The conical pendulum can be a tough topic. Mainly because it relies heavily on numerous past topics. In this episode we start by looking at the forces acting on the pendulum.
The conical pendulum can be a tough topic. Mainly because it relies heavily on numerous past topics. In this episode we start by looking at the forces acting on the pendulum (1:10). Once you apply what you know about forces, you can use knowledge gained about circular motion (4:22). Of course, trigonometry makes its obligatory cameo, and we boil it all down to two simplified equations (5:45).
The Question of the Day asks: (7:13)
If a pendulum is relocated to a planet with one fourth the gravitational acceleration of Earth, how will the period of its rotation compare to on Earth?
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Hi and welcome to the APsolute RecAP: Physics 1 Edition. Today’s episode will focus on the conical pendulum, a sometimes tricky concept.
Let’s Zoom out:
Unit 6 – Simple Harmonic Motion
Topic 6.1-6.2
Big Idea – Force Interactions and Conservation
Tetherball is a popular game played on many school playgrounds and summer camps in the United States. It involves hitting a volleyball tethered to a tall pole. When hit, the tethered ball sweeps out a conical shape as it rotates around the pole. In this episode we will aim to relate the time the ball takes to make a rotation to the angle made by the rope and the pole.
Let’s Zoom in:
For any object to move in a circle, there must be a centripetal force. You may think that the tension in the rope is the centripetal force for the tetherball, but actually it is the horizontal component of the tension that is the centripetal force. Grab a string and tie a mass to the end of it, anything works as long as it is a good amount of mass more than the string itself. Now, with one arm held straight upward in the air holding the open end of the string, slowly set the object in motion so that it rotates around your body. You will notice that the object makes a circle around you, but that the tension is located in the string and is actually at an angle compared to your arm. If that is hard to wrap your head around, it should at least be clear that when rotated slowly, the string is definitely not parallel to the floor, while the circle that the mass is traveling in is parallel to the floor. Now, and be careful here, spin object faster, and faster. You will notice that while the string does get closer to horizontal, it never makes it there.
Sketch the free-body diagram for the mass on the string from the perspective of an individual watching you twirl this mass around your body. Better yet, make a friend do the twirling and step back and watch the madness! It is important to sketch the free-body diagram when the mass would be on either side, not between you or your friend or behind the person swinging it. The FBD should have 2 forces acting on it, gravitational force downward and tension directed at an angle from vertical along where the path the string was located. If you break the tension force into its vertical and horizontal components, it is easy to see that only the horizontal component is along the radius of the circle made by the mass on the string. The vertical component is actually up, and counter to the gravitational force. In fact, it is exactly equal to the gravitational force. The fact that there is always gravitational force when you use a conical pendulum on Earth is why it can NEVER swing perfectly horizontal.
So, the vertical component of tension is equal to the gravitational force, and the horizontal component is then the only force remaining to act as the centripetal force. The gravitational force (m*g) is equal to tension times cosine of the angle the string makes with the outstretched arm. The horizontal component of tension is equal to the tension force times sine of that same angle and can be set equal to the equation for centripetal force. Centripetal force is mass multiplied by the mass’ linear speed squared divided by the radius of the circle made by the object on the string. Additionally, if you remember that constant linear speed can be found by distance divided by time, we can replace that with the circle’s circumference (the distance for one rotation) divided by the period (the time for one rotation).
All of this boils down to a nice equation that can predict the period of one tetherball rotation in terms of the radius of rotation, the gravitational acceleration of the planet where the game is being played, and the angle made by the string with vertical. I get period is equal to 2(Pi)multiplied by the square root of the radius divided by little “g” times tangent theta. PHEW! That was a mouthful.
There are some other fancy trigonometry things like substituting trig functions for the ratios they are equal to in the invisible triangle made by the string, plane of the circle and the vertical that can be done to simplify the equation. This way the only factors affecting the period are gravitational acceleration and the height of the triangle made by the string with the pole and the plane of the circle. I get that the period is equal to 2(Pi) times the square root of the height divided by the gravitational acceleration.
The real fun of Physics isn’t manipulating equations, it is when you have that ah-ha moment, when it all clicks. Those moments usually come when you see the physics in action. So, go online, obtain a flying pig toy that flies around in circles on a string, and use a ruler to make some measurements. You will truly see the physics come to life.
To Recap…
A conical pendulum, although somewhat of a nuanced topic, can show up on the Physics 1 exam. When it does, you should rely on your knowledge of drawing FBDs, uniform circular motion, and simple harmonic motion to be able to determine or use the period of the conical pendulum.
Coming up next on the APsolute RecAP Physics 1 Edition, we look at spring oscillators, and the factors that affect how quickly they oscillate.
Today’s Question of the Day focuses on the conical pendulum.
Question: If a pendulum is relocated to a planet with one fourth the gravitational acceleration of Earth, how will the period of its rotation compare to on Earth?