/*! 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 551 Two circular loop \(A \& B\)... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Two circular loop \(A \& B\) of radius \(r_{A}\) and \(r_{B}\) respectively are made from a uniform wire. The ratio of their moment of inertia about axes passing through their centres and perpendicular to their planes is \(\left(\mathrm{I}_{\mathrm{B}} / \mathrm{I}_{\mathrm{A}}\right)=8\) then \(\left(\mathrm{r}_{\mathrm{B}} / \mathrm{r}_{\mathrm{A}}\right)\) Ra is equal to... \(\\{\mathrm{A}\\} 2\) \(\\{B\\} 4\) \\{C \(\\}\) \(\\{\mathrm{D}\\} 8\)

Short Answer

Expert verified
The required ratio between the radii \(\left(\frac{r_{B}}{r_{A}}\right)\) is 2, which corresponds to option A.

Step by step solution

01

Write down the moment of inertia for a circular loop.

The moment of inertia (I) of a circular loop can be calculated using the formula: \[I = M r^2\] Here, M is the mass of the loop, and r is the radius of the loop.
02

Set up the equation for the ratio of moments of inertia.

According to the given problem, the ratio of the moments of inertia of loop B to loop A is 8. Therefore, we have: \[\frac{I_B}{I_A} = \frac{M_Br_B^2}{M_Ar_A^2} = 8\]
03

Use the mass of the uniform wire to express M in terms of the radius.

As the wire is uniform, the mass per unit length (m) is the same for both loops. We can express the mass of each loop in terms of its radius and the mass per unit length: \[M_A = m(2 \pi r_A)\] \[M_B = m(2 \pi r_B)\]
04

Substitute the mass expressions into the ratio equation.

Now, replace \(M_A\) and \(M_B\) in the equation for the ratio of moments of inertia: \[\frac{m(2\pi r_B) r_B^2}{m(2\pi r_A) r_A^2} = 8\]
05

Simplify the equation and solve for the ratio of radii.

The constant terms \(m\) and \(2\pi\) cancel out on both sides of the equation, leaving us with: \[\frac{r_B^3}{r_A^3} = 8\] Take the cube root of both sides of the equation to obtain: \[\frac{r_B}{r_A} = \sqrt[3]{8} = 2\] Hence, the required ratio between the radii \(\left(\frac{r_{B}}{r_{A}}\right)\) is 2, which corresponds to 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 91Ó°ÊÓ!

Key Concepts

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

Moment of Inertia in a Circular Loop
A circular loop is a simple geometric shape that is often used in physics to demonstrate various properties and calculations. The moment of inertia is one of these properties. It represents the rotational inertia of an object, essentially quantifying how resistant the object is to changes in its rotational motion. For a circular loop, understanding this concept requires recognizing how mass is distributed from the axis of rotation.
The moment of inertia of a circular loop is uniquely determined because all points on the loop are equidistant from the axis. The formula to calculate it is given by: \[ I = M r^2 \] where \( M \) is the mass of the circular loop and \( r \) is the radius of the loop. This expression highlights how the radius significantly influences the loop's moment of inertia. As the radius increases, the inertia grows, which means the loop becomes harder to rotate.
Understanding Mass per Unit Length
Mass per unit length, often denoted by \( m \), is a measurement that tells us how mass is distributed along a length. For a wire, this concept is key to evaluating how much mass is in a portion of its length. If a wire is uniform, it means that its mass per unit length is the same across its entire length.
In the case of circular loops created from a uniform wire, each loop, regardless of its size, will be directly proportional in mass to its circumference. The total mass \( M \) of the circular loop can be calculated as: \[ M = m(2 \, \pi \, r) \] where \( r \) is the radius of the loop, and \( 2\pi r \) is its circumference. This relationship is essential when comparing different loops made from the same wire, as seen in this exercise.
The Ratio of Radii in Circular Loops
In the context of comparing circular loops, the ratio of radii often comes into play, especially when determining proportions or scaling. When confronted with two loops made from the same material and having different sizes, their physical properties such as the moment of inertia can be compared using their radii.
In our exercise, the ratio is derived from the mathematical simplification of the loops' moments of inertia and uses the property that these moments are proportional to the cube of their radii: \[ \frac{I_B}{I_A} = \frac{r_B^3}{r_A^3} \] Given that the ratio of moments of inertia is 8, the ratio of radii can be found by solving the equation: \[ \frac{r_B}{r_A} = \sqrt[3]{8} = 2 \] Therefore, loop B has twice the radius of loop A.
Properties of Uniform Wire
A uniform wire is a type of wire that has a consistent mass per unit length throughout its length. This consistency is important because it allows for straightforward calculations of mass and other properties when the wire is shaped into various forms, like circles.
In exercises involving uniform wires, one can assume that every segment of the wire, irrespective of its position, has the same density and thickness. This uniformity simplifies the computation of quantities like total mass and moment of inertia, especially for shapes with continuous symmetry like circular loops.
When loops are made from a uniform wire, the mass of each loop is a function of its radius, meaning that larger loops will correspondingly have larger masses if the wire remains uniform. This straightly ties into the use of mass per unit length to express mass in terms of the radius, as employed in our exercise.

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

Let I be the moment of inertia of a uniform square plate about an axis \(\mathrm{AB}\) that passes through its centre and is parallel to two of its sides \(\mathrm{CD}\) is a line in the plane of the plate that passes through the centre of the plate and makes an angle of \(\theta\) with \(\mathrm{AB}\). The moment of inertia of the plate about the axis \(\mathrm{CD}\) is then equal to.... \(\\{\mathrm{A}\\} \mathrm{I}\) \(\\{B\\} I \sin ^{2} \theta\) \(\\{C\\} I \cos ^{2} \theta\) \(\\{\mathrm{D}\\} I \cos ^{2}(\theta / 2)\)

Two blocks of masses \(10 \mathrm{~kg}\) an \(4 \mathrm{~kg}\) are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives velocity of \(14 \mathrm{~m} / \mathrm{s}\) to the heavier block in the direction of the lighter block. The velocity of the centre of mass is : \(\\{\mathrm{A}\\} 30 \mathrm{~m} / \mathrm{s}\) \(\\{\mathrm{B}\\} 20 \mathrm{~m} / \mathrm{s}\) \(\\{\mathrm{C}\\} 10 \mathrm{~m} / \mathrm{s}\) \(\\{\mathrm{D}\\} 5 \mathrm{~m} / \mathrm{s}\)

The M.I of a disc of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) about an axis passing through the centre \(\mathrm{O}\) and perpendicular to the plane of disc is \(\left(\mathrm{MR}^{2} / 2\right)\). If one quarter of the disc is removed the new moment of inertia of disc will be..... \(\\{\mathrm{A}\\}\left(\mathrm{MR}^{2} / 3\right)\) \(\\{B\\}\left(M R^{2} / 4\right)\) \(\\{\mathrm{C}\\}(3 / 8) \mathrm{MR}^{2}\) \(\\{\mathrm{D}\\}(3 / 2) \mathrm{MR}^{2}\)

A gramophone record of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) is rotating with angular speed \(\mathrm{W}\). If two pieces of wax each of mass \(\mathrm{M}\) are kept on it at a distance of \(R / 2\) from the centre on opposite side then the new angular velocity will be..... \(\\{\mathrm{A}\\}(\omega / 2)\) \(\\{B\\}[m \omega /(M+m)\) \(\\{C\\}[M \omega /(M+m)]\) \(\\{\mathrm{D}\\}[(\mathrm{M}+\mathrm{m}) / \mathrm{M} \omega]\)

A wheel having moment of inertia \(2 \mathrm{~kg} \mathrm{M}^{2}\) about its vertical axis, rotates at the rate of 60 rpm about this axis. The torque which can stop the wheels rotation in one minute will be.. \(\\{\mathrm{A}\\}(\pi / 15) \mathrm{N}-\mathrm{m}\) \(\\{\mathrm{B}\\}(\pi / 18) \mathrm{N}-\mathrm{m}\) \(\\{\mathrm{C}\\}(2 \pi / 15) \mathrm{N}-\mathrm{m}\) \(\\{\mathrm{D}\\}(\pi / 12) \mathrm{N}-\mathrm{m}\)

See all solutions

Recommended explanations on English 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.