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Chapter 8: Problem 2

A pendulum with a disc as the pendulum bob is oscillating with period \(T\). You then stick a small piece of chewing gum onto the center of the disc. How does the gum affect the pendulum's period: an increase, no effect, or a decrease?

Short Answer

Expert verified
The addition of a small piece of chewing gum to the center of the pendulum bob (disc) has no effect on the pendulum's period, as both the mass and the length remain essentially unchanged. The period T is given by the formula: \[ T = 2\pi\sqrt{\frac{L}{g}} \] which does not include the mass of the pendulum bob.

Step by step solution

01

Recall the formula for the period of a pendulum

The formula for the period T of a simple pendulum is given by the equation: \[ T = 2\pi\sqrt{\frac{L}{g}} \] Where: - T is the period of the pendulum, - L is the length of the pendulum string from the pivot point to the center of mass of the pendulum bob (or disc), - g is the acceleration due to gravity. Notice that the equation does not include the mass of the pendulum bob.
02

The effect of chewing gum on the pendulum mass

The additional mass from the small piece of chewing gum is negligible compared to the mass of the pendulum bob (disc). Therefore, we can assume that the mass of the pendulum is essentially unchanged.
03

The effect of chewing gum on the pendulum's length

Since the small piece of chewing gum is placed at the center of the disc, it does not change the length of the pendulum string nor the distance between the pivot point and the center of mass of the bob. Therefore, the length L remains unchanged.
04

The effect of chewing gum on the pendulum's period

Since neither the mass nor the length of the pendulum are affected by the addition of the small piece of chewing gum, the period T also remains unchanged according to the formula: \[ T = 2\pi\sqrt{\frac{L}{g}} \]
05

Conclusion

The addition of a small piece of chewing gum to the center of the pendulum bob (disc) has no effect on the pendulum's period, as both the mass and the length remain essentially unchanged.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Simple Pendulum
The concept of a simple pendulum is central to understanding periodic motion in physics. A simple pendulum consists of a weight, known as the bob, suspended from a pivot point by a string or rod of negligible weight. When displaced from its equilibrium position and then released, the bob oscillates back and forth under the influence of gravity. This back and forth motion is termed as an oscillation.

The time taken for one complete oscillation is referred to as the period (\( T \) ) of the pendulum. The period depends on two key factors: the length of the pendulum (\( L \)) and the acceleration due to gravity (\( g \)), not on the mass or the material of the bob. The period can be mathematically expressed as:\[\begin{equation}T = 2\text{\pi}\sqrt{\frac{L}{g}}\end{equation}\]
Thus, when evaluating changes to the period of a pendulum, one only needs to consider alterations to the length of the string and the gravitational force acting on the bob.
Center of Mass
The center of mass is a crucial concept when dealing with the physical motion of objects. It can be defined as the point where the mass of an object is concentrated or balanced. For a uniform symmetrical object like a disc, the center of mass is at its geometric center. When we discuss pendulums, the focus is on the distance from the pivot point to the center of mass of the bob.

In the context of a pendulum's period, it is crucial to note that altering the position of the center of mass can change the effective length of the pendulum. However, in the case of our exercise, placing a small piece of chewing gum at the center of the disc does not affect its center of mass because the center remains in the same location. Therefore, while the center of mass is an essential factor for pendulum motion, in this particular situation, it remains unchanged and the period of the pendulum remains the same.
Acceleration due to Gravity
Acceleration due to gravity (\( g \)) is the rate at which an object accelerates when free-falling in a gravitational field such as Earth's. On the surface of the Earth, this value is approximately 9.81 meters per second squared (\( m/s^2 \)). The acceleration due to gravity plays a pivotal role in the period of a simple pendulum as it directly influences how fast the pendulum bob accelerates back toward the equilibrium position during its swing.

Since the acceleration due to gravity is a constant at a given location, it remains unchanged regardless of changes to the mass of the pendulum bob. Therefore, our original exercise's addition of a small piece of chewing gum to the pendulum bob does not impact the acceleration due to gravity, and consequently, does not affect the period of the pendulum. This reinforces the understanding that the simple pendulum's period is invariant to mass changes and depends solely on the length of the pendulum and the gravitational force being exerted on it.

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Most popular questions from this chapter

Imagine a wave with reduced wavelength \(\not \gamma\) on the surface of a fluid. a. Show that the dimensionless ratio \(R \equiv \frac{\text { potential energy due to gravity }}{\text { potential energy due to surface tension }}\) is, after ignoring dimensionless constants, the dimensionless group \(\rho g \not \gamma^{2} / \gamma\) that we used in Section \(8.4 .1\) to distinguish waves driven by gravity from waves driven by surface tension. b. For water, estimate the critical wavelength \(\lambda\) at which \(R \sim 1\).

In this problem you use easy cases to guess the tension in the connecting string. a. The tension \(T\), like the acceleration, depends on \(m_{1}, m_{2}\), and \(g\). Explain why these four quantities produce two independent dimensionless groups. b. Choose two suitable independent dimensionless groups so that you can write an equation for the tension in this form: group proportional to \(T=f(\) group without \(T\) ). c. Use easy cases to guess \(f\); then sketch \(f\). d. Solve for \(T\) using freebody diagrams and Newton's laws. Then compare that result with your guess in part (c), and use \(T\) to find \(a\), the downward acceleration of \(m_{1}\).

Explain why the following dimensionless ratio measures the importance of drag: \(\frac{\text { mass of fluid swept out }}{\text { mass of object }}\) An equivalent computation, except for a dimensionless factor of order unity, is \(\frac{\text { fluid density }}{\text { object density }} \times \frac{\text { distance traveled }}{\text { size of object }}\) where the size of the object is its linear dimension in the direction of travel. When this ratio is comparable to unity, drag significantly affects the trajectory. Apply either form of the ratio to a car during city driving, and find the distance \(d\) at which the ratio becomes significant (say, roughly 1). How does the distance compare with the distance between stop signs ortraffic lights on city streets? What therefore is the main mechanism of energy loss in city driving? How does this analysis change for highway driving?

The simplest model of the atmosphere is isothermal: The atmosphere has one temperature throughout it. A better approximation, the adiabatic atmosphere, relaxes this assumption and incorporates the adiabatic gas law: $$p V^{\gamma} \propto 1$$ where \(p\) is atmospheric pressure, \(V\) is the volume of a parcel of air, and \(\gamma \equiv c_{p} / c_{v}\) is the ratio of the two specific heats in the gas. (For dry air, \(\gamma=1.4\).) Imagine an air parcel rising up a mountain. As the parcel rises into air with a lower. pressure, it expands, and its volume and temperature change according to a combination of the adiabatic and ideal gas laws. a. What easy case of \(\gamma\) reproduces the isothermal atmosphere? b. For \(\gamma=1.4\) (dry air), will air temperature decrease with, increase with, or be independent of height?

Helium, when cold, turns into a liquid. When very cold, the liquid turns into a superfluid-a quantum liquid. Here is a dimensionless ratio determining how quantum the liquid is \(\beta \equiv \frac{\text { quantum uncertainty in the position of a helium atom }}{\text { separation between atoms }}\) a. Estimate \(\beta\) in terms of the quantum constant \(\hbar\), helium's density \(\rho\) (as a liquid), the thermal energy \(k_{\mathrm{B}} T\), and the atomic mass \(m_{\mathrm{He}}\) b. In the \(\beta \sim 1\) regime, helium becomes a quantum liquid (a superfluid). Thus, estimate the superfluid transition temperature.
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