Mass and inertia

Introduction

In this experiment we will !!!We'll put this in later... we need to motivate here why it is interesting and worth doing. One paragraph is fine here.!!!

Background theory

We're only going to deal with one dimension here so we'll neglect the vector nature of the force. Let's consider a force, $F$, which increases linearly with displacement, $\Delta x$. That is, the the force is directly proportional to the displacement
\begin{equation} F \propto \Delta x\,. \label{eq:fxprop} \end{equation} There will be a proporionality constant in order to turn (\ref{eq:fxprop}) into an equation so that (\ref{eq:fxprop}) becomes \begin{equation} F = -k \Delta x\,. \label{eq:fxeqn} \end{equation} The above equation will probably look familiar to you. It has exactly the form of Hooke's law \cite{ref:Hooke} for a linear spring. However, note that we are not assuming a spring. We are just assuming that whatever the force is, it varies linearly with displacement, as shown in (\ref{eq:fxeqn}). We are assuming that $k$ is a positive number here. Now, why exactly are we interested in this type of force? One reason is that it is easy to build devices which exert linear forces. You already know that many springs, as long as they're not stretched too much, exert linear forces. The inertial balance we'll use in this experiment, as long as the swing is not too large, also exerts a linear force. Another reason to consider linear forces is that they are mathematically easy to deal with. So, linear devices are easy to build and easy to analyze. Note that we can write Newton's second law in the case of constant mass as \begin{equation} F=ma=m \frac{d^{2}x}{dt^{2}}\,. \label{eq:nslaw} \end{equation} Let's assume from now on that the equilibrium point of our device (the point where the force is zero) is $x=0$. In this case we can replace $\Delta x$ just with $x$. This makes the math a bit easier and doesn't affect anything. It just means that we are using a ruler to measure displacement whose origin is at the equilibrium point. Then, setting (\ref{eq:fxeqn}) equal to (\ref{eq:nslaw}), and noting that the position is a function of time (that is $x$ is $x(t)$) we can write \begin{equation} m \frac{d^{2}x(t)}{dt^{2}} = -k\,x(t)\;\;\; \mbox{or }\;\;\; \frac{d^{2}x(t)}{dt^{2}} = -\frac{k}{m}\,x(t)\;.\label{eq:nslinear} \end{equation}

The above equation just tells us that the acceleration, $d^{2}x(t)/dt^{2}$, is equal to "`minus some number" ($-k/m$) times the position ($x(t)$). In the language of your calculus class it tell us that we are looking for a function, $x(t)$, whose second derivative is equal to minus some number times the function itself. You may recall from calculus that the functions $A\cos(\omega t + \delta)$ and $A\sin(\omega t+\delta)$ have this property (for $A$ and $\delta$ some constants, which we wont worry about too much for this lab... we'll talk about the constant $\omega$ in a sec.). That is \begin{equation} \frac{d^2}{dt^{2}}\left[A\cos(\omega t + \delta)\right] = -\omega^{2} A\cos(\omega t+ \delta)\;\;\; \mbox{and }\;\;\; \frac{d^2}{dt^{2}}\left[A\sin(\omega t+ \delta)\right] = -\omega^{2} A\sin(\omega t +\delta)\,, \end{equation} where $\omega^{2}$ is our "some number". Try it. Just take the function $x(t)=A\cos(\omega t + \delta)$ and put it in place of $x(t)$ in (\ref{eq:nslinear}) and notice that the left-hand-side of the equation will equal the right-hand-side of the equation as long as we set $\omega=\sqrt{k/m}$. That is, $\omega$ can't be just anything in our case. We must set it equal to $\sqrt{k/m}$. So, from this we have established that, for a linear restoring force, the position of the mass is given by the function $x(t)= A\cos(\sqrt{k/m} t + \delta)$ (we'll only use cosine from now on). In summary, we now know that the motion of our mass, subject to a linear restoring force, moves as follows as a function of time: \begin{eqnarray} \mbox{position:}\qquad &x(t)=A\cos(\sqrt{k/m}\,t + \delta)\,, \label{eq:pos} \\ \mbox{velocity:}\qquad &v(t)=\frac{dx(t)}{dt}=-\sqrt{\frac{k}{m}}A\sin(\sqrt{k/m}\,t + \delta)\,, \label{eq:vel} \\ \mbox{acceleration:}\qquad &a(t)=\frac{dv(t)}{dt}=-\frac{k}{m}A\cos(\sqrt{k/m}\,t + \delta)\,. \label{eq:accel} \end{eqnarray} For this experiment we will concentrate on the acceleration, (\ref{eq:accel}). In figure (\ref{fig:accelgraph}) we plot the acceleration (the right-hand-side of equation (\ref{eq:accel})) for some random values of the constants $k$, $m$, $A$ and $\delta$.

Recall from your physics class that the time interval shown as $T$ in the figure is known as the period of the motion. Also recall from your physics class that the period is the inverse of the frequency, $f$, of the motion (the number of oscillations per unit time). That is $T=1/f$. One other thing to recall from your physics class is that the frequency is given by \begin{equation} f= \frac{1}{T}=\frac{1}{2\pi} \omega = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \,. \label{eq:frequencyeq} \end{equation} By looking at the second term and last term in the above set of equalities, we note that \begin{equation} T = 2\pi \sqrt{\frac{m}{k}}\,, \end{equation} or, more important for us, by squaring the above relation: \begin{equation} T^{2} = \frac{4\pi^{2}}{k} m \quad \mbox{or} \quad T^{2} \propto m\,. \label{eq:tmreln} \end{equation} That is, the mass of the object is proportional to the period-squared. If we can measure the period of the motion, we can determine the mass, and this does not require weighing anything. It just requires measuring a time, $T$. We're going to use the above proportionality, (\ref{eq:tmreln}), to measure the mass of an object without weighing it. Specifically we're going to use what's called an inertial balance in order to determine the mass of various objects. An inertial balance is a device that looks something like what is depicted in figure \ref{fig:ib}.

You'll put a mass in the inertial balance tray and when you pull one end of the inertial balance to one side by a small amount the inertial balance will exert a linear restoring force on the mass. When you let go of the balance you'll notice it will move back-and-forth and this motion is described by the equations (\ref{eq:pos} - \ref{eq:accel}). By analyzing the period of this motion we can determine the mass $m$. There's one problem however. The problem is that the inertial balance itself also has some mass, so the mass that's moving back-and-forth is not just the lump that you're interested in, but also some of the balance itself. We'll have to figure out how to take this into account.

How we'll determine the mass (Procedure)

First we'll get the inertial balance ready. Clamp one end of the balance to the table as shown in figure \ref{fig:ibphoto}.

It is difficult to measure the position or velocity of the balance when it is in motion but, as fortune would have it, we have an accelerometer in the i/o Lab that can measure the acceleration. So, the first thing we need to do is mount the i/o Lab device to the inertial balance tray so that it can measure the acceleration. Stick the i/o Lab underneath the tray of the balance using double-sided tape as shown in the above photo and, to make sure it doesn't fall off, also strap it to the tray with the elastic cord. Now, what we want to do is put masses in the tray, set the tray swinging back-and-forth, and measure the period. Recall that by measuring the period we can determine the mass. Unfortunately it's not quite that straight-forward. First, note from (\ref{eq:tmreln}) that the period-squared is not equal to the mass, it is only proportional to it, so by measuring the period we can not determine the mass itself. Second, as stated previously, when we place a mass in the balance and swing it the mass we're interested in isn't the only mass that's swinging. The mass of the balance tray, as well as the mass of the i/o Lab, are (is?) also swinging along with our mass. So, not only do we not know what the proportionality constant is, we are not even measuring the period for the mass we're interested in, but instead for our mass plus the tray plus the i/o Lab. That is, if our mass is $m$, $m_{t}$ is the mass of the balance tray, and $m_{L}$ is the mass of the i/o Lab then the total mass we have swinging is \begin{equation} m_{\mbox{tot}}=m+m_{t}+m_{L}\,. \label{eq:mtot} \end{equation} One thing we could do is try to measure the constant $k$. For this we'd need a force sensor. We would pull the inertial balance to one side with the force sensor, maybe by about one centimeter (0.01m) and measure the force required to do this. This force, divided by 0.01m, gives us the constant $k$ in Newtons per meter (since, from (\ref{eq:fxeqn}), $k=F/x$). Once we know $k$ we could set the empty tray with the i/o Lab strapped to it swinging, and determine the mass of the tray+i/o Lab via the equality in (\ref{eq:tmreln}). This is problematic for us. First, as previously stated it is difficult to measure the sideways displacement of the tray accurately, even when it is not in motion. Second it requires a force sensor, (although we do have one with the i/o Lab). We're not going to do it this way, but you're welcome to try and compare your results with the method we're going to use. !!!Any other reasons for not doing this? We don't want students to think we're doing something in a complicated way just for the sake of making it complicated!!!

So how are we going to handle this if we don't know $k$? Well, we're going to need one more mass of known mass in order to do this. We're going to set the balance swinging without the unknown mass, and then again but now with the unknown mass. !!!I am going to assume the person doing this experiment has been tutored on how to use the i/o Lab and its software!!! OK let's get started to see how this helps us determine the mass of the i/o Lab plus tray.

With the i/o Lab mounted to the tray (but no other mass) set the the tray swinging back-and forth by displacing it about 1cm to one side and letting go. Measure the acceleration of the motion with the i/o Lab. From the graph you can determine the period of the motion. Record your data in the table \ref{table:t1}. Let's call this period $T_{1}$. Next, repeat what you just did but now with the known mass (we'll call the known mass $m_{k}$) and call this period $T_{2}$ and record the results in table \ref{table:t1}.

Now, notice that if we take the ratio of the two periods-squared, all the irrelevant constants drop out. That is, we don't need to know $k$ this way since in the ratio \begin{equation} \frac{(T_{1})^{2}}{(T_{2})^{2}} = \frac{\frac{4\pi^{2}}{k} m_{1}}{\frac{4\pi^{2}}{k} m_{2}} = \frac{m_{1}}{m_{2}}\, , \label{eq:massesratio} \end{equation} the $k$ (along with other constants) cancels out.

Now, recall that in this experiment $m_{1}$ is the mass of the i/o Lab ($m_{L}$) plus tray ($m_{t}$), and $m_{2}$ is the mass of the known mass ($m_{k}$) + i/o Lab + tray. That is, in this case we can write equation (\ref{eq:massesratio}) as \begin{equation} \frac{(T_{1})^{2}}{(T_{2})^{2}} = \frac{m_{1}}{m_{2}}=\frac{m_{t}+m_{L}}{m_{k}+m_{t}+m_{L}}\,. \label{eq:newmassratio} \end{equation} Algebraically manipulating the equality of the left and right parts of the above equation we get \begin{equation} m_{t}+m_{L} = \frac{m_{k}}{\left(\frac{T_{2}}{T_{1}}\right)^{2}-1}\,. \end{equation} Notice that everything on the right-hand-side of the above equation is known. Therefore, we now know the left-hand-side. That is, we know what the mass of the tray plus i/o Lab is. Record this value in the table. OK, now we're ready to actually determine the mass of unknown masses. In fact, this is really no different that what we just did.

 

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