/*! 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 32 Find the moment of incrtia of a ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the moment of incrtia of a triangular lamina of mass \(m\), base \(b\) and height \(h\) about its base.

Short Answer

Expert verified
The moment of inertia of the triangular lamina about its base is \( I = \frac{1}{6}mb^2 \).

Step by step solution

01

Understand the Problem

We need to calculate the moment of inertia of a triangular lamina about its base. This requires understanding the geometry of the triangle and how mass is distributed across it.
02

Formula for Moment of Inertia

The moment of inertia, denoted as \(I\), for a mass distribution is given by \( I = \int r^2 \, dm \), where \(r\) is the perpendicular distance from the axis of rotation to the element of mass \(dm\). Since we're dealing with a lamina, we can treat it as a 2D entity.
03

Density and Mass Distribution

Since the lamina is triangular, we assume a uniform mass distribution. Thus, the density \( \sigma = \frac{m}{A} \), where \( A = \frac{1}{2}bh \) is the area of the triangle. This allows us to express \( dm \) in terms of \( dx \) and \( dy \), which helps in the integration process.
04

Expressing dm for the Triangle

Here \( dm = \sigma \, dA = \sigma \, dx \, dy \). As \( y \) varies from \( 0 \) to \( h \), the width of the triangle linearly scales with height, so \( x \) ranges from \( 0 \) to \( \frac{b}{h}y \).
05

Integration to Find Moment of Inertia

With \( dm = \sigma dx dy \), you set up the integral \( I = \int_0^h \int_0^{\frac{b}{h}y} x^2 \sigma \, dx \, dy \). When you integrate with respect to \( x \) first, you find terms dependent on \( y \), then integrate those with respect to \( y \).

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 91Ó°ÊÓ!

Key Concepts

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

Triangular Lamina
A triangular lamina is a two-dimensional shape characterized by a base and a height. Picture a flat, thin piece of metal cut in the shape of a triangle. This is our lamina. Understanding the properties of a triangular lamina is crucial as it has unique geometric qualities that affect how mass is distributed across its surface. These surfaces are often analyzed in mechanics to find various physical properties, like the moment of inertia in this case, which affects how an object spins or rotates around an axis. With the base of the triangle as the axis of rotation, the calculations become more specific to this unique geometry.
Mass Distribution
In physics, particularly in the study of mechanics, mass distribution describes how mass is dispersed throughout an object. For a triangular lamina with a uniform mass distribution, the mass is spread evenly across its entire surface. The concept of density, which is the mass per unit area, plays a significant role here. For our triangular lamina, we calculate the density \( \sigma \) using the formula \( \sigma = \frac{m}{A} \), where \( m \) is the total mass and \( A \) is the area of the lamina.
  • The area of the lamina is given by \( A = \frac{1}{2}bh \).
  • This uniform density helps us express small elements of mass \( dm \) when we integrate over the lamina's surface in the calculation of the moment of inertia.
Recognizing how mass is distributed across the lamina is crucial for solving problems involving rotational dynamics.
Integral Calculus
Integral calculus provides the mathematical tools necessary to solve problems involving continuous mass distributions. When dealing with the moment of inertia of the triangular lamina, we use integration to sum up the contributions of each infinitesimally small mass element. Each of these elements is at a distance from the axis of rotation and influences the overall inertia. To integrate, we define \( dm = \sigma \cdot dA \) for small areas \( dA = dx \cdot dy \), which reflect the tiny slices of mass across the triangular shape. We set up a double integral, evaluating it first with respect to the variable \( x \), then \( y \), which are the coordinates across the base and height, respectively. The integration process culminates in deriving an expression that accounts for how each tiny portion of the triangle contributes to its rotational resistance about the base.
Rotational Dynamics
Rotational dynamics is the branch of physics that examines objects in rotation and applies similar fundamental principles as in linear dynamics but tailored to circular motion. The moment of inertia is a pivotal concept within this framework as it quantifies how resistant an object is to changes in its rotational state.
  • For a triangular lamina, the moment of inertia about its base tells us how the object's mass distribution affects its ability to rotate around that axis.
  • Consider it as the rotational equivalent to mass in linear motion; a higher moment of inertia indicates the lamina is harder to start or stop spinning.
Rotational dynamics is crucial because understanding these behaviors is key for designing and predicting the performance of various mechanical systems, from simple wheels to intricate machinery. The derivation using integral calculus, as applied to our triangular lamina, results in practical insights applicable in engineering and physics.

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 man spinning in free space changes the shape of his body, e.g. by spreading his amms or curling up. I3y doing this, he can change his (a) moment of inertia (b) angular momentum (c) angular velocity (d) rotational kinctic cnergy

\(\Lambda\) uniform disc of mass \(m\) and radius \(R\) is rotated about an axis passing through its centre and perpendicular to its plane with an angular velocity \(\omega_{0} .\) It is placed on a rough horizontal plane with the axis of the disc keeping vertical. Coefficient of friction between the disc and the surface is \(\mu\). Find (a) the time when disc stops rotating, (b) the angle rotated by the disc before stopping.

\(\Lambda\) solid sphere rolls without slipping on a rough horizontal floor, moving with a speed \(v\). It makes an elastic collision with a smooth vertical wall. \Lambdafter impact (a) it will move with a speed \(v\) initially. (b) its motion will be rolling without slipping. (c) its motion will be rolling without slipping initially and its rotational motion will stop momentarily at some instant. (d) its motion will be rolling without slipping only after some time.

\(\Lambda\) man standing on a platform holds weights in his outstretched arms. 'Ihe system rotates freely about a central vertical axis. If he now draws the weights inwards close to his body (a) the angular velocity of the system will increase (b) the angular momentum of the system will decrease (c) the kinetic energy of the system will increase (d) he will have to expend some energy to draw the weights in

A body is projected at an angle with the horizontal in the uniform gravitational ficld of thecarth, the angular momentum of the body about the point of projection as it proceds along its path (a) Remains constant (b) Increases (c) Deercases (d) Initially docreases and afier its highost increases

See all solutions

Recommended explanations on Physics Textbooks

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.