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Two pendulums have the same dimensions (length \(L\)) and total mass ( \(m\) ). Pendulum \(A\) is a very small ball swinging at the end of a uniform massless bar. In pendulum \(B\), half the mass is in the ball and half is in the uniform bar. Find the period of each pendulum for small oscillations. Which one takes longer for a swing?

Short Answer

Expert verified
Pendulum A takes longer for a swing.

Step by step solution

01

Define the Problem

We are tasked with finding the period of small oscillations for two pendulums with the same length \(L\) and total mass \(m\). Pendulum \(A\) has all its mass concentrated at a point (a simple pendulum), and pendulum \(B\) has its mass distributed equally between the ball and the bar (a physical pendulum).
02

Calculate the Period for Pendulum A

For pendulum \(A\), which behaves like a simple pendulum, the formula for the period \(T\) of small oscillations is given by:\[ T_A = 2\pi \sqrt{\frac{L}{g}} \]Where \(L\) is the length of the pendulum and \(g\) is the acceleration due to gravity.
03

Calculate the Moment of Inertia for Pendulum B

Pendulum \(B\) is a physical pendulum with half the mass in the ball and the other half evenly distributed along the bar. First, calculate the moment of inertia \(I\). The moment of inertia \(I_p\) of the bar (length \(L\), mass \(m/2\), pivoted at one end) is given by:\[ I_p = \frac{1}{3} \left(\frac{m}{2}\right)L^2 = \frac{mL^2}{6} \]For the ball (mass \(m/2\) at a vertex \(L\) from the pivot):\[ I_b = \left(\frac{m}{2}\right)L^2 = \frac{mL^2}{2} \]Adding these: \[ I = \frac{mL^2}{6} + \frac{mL^2}{2} = \frac{2mL^2}{3} \]
04

Calculate the Period for Pendulum B

The period \(T\) of a physical pendulum is given by:\[ T_B = 2\pi \sqrt{\frac{I}{mgd}} \]where \(d=L\) is the distance from the pivot to the center of mass of the ball. Substituting the calculated \(I\) gives:\[ T_B = 2\pi \sqrt{\frac{\frac{2mL^2}{3}}{mgL}} = 2\pi \sqrt{\frac{2L}{3g}} \]
05

Compare the Periods of Both Pendulums

We found that for pendulum \(A\):\[ T_A = 2\pi \sqrt{\frac{L}{g}} \]And for pendulum \(B\):\[ T_B = 2\pi \sqrt{\frac{2L}{3g}} \]Now, we compare:\[ \sqrt{\frac{L}{g}} \quad \text{and} \quad \sqrt{\frac{2L}{3g}} \]Notice that \(\sqrt{\frac{2}{3}} < 1\), thus \(T_B < T_A\).

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Key Concepts

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

Simple Pendulum
Let's delve into the concept of a simple pendulum, which is a classic example in physics for studying motion and oscillations. A simple pendulum consists of a small and heavy object, known as a "bob," attached to the end of a lightweight and rigid rod. The rod is assumed to be massless, so the bob alone contributes to the pendulum's inertia.

The motion of a simple pendulum can be described by its period, which is the time taken for one complete cycle of swing motion. For small angles, the period of a simple pendulum is approximately determined by the formula:
  • \( T = 2\pi \sqrt{\frac{L}{g}} \)
Here, \( L \) is the length of the pendulum, and \( g \) is the acceleration due to gravity (approximately \(9.81 \ ms^{-2}\) on Earth's surface). This formula indicates that the period is independent of the mass of the bob and depends only on the length of the pendulum and the acceleration due to gravity.

In the problem example, pendulum A exemplifies a simple pendulum where the mass is concentrated entirely at the end. It shows typical behavior for periodic motion in physics.
Physical Pendulum
A physical pendulum is a generalized form of a pendulum where the mass is distributed along its length, rather than concentrated at a single point. This distribution of mass makes calculations slightly more complex than those for a simple pendulum.

For a physical pendulum, the period of oscillation depends on the moment of inertia and the distance between the pivot point and the center of mass.

In our specific example, pendulum B is a physical pendulum. Its mass is split evenly between a rod and the bob at the end. This results in different mechanics than a simple pendulum. To calculate its period, we use:
  • \( T = 2\pi \sqrt{\frac{I}{mgd}} \)
Where \( I \) is the moment of inertia, \( m \) is the total mass, \( g \) is gravitational acceleration, and \( d \) is the distance from the pivot to the center of mass.

This setup increases the complexity but gives insights into how mass distribution affects oscillation in physical objects.
Moment of Inertia
The moment of inertia is a fundamental concept that plays a crucial role when studying a physical pendulum. It is a measure of how much torque is needed for a desired angular acceleration around an axis. In essence, it quantifies how the mass is spread out around a pivot, affecting the pendulum's resistance to changes in motion.

Calculating the moment of inertia for different shapes and configurations is key to understanding and predicting the motion of physical pendulums. For pendulum B in our exercise:
  • The rod's moment of inertia, when pivoted at one end, is \( \frac{mL^2}{6} \) for its portion of the mass.
  • The moment of inertia for the bob, concentrated at a distance \( L \), is \( \frac{mL^2}{2} \).
Adding these moments results in the total moment of inertia:
  • \( I = \frac{2mL^2}{3} \)
Understanding and calculating the moment of inertia reveals how the same mass distributed differently in this scenario affects the pendulum's periodic motion. It explains why pendulum B swings faster compared to pendulum A, despite having the same total mass and length.

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Most popular questions from this chapter

A 2.00-kg, frictionless block is attached to an ideal spring with force constant 300 N/m. At \(t\) = 0 the spring is neither stretched nor compressed and the block is moving in the negative direction at 12.0 m/s. Find (a) the amplitude and (b) the phase angle. (c) Write an equation for the position as a function of time.

A cheerleader waves her pom-pom in SHM with an amplitude of 18.0 cm and a frequency of 0.850 Hz. Find (a) the maximum magnitude of the acceleration and of the velocity; (b) the acceleration and speed when the pom-pom's coordinate is \(x = +\)9.0 cm; (c) the time required to move from the equilibrium position directly to a point 12.0 cm away. (d) Which of the quantities asked for in parts (a), (b), and (c) can be found by using the energy approach used in Section 14.3, and which cannot? Explain.

An object with height \(h\), mass \(M\), and a uniform cross-sectional area \(A\) floats upright in a liquid with density \(\rho\). (a) Calculate the vertical distance from the surface of the liquid to the bottom of the floating object at equilibrium. (b) A downward force with magnitude \(F\) is applied to the top of the object. At the new equilibrium position, how much farther below the surface of the liquid is the bottom of the object than it was in part (a)? (Assume that some of the object remains above the surface of the liquid.) (c) Your result in part (b) shows that if the force is suddenly removed, the object will oscillate up and down in SHM. Calculate the period of this motion in terms of the density \(\rho\) of the liquid, the mass \(M\), and the cross- sectional area A of the object. You can ignore the damping due to fluid friction (see Section 14.7).

A 10.0-kg mass is traveling to the right with a speed of 2.00 m/s on a smooth horizontal surface when it collides with and sticks to a second 10.0-kg mass that is initially at rest but is attached to a light spring with force constant 170.0 N/m. (a) Find the frequency, amplitude, and period of the subsequent oscillations. (b) How long does it take the system to return the first time to the position it had immediately after the collision?

A 50.0-g hard-boiled egg moves on the end of a spring with force constant \(k =\) 25.0 N/m. Its initial displacement is 0.300 m. A damping force \(F_x = -bv_x\) acts on the egg, and the amplitude of the motion decreases to 0.100 m in 5.00 s. Calculate the magnitude of the damping constant \(b\).

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