/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 50 There is a rod of length ' 1 ' a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 50

There is a rod of length ' 1 ' and mass ' \(m^{\prime}\). It is hinged at one end to the ceiling. The period of small oscillation is : (a) \(T=2 \pi \sqrt{\left(\frac{2 l}{3 g}\right)}\) (b) \(T=\pi \sqrt{\left(\frac{l}{3 g}\right)}\) (c) \(T=2 \pi \sqrt{\left(\frac{1}{3 g}\right)}\) (d) \(T=2 \pi \sqrt{\left(\frac{l}{g}\right)}\)

Short Answer

Expert verified
The period is \(T = 2 \pi \sqrt{\left(\frac{2 l}{3 g}\right)}\), which is option (a).

Step by step solution

01

Identify the Physical System

We are dealing with a physical pendulum, which is a rod that is oscillating around one of its ends that is fixed (hinged). It is not a simple pendulum because the mass is distributed along the length of the rod.
02

Apply the Formula for the Period of a Physical Pendulum

The general formula for the period of small oscillations of a physical pendulum is given by: \[ T = 2\pi \sqrt{\left(\frac{I}{mgd}\right)} \]where \(I\) is the moment of inertia about the pivot point, \(m\) is the mass of the rod, \(g\) is the acceleration due to gravity, and \(d\) is the distance from the pivot point to the center of mass of the rod.
03

Calculate the Moment of Inertia

The moment of inertia \(I\) of a rod of length \(l\) pivoted about one end is given by \[ I = \frac{1}{3} ml^2 \]This formula is derived from the parallel axis theorem and considering the mass is distributed along the length.
04

Determine the Distance to the Center of Mass

The center of mass of a uniform rod of length \(l\) is at its midpoint. Thus, the distance \(d\) from the pivot point to the center of mass is \[ d = \frac{l}{2} \]
05

Substitute Values into the Formula

Substitute \(I = \frac{1}{3}ml^2\) and \(d = \frac{l}{2}\) into the period formula:\[ T = 2\pi \sqrt{\left(\frac{\frac{1}{3}ml^2}{mg \cdot \frac{l}{2}}\right)} \]
06

Simplify the Expression

Simplify the expression as follows:\[ T = 2\pi \sqrt{\left(\frac{\frac{1}{3}ml^2}{mg \cdot \frac{l}{2}}\right)} = 2\pi \sqrt{\left(\frac{2l}{3g}\right)} \]Thus, the period \(T\) matches option \((a)\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Moment of Inertia
When we discuss a physical pendulum like a rod swinging around a pivot, we need to consider its moment of inertia. This is a measure of how much an object resists rotational motion around a certain axis. For a uniform rod of mass \(m\) and length \(l\), when pivoted about one end, the moment of inertia \(I\) is calculated using the formula \(I = \frac{1}{3} ml^2\).
This formula stems from the integration of infinitesimal mass elements along the rod's length, considering how far each element is from the pivot point.
  • The further the mass is from the pivot, the more it contributes to the moment of inertia.
  • This concept is crucial as it affects how the rod oscillates around the pivot point.
The higher the moment of inertia, the lesser the angular acceleration of the rod for a given torque, influencing the period of oscillation.
Small Oscillations
In a physical pendulum, small oscillations refer to the small-angle approximation. This means that the angle through which the pendulum swings is relatively small, typically less than 15 degrees. This condition allows simplifying the mathematical treatment of pendulum motion.
In this regime, the restoring torque is proportional to the angular displacement, similar to Hooke's Law. As a result, the pendulum's motion can be described by simple harmonic motion equations.
  • This approximation makes the problem analytically solvable and predictable.
  • The resulting period of the pendulum can be found using a formula that is less complex than in cases of large oscillations.
It is essential to stick to small angles to ensure that the derived formulas for the period of oscillation remain valid.
Center of Mass
The center of mass is a critical point in an object where its mass can be considered to be concentrated. For a uniform rod, this point is located at its midpoint. Understanding the center of mass is crucial when calculating how the rod will oscillate as a pendulum.
  • For a rod of length \(l\), the center of mass is at a distance \(d = \frac{l}{2}\) from the pivot.
  • This distance is used in formulas to determine the pendulum's period.
The center of mass's location affects the torque and rotational inertia of the pendulum. It is pivotal in accurately determining the period of oscillation for physical pendulum problems.
Period of Oscillation
The period of oscillation is the time it takes for the pendulum to complete one full swing back and forth. For a physical pendulum, like our rod, the period \(T\) is given by the formula: \[ T = 2\pi \sqrt{\left(\frac{I}{mgd}\right)} \]where \(I\) is the moment of inertia, \(m\) is the mass of the rod, \(g\) is the acceleration due to gravity, and \(d\) is the distance to the center of mass from the pivot.
With the given values for a rod, these simplify the period expression to \(T = 2\pi \sqrt{\left(\frac{2l}{3g}\right)}\).
  • Understanding this formula helps in predicting how a physical pendulum will behave.
  • The period is directly proportional to the square root of how far the center of mass is from the pivot.
Knowing the period assists in various applications, like designing clocks and understanding natural oscillatory systems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A block of mass \(1 \mathrm{~kg}\) is connected with a massless spring of force constant 100 \(\mathrm{N} / \mathrm{m}\). At \(t=0\), a constant force \(F=10 \mathrm{~N}\) is applied on the block. The spring 8 is in its natural length at \(t=0\). The speed of particle at \(x=6 \mathrm{~cm}\) from mean position is : (a) \(4 \mathrm{~cm} / \mathrm{s}\) (b) \(10 \mathrm{~cm} / \mathrm{s}\) (c) \(80 \mathrm{~cm} / \mathrm{s}\) (d) \(50 \mathrm{~cm} / \mathrm{s}\)

A particle executes S.H.M. along a straight line so that its period is 12 second. The time it takes in traversing a distance equal to half its amplitude from its equilibrium position is : (a) 6 second (b) 4 second (c) 2 second (d) 1 second

If \(\overrightarrow{\mathrm{s}}=a \sin \omega t \hat{\mathbf{i}}+b \cos \omega t \hat{\mathbf{j}}\), the equation of path of particle is : (a) \(x^{2}+y^{2}=\sqrt{a^{2}+b^{2}}\) (b) \(\frac{x^{2}}{\dot{b}^{2}}+\frac{y^{2}}{a^{2}}=1\) (c) \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) (d) none of these

Two point masses of \(3 \mathrm{~kg}\) and \(1 \mathrm{~kg}\) are attached to opposite ends of a horizontal spring whose spring constant is \(300 \mathrm{~N} / \mathrm{m}\) as shown in figure. The natural vibrational frequency of the system is of order : (a) \(4 \mathrm{~Hz}\) (b) \(3 \mathrm{~Hz}\) (c) \(2 \mathrm{~Hz}\) (d) \(1 \mathrm{~Hz}\)

A particle executing simple harmonic motion has amplitude of 1 metre and time period 2 second. At \(t=0\), net force on the particle is zero. The equation of displacement of particle is: (a) \(x=\sin \pi t\) (b) \(x=\cos \pi t\) (c) \(x=\sin 2 \pi t\) (d) \(x=\cos 2 \pi t\)
See all solutions

Recommended explanations on Physics Textbooks

Scientific Method Physics

Read Explanation

Classical Mechanics

Read Explanation

Kinematics Physics

Read Explanation

Fundamentals of Physics

Read Explanation

Solid State Physics

Read Explanation

Wave Optics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.