4.1 Examples, exercises and experiments

We first complete the drawing. As we need to think about the angular momentum, we need the position vector of $m$. Furthermore, we need to include all forces acting on $m$ in order to evaluate the effect of all torques on $m$.

Apart from gravity also a normal force from the slope on $m$ is present.

Note: in the figure we have now for the position vector, the velocity and both forces set the size of $m$ to zero. As a consequence, $\vec{r}$ and $\vec{v}$ are parallel during the entire motion

Next step, we develop our solution strategy. As this exercise is concerned with angular moment, we will use N2 for angular momentum as well as, of course, the definition of angular momentum $\vec{l} \equiv \vec{r} \times \vec{p}$.

Now we evaluate our ideas:

since, in this case $\vec{r} // \vec{v}$.

N2 for angular momentum reads as:

We know that the left hand side is zero: $\vec{l}$ is constant. Thus, the sum of all torques on $m$ must also be zero. We have identified two forces acting on $m$: gravity and the normal force. The latter does nothing but opposing the component of gravity perpendicular tot he slope (that is its $y$-component). We split gravity in its $x$ and $y$-component:

with this we compute the total torque on $m$

The first term is zero because $\vec{r} // \hat{x}$ and the second one due to cancelling of the two forces in the brackets.

If we assess our answer, we see that this is perfectly in line with N2 for angular momentum: no net torque, hence no change of the angular momentum!

We will exploit that the angular momentum of the earth and the comet are constant. The angular momentum is defined as $\vec{l} = \vec{r} \times \vec{p}$. At the perihelion and at the aphelion, the position vector and the velocity are perpendicular. Thus, the angular momentum simplifies to $\vec{l} = x_hmv \hat{z}$, with $x_h$ the $x$-coordinate of the perihelion or aphelion.

We can find these coordinates by realizing that at the perihelion and the aphelion, the $y$-coordinate is zero. Thus,

where we have labelled the $x$-coordinate of the perihelion as $x_{ph}$ and of the aphelion as $x_{ah}$.

The velocity ratio of $v_{ph}$ over $v_{ah}$ is now easily found, by using that the angular momentum is a constant:

Putting in the given numbers, we find:

Does this make sense? Yes: the earth orbits in an ellipse that is close to a circle. Hence we expect only a mild difference between the velocity at the perihelion and the aphelion. For Halley’s comet this is rather different. According to Kepler’s law of equal area, the comets sweeps through the same area close to the sun (i.e. around the perihelion) and far away from the sun (the aphelion). As the distance to the sun close to the perihelion is much smaller than that around the aphelion, the velocity much be much smaller around the aphelion to ensure equal areas in equal times. Note: our expressions are dimensionally correct.

The energy of the object is given by:

We can find the potential from the force:

where we have used that the force is central. Furthermore, we have taken as our reference point: infinity.

If the object follows a parabola, then its energy must be exactly zero. Thus:

a) This holds at any point of the trajectory. Thus for the point of closest distance to the force center we get:

b) Now that we have both the velocity and the position at one point on the parabola, we can compute the angular momentum. Since the force is central, we do know that the angular momentum is a constant. Hence, computing it for one particular point will give us the angular momentum for all points on the trajectory.

We use that at the point of closest approach the position vector and the velocity are perpendicular: $\vec{r}_0 \perp \vec{v}_{esc}$. Thus the angular momentum is:

Note: the minus sign is needed to have the angular momentum point into the drawing as $\hat{z}$ points towards us.

According to Kepler 3, the orbital period, $T$, and the length of the semi major axis, $a$ are related as:

with k a know constant, the same for all planets orbiting the sun. From this law, we immediately get that

Thus, if we want to find the relative distance to the sun of each planet (that is, we express $a$ in terms of $a_{earth}$ which is 1 A.U.), we get:

Data

Fun fact: Neptune’s orbital period is so large, that since its discovery in 1846, it has only completed 1 Neptune-year (first full orbit in 2011!).

First we make a sketch.

We have drawn gravity and the tension in the rod connecting the mass m tot the pivot point. From a physics point of view, the situation is as follows: the swing initially has only potential energy and it will gain kinetic energy on its way down, until all potential energy is converted into kinetic energy.

Two forces are acting on the swing, giving rise to two torques that cause the angular momentum to change. The velocity of the mass is always perpendicular to its position vector (a consequence of using a ‘clever’ choice for the origin of our coordinate system).

Now, we can set up the modeling for this problem.

Conservation of energy:

Angular momentum:

Change of angular momentum:

With this, we are ready for the develop phase. Conservation of energy gives us for the velocity at the lowest point:

The minimum angular momentum is of course $l_{min} = 0$ at the highest point where the swing has zero velocity and changes direction.

The maximum angular momentum is at the lowest point: $l_{max} = mL\sqrt{2gL}$

Finally, the torque comes from gravity: $\vec{\tau} = \vec{r} \times -mg\hat{y} = mgL\sin \theta \hat{z}$. The tension in the rod is always parallel to the position vector of $m$ and thus has zero torque.

a) The seesaw is balanced if: $\sum_i \vec{r}_i \times \vec{F}_i = 0$. Hence, we compute the net torque:

Thus the net torque is not zero and, indeed, the seesaw is unbalanced.

b) Where to place on the right hand side an additional mass $m$? We need to compensate $-mgL\hat{y}$ by adding $2m$ at some $x\gt0$. Thus, we require:

and we find that we need to place $2m$ at $x=\frac{1}{2}L$.

c) In the second situation, we need to rethink how to compute the total torque. It still requires to calculate the contribution of all small masses $dm$. We will label each small mass with index $i$. It has a distance to the origin of $x_i$. (Actually it does not have a single distance, but a varying one between $x_i$ and $x_i+dx$. But since $dx$ is very small (and the same for all masses $dm_i$), it is accurate enough to use $x_i$. In your calculus classes you will get the proof of this.)

Next, we realize that each mass $dm_i$ is given by

Thus, the net torque can be written as

This is a Riemann sum and if we let $dx \rightarrow 0$, we arrive at an integral:

In the final step, we will replace $\rho$ in terms of the mass of the bar. We do that by noting that the total mass of the bar is the sum of all small parts $dm_i$:

This is to be expected: the total length of the bar is $\frac{7}{4}L$ and with a constant density per unit length we get the above result.

Thus in terms of $m$ we find:

If we substitute this in the equation for the net torque, we find:

Indeed, the seesaw is unbalanced.
Note: we have an expression for the torque with the right dimensions!

d) Now we can find the position where we should place a mass $2m$ to balance the seesaw: we need to do that such that the net torque is zero.

Conclusion: by splitting up the real bar in infinitesimally small parts, we can still compute the net torque (with many thanks to Bernhard Riemann, 1826-1866, a German mathematician).

First, we complete the sketch by adding the force.

We note that the masses move under the influence of a central force. Hence, no torque is exerted and we conclude from N2 for angular momentum that the angular momentum is constant. From this, we can find the momentum and, hence, the velocity of each mass.
Subsequently, we can compute the kinetic energy in the initial and final situation. From that, we easily find the work done as $W_{12} = E_{kin,2} - E_{kin,1}$.
Our final step will then be: compute the force from the definition of work.

Here we go:

If we substitute the initial and final conditions, we get:

The work done is:

From the definition of work, we can find the force, using that it is central and of constant magnitude:

Thus, we get for $F$:

Assess: we have the right units. Moreover, the value we have found is negative. This makes sense: to get $m$ from a distance $R$ to $\frac{R}{2}$ we need an inwards pointing force. Thus, $F$ must be negative as the central force is written as $\vec{F} = F\hat{r}$. The direction of the force follows from the sign of $F$, negative to make it attracting to the origin!