When rolling on the carpet, all the Potential Energy has gone, but there is heat energy. This is caused by friction. In this experiment, two things are affected by friction.
The greater the friction there is between the ball and the surface upon which it is rolling, the more quickly it will stop moving. The smoothness of the carpet will also affect how long it will take the ball to stop rolling, for example, smooth sand compared to sand with bumps so make sure all surfaces are as level as possible. In this case the carpet is not very smooth so there will be more friction.
Prediction
My prediction will be that the more height there is from the carpet the further the cylinder travels. Therefore they will be proportional.
Science involved?
If the height doubles, the so does the distance rolled. So what I will use is the distance rolled divide by height. The reason is because when I do this, I predict the answer will be repeated all the time (e.g. if the height is 7 and the distance is 70, D/H would be 10, and if the height was 6, I’d expect the distance to be 60, because 60/6=10.)
A bit of Working out to Test the Prediction
P.E=Potential Energy, m=mass, g=gravity, h= height, f=force, d=distance.
P.E=mgh and P.E=fd
Therefore mgh=fd
I have predicted that h will be directly proportional to d, so mg must be proportional to f.
An example: M=0.1, G=10, H= 0.05cm, F=0.083333…
FD = 0.05 J
D = 0.05/0.083333…
D =0.60
If this example is the correct formula, then I have it.
Preliminary Results
These are the preliminary results:
I will collect 13 results. This means I will release the cylinder 13 times at regular intervals of 0.5cm (39 in total because I will collect the reading three times and take the average distance.
What have I done?
I obtained all the readings at 1.5cm intervals from the heights 0.8cm – 6.8cm, which covered a good range because it is measured right from the bottom of the ramp and from the top of the ramp (with the same angle).
What will I do?
I will obtain all the readings at 0.5cm intervals from the heights 0.8cm – 6.8cm, that covers a good range because it is measured right from the bottom of the ramp and from the top of the ramp (with the same angle).
Were there any problems? If so what will I do?
There were quite a few problems. Some problems that appeared were that:
- There was a drop in between the end of the ramp and carpet. This would not be a fair test because a drop would mean a sort of bounce, which would slow the cylinder down rapidly. What I will do overcome this problem was to place three equal size books, which ensures that there is no drop to the carpet.
- I also think there would have been some human error, but this would be a very minor, e.g. you may accidentally push the cylinder slightly.
- You may start the cylinder in slightly the wrong place. There a big range in the results so I will take the reading three times, and then I will work out the average.
What will I do with the Readings to test the prediction?
I will make a hand-drawn line graph to show Height v Distance. Distance will be on the x-axis because that is the thing I am trying to find out. On the y-axis will be the height because I can change that manually. I will then draw in the line of best fit. If the graph goes through 0,0 and is straight, then my prediction will be right.
I have done a computer graph just to test my prediction. But this is not the final Graph.
This is the Computer Graph:
It is almost a very straight graph and very nearly goes through 0,0. This is only the preliminary graph and it looks right (according to my prediction.)
From the evidence I have obtained, I can see straight away that the higher the cylinder is up the ramp, the further the cylinder goes on the carpet. For every 0.5 of a centimetre the height increases, the height increases by 5, 6 or 7 (Average of 6.1666666666….).
The speed that the Cylinder rolls down the ramp at can be found by firstly finding the kinetic energy (KE). The KE at the bottom of the ramp is the same as the PE at the top of the ramp. The mass of the cylinder was 250g. The height of the ramp was 6.8 cm. The length of it was 55 cm.
KE gained = PE lost
PE lost = mgh.
PE lost =0.25 x 10 x .068
PE lost =0.17J
KE gained =0.17J
KE =½mv²
0.17J =½ x 0.25 x v²
0.17J =0.125 x v²
V² =0.17 / 0.125
V² =1.36m/s
V = 1.36
VELOCITY =1.1661m/s
The acceleration can be worked out: (v²=u²+2as)
(a=v²-u²)
2s
S (Distance) :0.55m
U (Initial velocity) :0
V (Final velocity) :1.16m/s
A (Acceleration)* :?
T (Time) :
*Where a is a constant
v²=u²+2as
- = 0 + 2(A x 0.55)
- = 2(0.55 x A)
a = 1.36 / 1.1
ACCELERATION = 1.2 m/s² (1d.p)
Now, we can now work out the deceleration on the carpet.
S:0.82m
U:1.16m/s
V:0m/s
A:?
T:
v²=u²+2as
0²=1.16²+ 2(82 x a)
0 =1.36 + 1.64a
-1.36= 1.64a
a = -1.36 / 1.64
DECELERATION = 0.8296m/s²
The time taken for the cylinder to roll between the foot of the ramp and the place where the cylinder stops can be worked out:
S:0.82m
U:1.16m/s
V:0m/s
A:0.8296m/s²
T:
S=½(u+v)t
0.82=½ x (1.16 +0)t
0.82=0.58t
t=0.82/0.58
TIME=1.413s
It takes the cylinder 1.413s to stop.
That is when the cylinder is released at 6.8cm high
When the same cylinder is released at 3.3cm high, the velocity of the cylinder can be worked out:
PE lost = mgh.
PE lost =0.25 x 10 x .033
PE lost =0.0825J
KE gained =0.0825J
KE =½mv²
KE =0.0825
0.0825 =½ x 0.25 x v²
0.0825 =0.125 x v²
0.0825 / 0.125 =v²
v² =0.66m/s
VELOCITY =0.812 m/s
The acceleration can be worked out: (v²=u²+2as)
(a=v²-u²)
2s
S (Distance) :0.27m
U (Initial velocity) :0
V (Final velocity) :0.8m/s (1d.p)
A (Acceleration)* :?
T (Time) :
*Where a is a constant
v²=u²+2as, a=v²-u²
2s
a =0.66
0.44
ACCELERATION =1.2…m/s² (1d.p)
The deceleration on the carpet can be worked out
(a = v²-u²)
2s
S:0. 40
U:0.812m/s
V:0
A:
T:
a=v²-u²
2s
a= -0.659
0.8
DECELERATION= 0.8m/s² (1d.p)
The time for this need to be worked out
S:0.40
U:0.81m/s
V:0
A:0.8
T:?
S=½(u+v)t
- = ½ x (0.81+0)t
0.4 =0.405t
t =0.4/0.405
TIME =0.99s
Another example is when the cylinder was released at 1.8cm high. The velocity can be worked out:
PE lost = mgh.
PE lost =0.25 x 10 x 0.018
PE lost =0.045J
KE gained =0.045J
KE =½mv²
0.045 =½ x 0.25 x v²
0.045 =0.125v²
v² =0.045 / 0.125
v² =0.36
VELOCITY =0.6m/s
The acceleration on the ramp can be worked out:(v²=u²+2as)
Or: a = v²-u²
2s
S:0.15m
U:0m/s
V:0.6m/s
A:?
T:
a=0.36
0.3
ACCELERATION = 1.2m/s
The deceleration on the carpet can be worked out:
S:0.22
U:0.6m/s
V:0
A:?
T:
a = v²-u²
2s
a = -0.36
0.44
DECELERATION= 0.8m/s² (1.d.p)
The time for this can also be worked out:
S=½(u+v)t
S:0.22
U:0.6m/s
V:0
A:0.8m/s²
T:
0.22 =½(0.6+0)t
0.22 =0.3t
t =0.22/0.3
TIME= 0.73333….s
The most obvious things to spot are that the acceleration and the deceleration are the same all the time. This is because there is the same gradient slope is the same and the force of gravity is the same.
To get the velocity for a certain height, I have created a formula:
It is the height at which the cylinder falls from, divided by 5, and then square root that number
Or: (h/5) = velocity
To find the time taken to stop, you do: height at which the cylinder falls from, divided by 5, and then square root that number and then multiply it by 1.2
1.2 (h/5)
So, I predict that on a cylinder where the height of where it falls is 4.3 cm, the speed of the cylinder at the of the ramp is:
4.3/5=0.86
-
=0.927m/s
The time taken for it to stop on the carpet is
0.927 x 1.2= 1.112s.
The results that I got agree with my earlier prediction because I said, “My prediction will be that the more height there is from the carpet the further the cylinder travels. Therefore they will be proportional. If the height doubles, the so does the distance rolled. So what I will use is the distance rolled divide by height. The reason is because when I do this, I predict the answer will be repeated all the time (e.g. if the height is 7 and the distance is 70, D/H would be 10, and if the height was 6, I’d expect the distance to be 60, because 60/6=10.) I was right about this because:
When the cylinder was dropped at 6.8cm, it went 82cm
When the cylinder was dropped at 2.3cm, it went 28
82/6.8 = 12.1 (1d.p)
28/2.3 = 12.2 (1d.p)
They are not exact, but this is de to slight human error, perhaps placing the cylinder 1 or 2mm from the actual starting place.
From my graph (Distance travelled by cylinder on carpet versus Height of the Ramp), I saw that the points would not join in a straight line. But then I drew in the line of best fit. This line goes through 0,0 and is straight, so my prediction was right.
My conclusion is that for every single millimetre that the cylinder gets moved higher on the ramp, as long as there is a slope, the more distance the cylinder will travel on the carpet. This is because of gravity, which has a weight of 10N (9.8N). There is friction that is caused by the rolling of the cylinder on the carpet. The Kinetic Energy at the bottom is the same as the Potential Energy at the top. Potential Energy is the same as mass x gravity x height, force x distance. Kinetic Energy is the same as ½mass x velocity²
By working out the Kinetic energy you could work out a lot of other things such as:
- Velocity at the end of the Ramp
- Acceleration on Ramp
- Deceleration on carpet
- Time taken to stop on the carpet
To sort out the drop in between the end of the ramp and carpet, the way to get out of that hurdle was simple. You had to place some things (I used 3 books) underneath the carpet, and should make the carpet in line with the bottom of the ramp. You should use something hard and flat (for smoothness).
I have found out that the acceleration and, on the same slope, (deceleration on a flat surface), with the same cylinder, will always be the same in m/s².
The further down a cylinder is on a slope, the faster it goes.
Although the points on the graph are not exactly in a straight line, they are very close to being on a straight line. There are 3 possible inaccuracies in this experiment:
- While one is releasing the cylinder on the ramp, he can inadvertently push or hinder the cylinder. Also, one may not place the cylinder on the ramp in exactly a straight line
- The distance is only measured to the nearest centimetre and is subject to rounding up or down
- The surface where the distance is measured may not be uniform so there is excessive or reduced friction.
I think the method I used was good because the graph shows an almost perfect fit.
It would be unlikely that these results would just be a coincidence.
Conclusion
Through this evidence, a firm conclusion can be drawn, especially from the graph.
The higher the cylinder is dropped from, the further the cylinder rolls on the carpet. The evidence of this is firmly supported by the graph. The graph was a straight line going, (or very close to) through 0,0.
Improvements
The follow improvements would be recommended:
- There should be a mechanism that holds the cylinder in place on the ramp, and at the intended height, the cylinder can be released without aiding or hindering it.
- The ground surface must be uniform.
- There should be a mechanism, which ensures that the cylinder is in a straight line at the release of it.
- A computerised device that measures the distance in millimetres, or even micrometers could measure the distance that the cylinder travels.
- The cylinder should be a perfect cylinder with perfect grooves or dents.
Further work
One could:
- For each point on the ramp, repeat the experiment ten times, and then take the average.
- One could do the experiment with different sorts of cylinders
- Wooden cylinders
- Different metal cylinders
- Hollow cylinders
- Different sized cylinders
- One could repeat the experiment but instead of cylinders, one could use different spheres.
