Derivation of time period for physical pendula without calculus

  • #1
danpendr
2
0
Thread moved from the technical forums to the schoolwork forums
TL;DR Summary: I'm stuck trying to find the equation for time period T of a physical pendulum without any calculus using torque.

Hello all.

I am currently writing my IB Physics HL IA (high school physics lab report).

I am investigating the effect of length on the time period of a uniform rod pendulum.

I need to derive the following equation, ideally without using calculus:
1697396880317.png

This website has a good derivation but skips an important step at the end, when stating "This is identical in form to the equation for the simple pendulum and yields a period: EQUATION ABOVE". I was wondering if there was a way to arrive to the equation without jumping through hoops. If anyone could help me continue my derivation I'd be very appreciative. I got as far as this:

1697397217475.png
1697397158902.png


Kind regards
Dan
 
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  • #2
Consider a point on the periphery of a disc rotating at constant speed. What is the relationship between the disc's rotation angle at some instant, the x coordinate of the point and the component of its acceleration in the x direction?
 
  • #3
haruspex said:
Consider a point on the periphery of a disc rotating at constant speed. What is the relationship between the disc's rotation angle at some instant, the x coordinate of the point and the component of its acceleration in the x direction?
haruspex, thank you for your response, but I don't seem to understand. What do you mean by x-coordinate?

Could you show your working out?

Many thanks
Dan
 
  • #4
danpendr said:
What do you mean by x-coordinate?
Take a disc radius r to be rotating about the origin in the XY plane at angular velocity ω. For a point on the perimeter, what is the relationship between its x coordinate and the x component of its acceleration?
 

FAQ: Derivation of time period for physical pendula without calculus

What is a physical pendulum?

A physical pendulum is a rigid body that is suspended from a pivot point and can oscillate back and forth under the influence of gravity. Unlike a simple pendulum, which is an idealized point mass at the end of a massless string, a physical pendulum has an extended shape and mass distribution.

What is the time period of a physical pendulum?

The time period of a physical pendulum is the time it takes for the pendulum to complete one full oscillation. It depends on the distribution of mass in the pendulum, the distance from the pivot point to the center of mass, and the acceleration due to gravity.

How can the time period of a physical pendulum be derived without using calculus?

The time period of a physical pendulum can be derived using the small-angle approximation and principles of rotational motion. The formula for the time period \( T \) is given by:\[ T = 2\pi \sqrt{\frac{I}{mgd}} \]where \( I \) is the moment of inertia of the pendulum about the pivot point, \( m \) is the mass of the pendulum, \( g \) is the acceleration due to gravity, and \( d \) is the distance from the pivot point to the center of mass.

What is the small-angle approximation?

The small-angle approximation assumes that the angle of oscillation is small enough that the sine of the angle can be approximated by the angle itself when measured in radians. Mathematically, this means \( \sin(\theta) \approx \theta \) for small \( \theta \). This simplifies the equations of motion for the pendulum.

What is the moment of inertia and how does it affect the time period of a physical pendulum?

The moment of inertia \( I \) is a measure of how the mass of the pendulum is distributed relative to the pivot point. It affects the time period because it determines the pendulum's resistance to rotational motion. A larger moment of inertia results in a longer time period, as the pendulum takes more time to complete one oscillation. The moment of inertia depends on the shape and mass distribution of the pendulum.

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