mouse trap report

You must also include the theoretical calculations (see later section) in your lab report, as an appendix.
Assessment Criteria
Clear, concise descriptions·
Justification of design based upon theory·
Correct calculations·
The aims of the exercise are:
to design and build a high-performance (high speed) vehicle to given· specifications based upon fundamental principles of mechanics;
to determine the most successful design from the results of a race against other vehicle designs;·
to calculate the performance of a theoretical vehicle and describe differences that exist between·
theory and practice.
MECH117 Lab 2 Draft Report Sheet – 1 –
The intended learning outcomes are that by the end of the exercise you should be able to:
Apply some fundamental principles of mechanics to a simple vehicle;·
Build and test a vehicle within the allocated time and with the resources available;·
Demonstrate effective team working skills.·
Mouse-Trap Vehicle
A mouse-trap vehicle is one that is powered by the energy which can be stored in a wound up mouse-trap spring. The most basic design is as follows: a string is attached to a mouse-trap’s lever arm and then the string is wound around a drive axle causing the mouse-trap’s spring to be under tension. Once the mouse-trap’s arm is released, the tension of the mouse-trap’s arm pulls the string off the drive axle causing the drive axle and the wheels to rotate, propelling the vehicle.
The vehicle needs to be built for speed as it will take part in a race, but it must also be able to perform over a long enough distance to reach the finish line. The speed performance of the car will be affected by a number of parameters that you can address. These are:
1. Position of the mouse trap in relation to the drive axle. The closer that the mouse trap is to the drive axle the faster the vehicle will travel. If the drive wheels slip it is too close to the drive axle.
2. Maximise traction force between the wheels and the track by using rubber bands or section of a balloon.
3. Decrease the rotational inertia of the wheels. This can be done by removing mass from the inside of the wheels or using smaller diameter wheels.
4. Adjusting the wheel-to-axle diameter ratio by adding or removing tape on the drive axle.
5. Ensure that string alignment is attached directly over the drive axle when the lever arm is in the fully wound position.
There are five appendices that provide further information on the theoretical basis for the above design issues.
Theoretical Calculations
Table 1 contains results from an imaginary test to determine the spring constant and hence the stored energy of a mouse-trap spring.
The spring constant, k, is related to the spring lever angle (in radians) and the torque by
is the angle measured from its zero torque position (in radians).q , where q k=T
MECH117 Lab 2 Draft Report Sheet – 2 –
For the data contained in Table 1 calculate the torsional spring constant, in units of Nm/rad.
The potential energy stored in the spring is given by P.E 2 ,q 1 k= qT dò =.
where θ is measured from zero torque position (measured in radians)
(a) Using the equations provided, calculate the stored (potential) energy of the mouse trap spring which has an initial extension of 170°.
Table 1: Imaginary test data
Angle of spring lever
Measured torque T (Nm)
(b) Assuming that 85% of the potential energy is converted into kinetic energy and the vehicle has a mass of 150 grams, calculate the velocity of the vehicle when the lever angle reaches 0°. Also, what would be the theoretical time for the vehicle to travel 5 m (based on the assumption that lever angle reaches 0° at 5m and acceleration is constant).
Figure 1: Velocity Displacement Curve
s (m)
(c) Assuming the vehicle reaches the halfway distance of 2.5 m when the mousetrap is half closed (when θ=85o), calculate the velocity at the half way point. Calculate the constant acceleration and time for each half stage and compare the total time (to travel 5m) with question part (b) above.
(d) Finally, plot (or sketch) the vehicle velocity-displacement graph for part (a) and part (b) on the same plot, commenting on why it differs from the graph shown in Figure 1. Give full justification of your calculations, defining all variables used.
If the pulling wheels do not have enough traction, the problem may be because of the distribution of the weight of the car.
Have you ever heard that front-wheel drive cars are better in snow and ice than rear-wheel drive vehicles? Front-wheel drive cars have the engine located directly above the drive wheels; this helps increase the traction on the front wheels by increasing the normal reaction between the drive wheels and the road. The closer the centre of mass is located to the drive wheels, the more traction that will result.
The balance point of your car, which represents the centre of mass, should be located as close as possible to the drive wheels. If your car continues to slip on the start, you may want to try adding some mass over or near the drive wheels in order to shift the centre of mass towards the drive train.
If your car is slipping, try adding a small amount of clay over the drive wheels. Your car will press harder on the ground, increasing traction, which is one of the keys to greater acceleration.
MECH117 Lab 2 Draft Report Sheet – 3 –
v (ms-1)

Friction is a force that always opposes motion in a direction that is opposite to the motion of the object. Friction occurs when two surfaces slip, slide, or move against one another.
Friction between two surfaces is actually what causes a ball or a wheel to roll. If it were not for friction the ball or wheel would slide or skid. Therefore friction between the wheel and the surface is desirable! But friction between the axle and its bearing is not.
Just as an object at rest tends to stay at rest and an object in motion tends to stay in motion, an object rotating about an axis tends to remain rotating about the same axis unless an external force or torque acts on it. The property of an object to resist changes in its rotational state of motion is called rotational inertia and is a restatement of Newton’s First Law of Motion.
Rotational inertia is the resistance an object has to changes in rotation. Just as inertia for linear motion depends on the mass of an object, so does rotational inertia. But rotational inertia also depends on one more element – the location of the mass with respect to the axis of rotation. The greater the distance between the bulk of an object’s mass and its axis of rotation, the greater the rotational inertia.
It is best to use wheels that have as little rotational inertia as possible. The lower the rotational inertia of your wheels, the less force that is needed to turn or accelerate your wheels. Wheels with a large amount of rotational inertia will have a greater coasting distance, but this performance will be offset by the increased amounts of force and energy required to accelerate the vehicle off the start. Always pick wheels that are lightweight.
Removing mass from the inside of your wheels will improve both speed and distance performance. This idea is extremely helpful in an effort to get cars to go faster off the start line or to get them to travel farther using less torque.
A large axle size means a larger force is transferred to the ground, causing greater acceleration. By decreasing the wheel-to-axle ratio, you will increase the tractive force but at the cost of decreasing the distance that the force is being applied. To achieve quicker accelerations with a speed car, use a wheel or wheels with a large axle or a smaller wheel-to-axle ratio than with your distance car
A perpendicular push or pull provides the greatest amount of rotation for the least amount of effort; for this reason, it is important that the drive axle and lever arm are
correctly positioned for the start. When the string is fully
wound around the drive axle, the position where the string is
tied to the lever arm should be directly above the drive axle. At this point the string will be pulled perpendicularly from the lever arm.

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