YouTube Transcript: Young Modulus - Physics A-level Required Practical
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This content details a practical physics experiment to determine the Young's Modulus of a copper wire, a crucial material property indicating its stiffness and resistance to deformation.
hello i'm mrs wilkins and welcome to
marsbury science
today we're going to look at an a-level
physics-required practical how to
determine the young modulus of a
material and specifically a copper wire
the young modulus is a really important
property in engineering as it tells us
how easily a material will stretch or
deform the young modulus is defined as
the ratio of tensile stress to tensile
strain where stress is the force applied
per unit area and the strain is the
extension relative to original length
the young modulus is given the letter e
and this is equal to f l
over a delta l this is the setup that
we're going to use today we have taken a
fairly long piece of copper wire it is
over two meters because the extensions
are so small you do want fairly long
original length this is just one example
of a setup you can also hang some wires
sometimes steel is the best vertically
suspended from a beam but it depends if
your school laboratory has that sort of
infrastructure that enables you to do it
so this is the best option for us in
this investigation there are two safety
precautions to consider the first is
that if the wire breaks and it may well
do it could snap across the surface of
the eye causing damage so it is really
important to wear eye protection in the
form of safety goggles this second is
that if the wire snaps of course the
slot masses will force the ground
be careful not to have your foot or a
knee underneath the slot masses and
perhaps also place a carpet or a tray of
sand underneath the slot masses to
protect the floor when they fall so the
fourth applied is the tension that we
apply to the wire as you can see we have
clamped the wire at the far end of the
bench and then we've run the wire over a
pulley and attached it to a vernier
scale at the end of the vernier scale we
have the hanger and we are going to
attach slot masses in increments of 100
grams and that will provide the tension
which is mg you will notice that we
actually have two wires attached and
this is because it's important to have a
test wire that we apply the tension
force to and also a comparison wire this
allows us to give us a reference point
and also if there are any changes in the
ambient atmosphere for example if the
wire extends due to temperature it will
happen to both and we can find the
relative extension of the test wire the
next step is to find the diameter of the
wire and for this the best equipment is
a micrometer and this will give us a
resolution to a hundredth of a
millimeter the wire may not be perfectly
uniform throughout and so it's a good
idea to take the diameter measure the
diameter at three separate points and
then calculate the mean i'm going to
take it here you
you
turn the small dial until you hear the
first click
and i can see the reading
to be
0.28 millimeters i then measured the
diameter in the middle of the wire and
at the far end of the wire
the first two readings were the same the
diameter was 0.28 millimeters but the
third was 0.27 millimeters however when
i calculated the mean you still have to
give the final result to two significant
figures and so it still averages out
2.28 millimeters cross sectional area
equals pi d squared over four the next
measurement we require is the original
length of the wire and for this we used
a series of meter rules and found the
original length to be 2.46 meters we
have of course already applied a small
tension to the wire to ensure that the
wire is taught when we took the readings
of diameter and original length and this
was supplied by the hangers already
attached to the vernier scale
however before we add the additional 100
grams we have to make sure that our
vernier scale is perfectly zeroed so if
we go back to our original equation e
equals f l over a delta l we've
accounted for the force we've measured
the original length we've calculated the
cross-sectional area by measuring the
diameter so now we can start to measure
the extension under an applied force by
attaching the slot masses and we will
measure the extension on the vernier
scale so i'm going to start by adding my
first 100 grams because this is our
reference point of zero as i mentioned
before the extensions are very small and
so far i have not seen a significant
extension so i'm going to add another
100 grams i've taken a few readings now
and i can see that adding 500 grams is
now ascended by 1.4 millimeters if
you're not sure how to read vernier
scales remember that there are two
scales the first reading you see where
the zero on the sliding scale where it's
between on the fixed scale so i can see
it's between one and two millimeters so
i know it's one point something
millimeters i then get the next decimal
point by seeing which is the first line
that lines up with a line on the fixed
scale and i can see here that the fourth
line lines up with the fixed scale and
therefore i can say it's 1.4 millimeters
in this investigation although we're
interested in extension in meters our
vernier scale gives us an extension in
millimeters so don't forget to convert
it to meters when plotting your graph
when you have a full set of data we can
now plot the graph there are various
ways of plotting the graph and you could
plot the stress versus the strain but
this is quite a complicated way of doing
it it's more standard to plot the force
against extension however plotting a
force when you have to times the mass by
g 9.81 the values aren't particularly
easy to plot on a graph so we're going
to stick with the mass and the extension
our preferred method is to plot the mass
on the y-axis and the extension on the
x-axis because this gives us quite a
typical stress-strain curve that you'd
be familiar with if you plot the
extension on the y-axis and the mass on
the x-axis which you may also see it
does of course give you the inverse for
the gradient plotting it this way our
gradient gives us the mass divided by
delta l we can then say that the young
modulus e is equal to the gradient
times g times the original length
divided by the cross-sectional area
when measuring the gradient it's really
important that you take the gradient
from the linear part of the graph
your graph may show a linear part and
then it may curve off in which case
that's great because you've shown that
the wire behaves elastically and then
starts to behave plastically
however for determining young modulus
it's really important to only take the
gradient in the linear part and not
include the part after it's gone beyond
the limit of proportionality it's also
important on your graph to plot the mass
in kilograms our gradient is 357
kilograms per meter times that by g 9.81
times it by our original length which
was 2.46 meters and divided by our
cross-sectional area which was 6.2 times
10 to the minus 8 meter squared and we
get a value for the young's modulus e of
1.39 x 10 to the 11 pascals
or 139 giga pascals we can compare this
to the known value of the young modulus
of copper which is
gigapascals we can now take a percentage
error in our value compared to the
theoretical value which gives us a
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