\\ v_{tot}& =\sqrt{{v_x}^2+{v_y}^2}\\ & =4.74437\text{ m/s}\end{array}$$

Energy Calculations §

Initial §

Initial PE §

$$\begin{array}{rl}P{E}_0& =mg{h}_0\\ & =(0.06691)(9.81)(1.257)\\ & \approx 0.82508\text{ J}\end{array}$$

Initial KE §

$$\begin{array}{rl}K{E}_0& =\frac{m{v}^2}{2}\\ & =\frac{m(0{)}^2}{2}\\ & =0\text{ J}\end{array}$$

Table §

Table PE §

$$\begin{array}{rl}P{E}_t& =mg{h}_t\\ & =(0.06691)(9.81)(0.917)\\ & \approx 0.60191\text{ J}\end{array}$$

Table KE §

To find the total KE at the table, we need to consider the horizontal velocity of the ball, its mass, and the rotational kinetic energy, which is calculated below but we will factor it in now. So:

$$\begin{array}{rl}K{E}_t& =K{E}_{r_t}+\frac{m{v_x}^2}{2}\\ & =0.06045368951+\frac{(0.06691)(2.125447{)}^2}{2}\\ & =0.2115874867\text{ J}\end{array}$$

Table Rotational KE §

We first have to calculate the inertia of our sphere, $I$.

$$\begin{array}{rl}I& =\frac{2}{5}M{R}^2\\ & =\frac{2}{5}(0.06691)(0.012{)}^2\\ & =3.854016\times 10^{-6}\text{ kg(}\text{m}{)}^2\end{array}$$

Then we need to find the angular speed of the ball at the table. The angular speed is only based off the x component of the velocity vector, which we calculated at the start of this whole shebang.

$$\begin{array}{rl}{\omega }_t& =\frac{v_x}{r}\\ & =\frac{2.12545}{0.012}\\ & =177.1208333\end{array}$$

Then we put those into the $K{E}_r$ equation:

$$\begin{array}{rl}K{E}_{r_t}& =\frac{I{\omega }^2}{2}\\ & =\frac{(3.854016\times 10^{-6})(177.1208333{)}^2}{2}\\ & =0.06045368951\text{ J}\end{array}$$

Final §

Final PE §

$$\begin{array}{rl}P{E}_f& =mg{h}_f\\ & =mg(0)\\ & =0\text{ J}\end{array}$$

Final KE §

$$\begin{array}{rl}K{E}_f& =K{E}_{r_f}+\frac{m{v}^2}{2}\\ & =0.06045368951+\frac{(0.06691)(4.74437{)}^2}{2}\\ & =0.8134933587\text{ J}\end{array}$$

Final Rotational KE §

This is the same as the rotational kinetic energy at the table, because $I$ does not change and $v_x$ is constant, which makes $\omega$ constant.

$$\begin{array}{rl}K{E}_{r_f}& =\frac{I{\omega }^2}{2}\\ & =0.06045368951\text{ J}\end{array}$$

Efficiency §

$$\begin{array}{rl}\text{Efficiency}& =\frac{\text{ Final ME}}{\text{ Initial ME}}\times 100\\ & =\frac{P{E}_0}{K{E}_f}\times 100\\ & =\frac{0.824592}{0.8134933587}\times 100\\ & =98.65404451\%\end{array}$$

Discussion §

How did the ball’s potential, kinetic, and total energy change as it moved along the system? §

At the top of the ramp, the ball was not yet in motion, which made the kinetic energy 0. All of its energy comes from its potential energy, which does not factor in velocity and only scales off mass, gravity and height. This means that the mechanical energy was equal to the potential energy.

At the edge of the table, before its projectile motion began, the ball’s potential energy had decreased because its height relative to the floor decreased, but not by too much. Because the ball is now in motion, the ball now has kinetic energy on top of its potential energy, part of which is rotational kinetic energy. However, the potential energy is still larger than the kinetic energy. The total/mechanical energy appears to have decreased, but only very slightly, by 0.01 or so for most trials.

At the point where the ball hit the floor, its height was zero relative to the floor, so its potential energy was reduced to zero and all of its total energy was actually just kinetic energy. This fits into what we know about potential and kinetic energy— they switch! It also appears that we’ve lost just the tiniest bit of total energy when comparing our final mechanical energy to the mechanical energy at the edge of the table, but this difference is even less than before and seems to be statistically insignificant. Compared to the initial mechanical energy, our retention is pretty darn good, with an average of $99.166$% efficiency.

What are two built-in problems that could reduce our efficiency? §

In an ideal system, the kinetic energy at the table for point 1 would be equal to the original potential energy minus the potential energy at the table, which should be $0.22317$. Our actual kinetic energy ends up being $0.2115873161$! The same loss occurs in our final kinetic energy vs our initial potential energy, which in theory should have been equal. We only lose a little bit of mechanical energy, but that still doesn’t align with how it should ideally be, so there’s obviously some kind of problem going on.

However, we see something weird occur when we compare the efficiency throughout trials. Every single trial between 1 and 4 seems to get more and more efficient. Interestingly enough, trial/point four seems to be almost perfect in terms of energy retention, with only very, very slight differences that are for the most part insignificant. While this trend dips a bit from trial 4 to 5 with a 0.3% decrease in efficiency, the last trial’s efficiency is still over a whole percentage point better than the first trial’s.

There are two problems that I believe could be contributing to this the most. The first problem is that our racecar track setup wasn’t completely smooth, with tape and the links between the tracks causing slight bumps that could have contributed to this tiny loss. As we go further and further down the track, there are fewer and fewer bumps to mess with conservation of energy. The second problem is really just the fact that we can’t make perfectly precise measurements with the tools that we have, which would be necessary to perfectly calculate the nearly imperceptible differences between some decimal point level measurements. We probably got a little better at this as we went on, which could have contributed to the rise in efficiency alongside the lack of bumps.