Problem 49 - University Physics with Modern Physics

Question

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 50.0 \(\mathrm{cm} .\) She finds that the pendulum makes 100 complete swings in 136 s. What is the value of \(g\) on this planet?

Step-by-Step Solution

Answer

The value of \( g \) on this planet is approximately 5.34 \( \text{m/s}^2 \).

Step 1: Identify the pendulum formula

The period of a simple pendulum is related to the acceleration due to gravity by the formula \( T = 2\pi\sqrt{\frac{L}{g}} \), where \( T \) is the period, \( L \) is the length of the pendulum, and \( g \) is the acceleration due to gravity.

Step 2: Calculate the period of the pendulum

Since the pendulum makes 100 complete swings in 136 seconds, we first calculate the period of one complete swing: \( T = \frac{136\, \text{s}}{100} = 1.36\, \text{s} \).

Step 3: Rearrange the formula to solve for g

Rearrange the pendulum formula to solve for \( g \): \( g = \frac{4\pi^2L}{T^2} \).

Step 4: Substitute the known values

Substitute \( L = 0.50 \) cm and \( T = 1.36 \) s into the formula: \( g = \frac{4\pi^2 \times 0.50}{(1.36)^2} \).

Step 5: Calculate g

Calculate \( g \) using the substituted values: \[ g \approx \frac{4\pi^2 \times 0.50}{1.8496} \approx \frac{19.7392 \times 0.50}{1.8496} \approx \frac{9.8696}{1.8496} \approx 5.34\, \text{m/s}^2 \].

Key Concepts

Acceleration Due to Gravity

The acceleration due to gravity, symbolized as \(g\), is a crucial concept in physics that describes how fast objects accelerate towards the center of a planet due to gravity. On Earth, \(g\) is approximately \(9.81 \, \text{m/s}^2\). It means that an object will increase its velocity by \(9.81 \, \text{m/s}\) every second, a principle first described by Sir Isaac Newton.

This value of \(g\) can vary depending on where you are in the universe. For instance, on the Moon or Mars, \(g\) is lower because these celestial bodies are smaller and exert a weaker gravitational force. When you measure \(g\) on an unfamiliar planet, you essentially want to understand how strong gravity is on that planet compared to Earth.

In experiments like the one done by our space explorer, \(g\) is calculated by observing a pendulum's movement, which provides a reliable way to gauge the planet's gravitational pull.

Period of a Pendulum

The period of a pendulum, denoted as \(T\), refers to the time it takes for the pendulum to complete one full back-and-forth swing. Understanding the period is essential because it links directly to the pendulum's length and the gravity affecting it, through the pendulum formula. It's a concept that helps us observe how gravity influences time and motion.

To find the period, you calculate the time it takes for multiple swings and then divide this by the number of swings. In our explorer's experiment, the pendulum made 100 complete swings in 136 seconds. Thus, the period \(T\) is \(1.36 \, \text{seconds per swing}\).

A pendulum's period reveals not only its speed of swinging but also the gravitational force at play. This period will be longer if gravity is weaker or the pendulum is longer, and shorter with stronger gravity or a shorter pendulum.

Pendulum Formula

The pendulum formula is a mathematical expression that gives us insight into the relationship between a pendulum's period \(T\), its length \(L\), and gravity \(g\). The formula is expressed as \( T = 2\pi\sqrt{\frac{L}{g}} \).

This equation is derived from Newtonian mechanics and shows that the period of a pendulum is:

• Directly proportional to the square root of its length \(L\).
• Inversely proportional to the square root of the acceleration due to gravity \(g\).

The simpler interpretation of this is that a longer pendulum or lower gravity results in a longer period, and vice versa.

In practical applications, if you know the period of a pendulum and its length, you can rearrange this formula to solve for \(g\), thereby finding the gravitational acceleration, just as the explorer did. This is done using the rearranged formula: \(g = \frac{4\pi^2L}{T^2}\). By incorporating known values of \(L\) and \(T\), one can effectively calculate \(g\) and gain insights into the gravitational characteristics of an unfamiliar environment.