8
$\begingroup$

Suppose you have a cylindrical mass you used as a pendulum bob. When is it possible to approximate the cylinder as a point particle?

$\endgroup$
1
  • 2
    $\begingroup$ It's always possible, the question becomes how good of an approximation do you need? The approximation will always be wrong, it's just a matter of how wrong. $\endgroup$ Aug 16 at 14:22
10
$\begingroup$

It depends on how small the size of the cylinder is compared to the rope/string. You want the rope as long and light as possible, and the bob as small and heavy as possible.

You also want the bob to be much heavier than the string.

Mathematically, you want the moment of inertia of the bob about the rotation point to be as close to $mL^2$ as possible, where $m$ is the bob mass and $L$ is the string length.

Next, you want the moment of inertia of the rope about the center of rotation to be as close to zero as possible (which you can achieve by using a very light rope).


For some more details, my conclusions above were based on:

1) Moment of inertia. A perfect point-like pendulum has a moment of inertia about the rotation point of $I=mL^2$ where $L$ is the rope length and $m$ is the bob mass. Since the net torque is $\tau = I\alpha$, if we can get the moment of inertia $I$ as close to that of a point particle as possible, we will most likely witness simple pendulum behavior, since $\alpha$ will be similar.

A physical pendulum bob (i.e. real-life, not simple pendulum) has a moment of inertia about the rotation point of $I_B=mR^2+I_\text{CoM}$ where $I_\text{CoM}$ is the moment of inertia the bob's center of mass (which would go to zero if it were a point mass). We will approximate the moment of inertia of the bob about its center of mass as $I_B\approx md^2$ where $d$ is some geometric parameter relating to length (i.e. the smaller $d$, the smaller is the bob).

Since the rope has weight, its moment of inertia about the rotation point is $I_R=\dfrac13 ML^2$. The rope's mass is related to its length. Let's assume that we use a rope with constant linear density $\mu=\dfrac ML$ such that $I_R=\dfrac 13 \mu L^3$.

Since, in real life, both the pendulum bob and rope make up our rotating system, the total moment of inertia about the rotating point is $$I=mL^2 + md^2 + \dfrac13\mu L^3$$ and we want to get this as close to $mL^2$ as possible (to resemble the moment of inertia of a simple pendulum).

We are therefore minimizing $md^2 + \frac13\mu L^3$ to get it close to $0$. Some algebra later, we are trying to get $m\left(\dfrac dL \right)^2 + \dfrac13 \mu L$ close to zero. We can decrease the first term by making $L$ very long and $d$ very small. We can decrease the second term by making $\mu$ very small (which lets us increase $L$ to minimize the first term).

These changes correspond to: using a long rope (high $L$), a small object (low $d$), and a light rope (low $\mu$).

However, this does not explain why $m$ should be large. For that, we need to look at:

2) Gravity. Gravity acts through the center of mass of an object. In the case of a simple pendulum, gravity would act on the point-like bob, where all the pendulum mass is concentrated. In real life, however, the rope has mass, which shifts the center of mass of the rope + bob system away from the pendulum bob, towards the rotation point. By using a light rope and heavy mass, the center of gravity will shift back towards the bob, and remain on the pendulum bob. Thus, gravity will act around the same point in the physical case as it would in the simple case, such that the net torque $\tau$ is similar.

Also, a large $m$ helps minimize air resistance, which will also impact your motion.

Hope this clarification helps.

$\endgroup$
4
$\begingroup$

Use the center of the bob, the approximation as a point mass ist bad for a short pendulum and gets better the longer it's overall length is, compared to the length of the bob.

$\endgroup$
2
$\begingroup$

There are two ways I can interpret the question:

  1. You have perforated the cylinder and put a pin through it can turn around it.

  2. You have a cylinder dangling from a string.

In the first case you have what is called a physical pendulum (Hypertext Summary), only slightly different from an idealized pendulum.

In the second case you could treat it as a pendulum with a top at the end. Whether or not you should consider the rotational degrees of freedom of the cylinder depend on how you attach it (at what point longitudinally speaking). It will also matter if you add spin in the beginning.

For example, if you attach the string at the end of the cylinder you have effectively a double pendulum(see chaotic motion section), the second being a physical pendulum. If the movement is not confined to a plane then you have a (or two) rotating pendulum(i).

On the other hand, if you attach it closer to the middle you have to consider the possibility of the cylinder acting like a Lagrange Top. For example

Qualitatively speaking though, the less moment of inertia the cylinder has (shorter it is), the better the approximation is.

$\endgroup$
0
$\begingroup$

Let the system have inertia $I$, the rope have negligable mass, and be put to oscillate in a uniform gravitational field.

Just write down the basic equations:

$$I \alpha =\sum_i \vec{r_i} \times \vec{dF_i} \tag{1}$$

In a gravitational field, the force vector on each mass is given as $ d \vec{F}_i = dm_i g \hat{j}$, we can write (1) as:

$$ I \alpha = \left[ \sum_i \vec{r}_i dm_i \right]\times (g \hat{j})$$

But what is $\sum \vec{r}_i dm_i$? That's the definition of centre of mass. Hence,

$$ I \alpha =M r_{com} \times g \hat{j} \tag{2}$$

Point of interest: The right side for a rigid body of equation (2) and the bob equation is of the same, only the left side differs ( Inertia expression). Hence, we can take time period of pendulum case and replace with inertia of the rigid body system to find time period of this new system.

A way to think about it is, for a uniform gravitational field, the 'push' is independent of geometry but how the body rotates according to the push is different.

One may compare this with translation for a rigid body vs a point object, where both the point mass and rigid body will follow the same law given their force laws acting on them is the same.


On the mention of approximation

The inertia of a rigid body is given as the inertia of a point mass + a correction term for the geometry:

$$ I_{net} = Md^2 + Mk^2$$

The $Mk^2$ is given by the integral $\int r^2 dm$, for the small body limit, we can approximate $ r \to 0$ when body is tiny and hence the integral vanishes and we return to the regular point mass case.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.