15.4 Pendulums  (Page 3/7)

Once again, the equation says that the second time derivative of the position (in this case, the angle) equals minus a constant $\left(-\frac{mgL}{I}\right)$ times the position. The solution is

where $\text{Θ}$ is the maximum angular displacement. The angular frequency is

The period is therefore

Note that for a simple pendulum, the moment of inertia is $I={\displaystyle \int r^{2}dm}=m{L}^2$ and the period reduces to $T=2\pi \sqrt{\frac{L}{g}}$ .

Reducing the swaying of a skyscraper

In extreme conditions, skyscrapers can sway up to two meters with a frequency of up to 20.00 Hz due to high winds or seismic activity. Several companies have developed physical pendulums that are placed on the top of the skyscrapers. As the skyscraper sways to the right, the pendulum swings to the left, reducing the sway. Assuming the oscillations have a frequency of 0.50 Hz, design a pendulum that consists of a long beam, of constant density, with a mass of 100 metric tons and a pivot point at one end of the beam. What should be the length of the beam?

Strategy

We are asked to find the length of the physical pendulum with a known mass. We first need to find the moment of inertia of the beam. We can then use the equation for the period of a physical pendulum to find the length.

Solution

1. Find the moment of inertia for the CM:
2. Use the parallel axis theorem to find the moment of inertia about the point of rotation:
   
   $I=I_{\text{CM}}+{\frac{L}{4}}^{2}M=\frac{1}{12}M{L}^2+\frac{1}{4}M{L}^2=\frac{1}{3}M{L}^2.$
3. The period of a physical pendulum has a period of $T=2\pi \sqrt{\frac{I}{mgL}}$ . Use the moment of inertia to solve for the length L :
   
   $\begin{array}{ccc}\hfill T& =\hfill & 2\pi \sqrt{\frac{I}{MgL}}=2\pi \sqrt{\frac{\frac{1}{3}M{L}^2}{MgL}}=2\pi \sqrt{\frac{L}{3g}};\hfill \\ \hfill L& =\hfill & 3g{\left(\frac{T}{2\pi }\right)}^2=3\left(9.8\frac{\text{m}}{{\text{s}}^2}\right){\left(\frac{2\text{s}}{2\pi }\right)}^2=2.98\text{m}\text{.}\hfill \end{array}$

Significance

There are many ways to reduce the oscillations, including modifying the shape of the skyscrapers, using multiple physical pendulums, and using tuned-mass dampers.

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Torsional pendulum

A torsional pendulum    consists of a rigid body suspended by a light wire or spring ( [link] ). When the body is twisted some small maximum angle $\left(\text{Θ}\right)$ and released from rest, the body oscillates between $\left(\theta =+\text{Θ}\right)$ and $\left(\theta =\text{−}\text{Θ}\right)$ . The restoring torque is supplied by the shearing of the string or wire.

The restoring torque can be modeled as being proportional to the angle:

The variable kappa $\left(\kappa \right)$ is known as the torsion constant of the wire or string. The minus sign shows that the restoring torque acts in the opposite direction to increasing angular displacement. The net torque is equal to the moment of inertia times the angular acceleration:

This equation says that the second time derivative of the position (in this case, the angle) equals a negative constant times the position. This looks very similar to the equation of motion for the SHM $\frac{d^{2}x}{d{t}^2}=-\frac{k}{m}x$ , where the period was found to be $T=2\pi \sqrt{\frac{m}{k}}$ . Therefore, the period of the torsional pendulum can be found using

The units for the torsion constant are $\left[\kappa \right]=\text{N-m}=\left(\text{kg}\frac{\text{m}}{{\text{s}}^2}\right)\text{m}=\text{kg}\frac{{\text{m}}^2}{{\text{s}}^2}$ and the units for the moment of inertial are $\left[I\right]={\text{kg-m}}^2,$ which show that the unit for the period is the second.