Curling!

a) Use the Work-Energy Theorem to relate the distance, the velocity, and the coefficient of friction. Newton’s Second Law may also be useful.

b) Use the relationships found in part (a), and divide the forces due to friction to see if anything can be cancelled.

c) Same hint as in part (a).

a) The Work-Energy Theorem states that:

\begin{equation*}
\Delta K = W_{tot},
\end{equation*}

where the work is done by the force due to friction: \(W_{tot} = -f_r d\), and \(f_r = \mu N\). To solve for the normal force, we can solve Newton’s Second Law in the \({y-}\)direction to obtain:

\begin{equation*}
N-mg=0.
\end{equation*}

Note that the final velocity is equal to zero, so \( K_f =0 \). Substituting everything into the first equation, and solving for \mu, we get:

\begin{equation*}
\mu = \frac{v_i^2}{2 g d}.
\end{equation*}

For the case with the sweepers, \(d= 15 \, \text{m}\), we can write:

\begin{equation*}
\mu_{sweep} = 0.014,
\end{equation*}

and with no sweepers, \(d= 10 \, \text{m}\), we can write:

\begin{equation*}
\mu_{sweep} = 0.020.
\end{equation*}

b) Regarding the ratio between the forces due to friction, and knowing that \(\mu\) is inversely proportional to \(d\), we get:

\begin{equation*}
\frac{f_{r_{\text{sweep}}}}{f_{r_{\text{NonSweep}}}} = \frac{d_{\text{NoSweep}}}{d_{\text{sweep}}},
\end{equation*}

which, with numerical values, is:

\begin{equation*}
\frac{f_{r_{\text{sweep}}}}{f_{r_{\text{NonSweep}}}} = \frac{2}{3}.
\end{equation*}

c) Since \(f_r=\mu N\) and \(N=mg\) in this situation, the force due to friction is directly proportional to the mass. If the mass is doubled, then \(f_r\) is also doubled, which gives:

\begin{equation*}
2 f_r = \mu (2 m) g.
\end{equation*}

For a more detailed explanation of any of these steps, click on “Detailed Solution”.

a) To find the ice kinetic friction coefficient, we should start by relating the friction force produced by the ice to the other variables. Notice that we know the initial speed (it is 2 m/s), and we also know that the final speed is zero (since the stone eventually stops). Hence, we can easily compute the change in kinetic energy, which, from the work-energy theorem, must be equal to the total work over the stone. So using that theorem and the change of kinetic energy, we can find the work done by the friction force, and with that find the friction coefficient.

The work-energy theorem states that

\begin{equation}
\label{Curling_teoremaTrabajoEnergia}
\Delta K = W_{tot},
\end{equation}

where \(\Delta K\) is the change in kinetic energy (\(\Delta K = K_f-K_i\) ), and \(W_{tot}\) is the total work on the stone. Now, to find the total work, let’s start by identifying the forces on the stone.

Once it is sliding, there are three forces on the stone, namely, the weight, the normal produced by the floor and the kinetic friction, as illustrated in figure 1 on the force diagram (notice that we use a coordinate system with X pointing in the direction of motion of the stone):

Figure 1: Free-body diagram for the stone with three forces exerted on it: the contact force with the ground \(\vec{N}\), the weight \(\vec{W}\) and the friction \(\vec{f}_r\). The coordinate system is chosen such that the positive X axis is parallel to the stone’s trajectory. We also included the displacement \(\vec{d}\) of the stone, because we will need it to find the work.

Now, to find the total work, we need to find the work produced by each one of these forces. In general, the work is given by

\begin{equation}
W = \vec{F} \cdot \vec{d},
\end{equation}

where \(\vec{F}\) is the force and \(\vec{d}\) the displacement vector. One can write the dot product as

\begin{equation}
W = F d \cos \theta,
\end{equation}

where \(\theta\) is the angle between the two vectors, and \(F\) and \(d\) their magnitudes (we know \(d\), it is 15 meters for the case of sweepers and 10 meters without sweepers). So, with this equation, we can find the force performed by each force. From the diagram, we clearly see that both the normal force and the weight are perpendicular to the displacement, thus \(\theta\) is \(\pi/2\). We also know that \( \cos \pi/2 = 0\), which means that these forces do not do any work on the stone (this is always true: forces perpendicular to the displacement of the body do not exert any work). Hence, the only other force that can do work is friction.

From the force diagram, we see that the friction is anti-parallel to the displacement (they point in opposite directions), meaning that the angle between them is \(\pi\). And \( \cos \pi = -1\). Hence,

\begin{equation}
W_{f_r} = – f_r d.
\end{equation}

The fact that the work of friction is negative means that the body is loosing energy due to the action of that force, which in turn means that the speed of the body is decreasing due to the force.

So, since the only force that produces work is the friction, we have

\begin{equation}
W_{tot} = W_{f_r},
\end{equation}

and so

\begin{equation}
W_{total} = – f_r d.
\end{equation}

Let’s use this in equation \eqref{Curling_teoremaTrabajoEnergia} to get

\begin{equation}
\label{Curling_cineticaFriccion}
\Delta K = – f_r d.
\end{equation}

To continue, we need to find \(f_r\). We know that the dynamic friction is given by

\begin{equation}
\label{Curling_friccion}
f_r = \mu N,
\end{equation}

where \(N\) is the magnitude of the normal force produced by the surface, and \(\mu\) is the coefficient of dynamic friction. To find \(N\), let’s use Newton’s second law along Y. Given our force diagram, it is clear that the weight is negative in Y and the normal is positive, and so we have

\begin{equation}
N \, \hat{\textbf{j}} – W \, \hat{\textbf{j}} = m a_y \, \hat{\textbf{j}}.
\end{equation}

But there is no acceleration in Y (only in X), and so

\begin{equation}
N \, \hat{\textbf{j}} – W \, \hat{\textbf{j}} = 0 \, \hat{\textbf{j}}.
\end{equation}

If we use that \(W=mg\), and focus on the magnitudes, we get

\begin{equation}
N – mg = 0.
\end{equation}

Hence,

\begin{equation}
\label{Curling_normal}
N = mg.
\end{equation}

Use this in equation \eqref{Curling_friccion} to get

\begin{equation}
\label{Curling_friccionConMasa}
f_r = \mu (mg).
\end{equation}

Now we can use this in equation \eqref{Curling_cineticaFriccion} to get

\begin{equation}
\Delta K = – (\mu mg) d.
\end{equation}

We finally have an expression where the only unknown variable is \(\mu\). To use it, let’s write \(\Delta K \) explicitly:

\begin{equation}
\frac{1}{2} m v_f^2 – \frac{1}{2} m v_i^2 = – \mu m g d.
\end{equation}

However, the final speed is zero (the stone stops), and so we have

\begin{equation}
– \frac{1}{2} m v_i^2 = – \mu m g d.
\end{equation}

If we divide by \(-mgd\), we get

\begin{equation}
\frac{-mv_i^2}{-2m g d} = \mu.
\end{equation}

The \(m\) and the \(-1\) cancel and we get

\begin{equation}
\label{Curling_mu}
\frac{v_i^2}{2gd} = \mu.
\end{equation}

Finally, insert the numerical values here for the case with sweepers and the case without sweepers. For sweepers, \(d\) is 15 meters, and so we have

\begin{equation}
\mu_{\text{sweep}} = \frac{(2 \, \text{m/s})^2}{2(9.8 \, \text{m/s}^2)(15 \, \text{m})},
\end{equation}

to get

\begin{equation}
\mu_{\text{sweep}} = 0.014.
\end{equation}

If there are no sweepers, we have

\begin{equation}
\mu_{\text{NoSweep}} = \frac{(2 \, \text{m/s})^2}{2(9.8 \, \text{m/s}^2)(10 \, \text{m})},
\end{equation}

to get

\begin{equation}
\mu_{\text{NoSweep}} = 0.020.
\end{equation}

These results are intuitive: the whole purpose of the sweepers is to reduce the friction between the stone and the ice so that the stone can move for a longer distance. And it takes a lot of training to figure out how much to sweep to control the distance travelled by the stone to a high precision!

b) Now, we want to find the ratio between the friction force with sweepers and the friction force without sweepers. That is, the ratio we want to find is

\begin{equation}
\label{Curling_cocientes}
\frac{f_{r_{\text{sweep}}}}{f_{r_{\text{NonSweep}}}}.
\end{equation}

As we explained before, \(f_r= \mu N\). Notice that the normal force does not change between the two cases, since the normal force is given by the weight (see equation \eqref{Curling_friccion} above) and the weight is not changed by the sweepers (the mass of the stone is not changed). So the only thing that changes in the two cases is \(\mu\) :

\begin{equation}
f_{r_{\text{sweep}}} = \mu_{\text{sweep}} N,
\end{equation}

and

\begin{equation}
f_{r_{\text{NonSweep}}} = \mu_{\text{NoSweep}} N.
\end{equation}

If we use these expressions in \eqref{Curling_cocientes}, we get

\begin{equation}
\frac{f_{r_{\text{sweep}}}}{f_{r_{\text{NonSweep}}}} = \frac{\mu_{\text{sweep}}N}{\mu_{\text{NoSweep}}N}.
\end{equation}

The normal force cancels and we get

\begin{equation}
\frac{f_{r_{\text{sweep}}}}{f_{r_{\text{NonSweep}}}} = \frac{\mu_{\text{sweep}}}{\mu_{\text{NoSweep}}}.
\end{equation}

At this point, the reader might be tempted to use the numerical values of the coefficients found in point (a), but the result would not be as precise because there are a lot of decimal digits that we have to ignore. Indeed, the reader could easily verify that the result of using the numerical values in this equation is \(0.014/0.020=0.7\), whereas the exact result, as we will see now, is \(2/3\). For the exact method, use the results from equation \eqref{Curling_mu} with the corresponding distances to get

\begin{equation}
\frac{f_{r_{\text{sweep}}}}{f_{r_{\text{NonSweep}}}} = \frac{\frac{v_i^2}{2gd_{\text{sweep}}}}{\frac{v_i^2}{2gd_{\text{NoSweep}}}}.
\end{equation}

Most terms on the right cancel, and we get

\begin{equation}
\frac{f_{r_{\text{sweep}}}}{f_{r_{\text{NonSweep}}}} = \frac{d_{\text{NoSweep}}}{d_{\text{sweep}}}.
\end{equation}

If we insert the numerical values, this gives us

\begin{equation}
\frac{f_{r_{\text{sweep}}}}{f_{r_{\text{NonSweep}}}} = \frac{(10 \, \text{m})}{(15 \, \text{m})},
\end{equation}

and the result is

\begin{equation}
\frac{f_{r_{\text{sweep}}}}{f_{r_{\text{NonSweep}}}} = \frac{2}{3}.
\end{equation}

Clearly, the friction is reduced.

c) What would happen to \(\mu\) if the mass of the stone doubles?

Well, a way to answer this is by going back to equation \eqref{Curling_mu}. Let’s write it here again

\begin{equation}
\label{Curling_muSegundo}
\mu = \frac{v_i^2}{2gd}.
\end{equation}

Notice that the coefficient does not depend on the mass, and so it will not change if the mass is doubled. But we actually did not even need to use equation \eqref{Curling_mu} to answer this. Conceptually, the coefficient of dynamic friction does not depend on the mass of an object but it only depends on the type of surfaces (the materials) that are in contact  (in more advanced cases, it can also depend on the temperature of the surfaces). Indeed, the coefficient is useful in the design of roads precisely because one can estimate it without having to take into account the particular mass of each car that goes over it!

Now, if the mass doubles, it is easy to see that the magnitude of the friction doubles. Just look at equation \eqref{Curling_friccionConMasa} that shows that

\begin{equation}
f_r = \mu mg.
\end{equation}

If the mass is doubled here, then \(f_r\) is doubled as well:

\begin{equation}
2 f_r = \mu (2 m) g.
\end{equation}

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