How Hot Are Your Hot Wheels?
How can you apply the conservation of energy concept to 1) predict from where to
release a car on a loop-the-loop track to just retain contact with the track through the loop and 2) predict how far and how high a car will travel from a ramp inclined at some angle?
Hot Wheels set with loop-the-loop and ramp, small car, meter stick, timing device (Computer interface photogates, if possible).
Procedure A: Loss of Energy Due to Friction
The illustration shows the track setup for Procedure A. Elevate both ends of the track to a height of about 1m above the floor or lab table.
Release the car from the starting point A. Record this height (h1) and the height to which the car rises (h2) on the other end of the track. Please make several trials. It is important to secure the track to the floor so some of the car's energy is not transformed into motion of the track.
The ratio of the final height to the initial height is equal to the ratio of potential energy transferred back into the car going up the left ramp to the potential energy the car had at the top of the right ramp at point A. Call this ratio of h2 to h1 the "efficiency" of the system for the car going from point A to point B on the track. The distance along the track from A to B is called the "standard length" so the ratio of h2 to h1 (efficiency) represents the fraction of the initial energy the car will have when it reaches point B on the track. This ratio is used in later experiments to determine how high the car must start, so it will have a predictable total amount of energy when it reaches certain points on the track.
1. What is the average height ratio, or efficiency?
2. In what unit is efficiency measured?
Procedure B: Loop the Loop
The next illustration shows a loop-the-loop section placed in the track at about the
standard length from the starting end of the track. The problem is to predict the minimum height from which the car must start, so it will successfully travel all the way around the loop without falling away from the track. In order for this to take place, there must be centripetal force acting on the car as it goes around the loop. This force Fc must be equal to the weight of the car.
The equations for centripetal force and weight appear below.
W = F 0 ....W = mg .... F 0 = mv 2 / r
Combine these equations and solve for v 2
What is the minimum speed the car can have at the top of the loop to perform a
At the top of the loop, the car has both kinetic and potential energy. This energy total was supplied by the loss of potential energy the car had at a point a. This means:
KE b + PE b = PE a
1/2mv 2 + 2 mgr = mgh
Substitute the value of v 2 you found above into the kinetic energy equation and then solve the total energy equation for h. this means that if there were no friction, the car should start from a height, h, to make it around the loop. Since the system does have friction, the car must start from a point just higher to make up for the frictional loss. Therefore the car must start from a height of h divided by the efficiency of the system. Write a report on your findings in this experiment.
1. State in your own words the energy changes as you lift the car to point A, until it
completes the loop-the-loop.
2. Do your results show that energy is conserved? Explain your answer.
Procedure C: The Dare Devil Jump (Optional)
The illustration below shows the jump-ramp set up. The problem here is to calculate the range of the car's jump when you release it from a chosen height (h) above the launch point of the ramp.
The starting point of the jump should be at the "standard length" from the release point of the car. The kinetic energy of the car at the launch point on the ramp equals the loss in gravitational potential energy from the starting point on the track, or
1/2 m v 2 = mgh,
assuming the track is 100% efficient. From this statement of conservation of energy, you can begin to solve the problem posed above. Calculate the loss in gravitational potential energy your car would undergo from the release point to the top of the launch ramp. Now calculate the actual energy at launch by using the efficiency factor you calculated in Procedure A. From this kinetic energy, you can calculate the launch velocity of the car.
You can calculate or measure the initial launch angle of the car, and use this to calculate the vertical and horizontal components of the launch velocity. From your knowledge of projectile motion, you should now be able to calculate the total jump time and the jump range. Now try out your predictions! How did you do? If time permits calculate the maximum height of a "fence" the car could clear as it jumps from ramp to ramp.
1. Describe the energy changes as a car moves down the track and completes the jump.
2. Explain whether or not energy was conserved in this activity.
The principle of conservation of energy is one of the most powerful and widely used concepts in solving science related problems. This activity is an example of the type of problem frequently found as an end-of-chapter numerical exercise. Students should find this type of application more interesting than the usual end-of- chapter exercise. Whereas textbook problems often neglect friction in word problems, students deal with it here in a practical way and apply it to make predictions of motion of the Hot Wheels car.
The mathematics in Procedure C is challenging. You may wish to make Procedure C optional, or provide extra help for struggling students.
If you do not have a Hot Wheels set in your school, it is well worth asking students to bring in enough Hot Wheels sets so you have at least one for every two to three students. For most predictable results secure the track, loop-the-loop and ramp with clamps or tape.
Sample Observations/Calculations Procedure A:
h 1 = 40cm h 2 = 32cm
eff = 32 cm / 40 cm = 0.80 = 80%
Summing Up Procedure A:
1. Friction reduces the effect of useful work and basically drains off some of the energy intoa heating effect. To compare the initial height, h 1, of the car with the final height, h 2 ,is to compare the initial potential energy of the car with its potential energy on the other side of the track (because m and g are constant). An efficiency of 80% means that 80% of the initial potential energy at h 1 transfers to potential energy at h 2 or that 20% of the initial potential energy was lost to frictional forces. This efficiency applies only to the "standard length" of track. For a longer track, the difference in elevations would be greater.
2. Efficiency is a ratio, so it has no units. The units always divide out.
Sample Observations/Calculations Procedure B:
1. They should take several sets of data, calculate the efficiency for each set, and then calculate the average efficiency. Students will use this average in Procedures B and C.
2. Efficiency is a ratio, so the units have divided out. Since the weight of the car will be equal to the centripetal force at the top of the loop we can write
W = F c
W = mg, .... and F c = mv 2 / r
mg = mv 2 / r
v 2 = gr
Typical values might be:
d = 20 cm, r = 10 cm, and
v 2 = 9.8 m/s/s * 0.10m = 1.0 m/s
Summing Up Procedure B:
In the equation, v 2 = gr , v ...is the minimum speed of the car at the top of the loop for the wheels to just touch the tracks. If the energy is conserved in this event, then the potential energy at A should equal the sum of all the energies at some other point such as at the top of the loop. In the form of an equation, we write
PE a = KE b + PE b + E friction
Recognizing that we will adjust for the energy lost to frictional forces in our release height we have by substitution:
mgh a =1/2 mv 2 + mgh b
gh a = 1/2 g r + g (2r) .....where v 2 = g r ...
h b = 2 r .... and h a = 1/2 r + 2 r = 2.5 r
As pointed out in the student sheet, we must now concern ourselves with the effect of frictional loss. Measuring our "standard length" from B back to A we should elevate point A such that its height h is found as:
0.80 h = 2.5 r .... or ... h = 3.125 r
Note that the mass of the car does not affect the height from where it should start. Check to determine that all forms of energy are accounted for in the conservation equation. The KE B term represents linear kinetic energy so if you use a steel ball instead of a car then some of the energy transforms into rotational energy, which for a solid sphere would be mv 2 /5. . Add this term to the other terms in equation 2 and 3. In applying equation 2 and 3 to the car we will assume that the wheels take on a negligible amount of rotational kinetic energy.
1. When the car is lifted, work is done on it, increasing its potential energy. The
work done to lift it to point A equals its potential energy at point A. As the car is
released, it rolls down the ramp, losing some of its potential energy, but gaining in
kinetic energy. The loss in potential should equal the gain in kinetic. At the
bottom, the potential is zero, but the kinetic should be equal to what the potential
was at point A. As the car enters the loop-the-loop, it loses kinetic energy but
gains potential. The opposite happens on the way down. At any point, the total
energy should equal the work done in lifting the car to point A.
2. ... m = 30 g = 0.03 kg
At point A PE = mgh = ( 0.03 kg ) ( 9.8 m/s/s )( 0.31 m ) = 0.091 J
At top of loop KE = 1/2mv 2 = ( 0.5 ) ( 0.03 kg ) ( 1.0 m/s )2 = .015 J
PE = mgh = . ( 0.03 kg ) ( 9.8 m/s/s ) ( 0.2 m ) = .059 J
KE . and PE = 0.015 J + 0.059 = 0.074 J
Energy lost = 0.091 - 0.074 = 0.017 J
Energy lost went into heat produced by friction.
Sample Observations/Calculations Procedure C:
These sample calculations use a release height ,h, of .40 m above the launch point and a track efficiency of 80%. The launch angle is 30 0 and mass of the car is .03 kg. Gravitational potential energy = mgh = ( .03 kg ) ( 9.8 m/s/s ) ( .40 m ) = .12 J. Actual kinetic energy = efficiency x potential energy = (.80)(.12 J) = .094 J.
Because kinetic energy =1/2mv 2 . then v 2 = 2 * KE/m . or
v 2 = (2) (.094 J) / (.03 kg) = 6.3 m 2 / s 2 or v . = 2.5 m/s.
The vertical component (vy ) of launch speed is used to calculate the time the car spends in the air after launch. V y = v sin .q = ( 2.5 m/s ) ( 0.5 ) = 1.25 m/s. Calculate the time in the air from the definition of acceleration, or t = v/a assuming the vertical velocity at the peak is zero. Then the time for one-half of the flight = 1.25 m/s 9.8 m/s/s = .13 s, or the total time of the flight is .26 s. Calculate the range by first finding the horizontal component (v h) of the launch velocity, v h = V cos .q = (2.5 m/s)(.866) = 2.2 m/s. This horizontal velocity component is constant and not affected by gravity. Therefore the range is simply the product of the total time in the air and the horizontal velocity. Range = ( 0.26 s ) ( 2.2 m/s ) = 0.56 m.
Summing Up Procedure C:
1. Answers could be very similar to Procedure B, #1.
2. Within experimental error, energy is conserved. Work was done in lifting in the car. From that point on, no more work was done on the car (except friction) to change the total energy.