Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Two disc of same thickness but of different radii are made of two different materials such that their masses are same. The densities of the materials are in the ratio \(1: 3\). The moment of inertia of these disc about the respective axes passing through their centres and perpendicular to their planes will be in the ratio. \(\\{\mathrm{A}\\} 1: 3\) \\{B\\} \(3: 1\) \\{C\\} \(1: 9\) \(\\{\mathrm{D}\\} 9: 1\)

Short Answer

Expert verified
The ratio of the moments of inertia of the two discs is \({B}\\ \, 3:1 \).

Step by step solution

01

Determine the formula for the moment of inertia of a disc

The moment of inertia of a disc about an axis passing through its center and perpendicular to its plane is given by the formula: \[I = \frac{1}{2}MR^2\] Where: \(I\) is the moment of inertia \(M\) is the mass of the disc \(R\) is the radius of the disc
02

Set up the relationship between the given densities and radii

The density \(\rho\) of a material is defined as the mass per unit volume: \(\rho = \frac{M}{V}\). For a circular disc, the volume can be expressed in terms of the radius and thickness: \(V = \pi R^2 h\), where \(h\) is the thickness. Since both discs have the same mass, we can write: \[\frac{\rho_1}{\rho_2} = \frac{3}{1} = \frac{M_1 R_1^2 h}{M_2 R_2^2 h}\] Since \(M_1 = M_2\) and the thicknesses are equal, we can cancel them out: \[\frac{3}{1} = \frac{R_1^2}{R_2^2}\]
03

Find the ratio of the moments of inertia of the two discs

We want to find the ratio \(\frac{I_1}{I_2}\), where \(I_1 = \frac{1}{2}M_1R_1^2\) and \(I_2 = \frac{1}{2}M_2R_2^2\). Using the known ratio of densities, we can calculate the ratio of moments of inertia: \[\frac{I_1}{I_2} = \frac{\frac{1}{2}M_1R_1^2}{\frac{1}{2}M_2R_2^2} = \frac{R_1^2}{R_2^2}\] Using the relation obtained in Step 2 for \(\frac{R_1^2}{R_2^2}\), we can write: \[\frac{I_1}{I_2} = \frac{3}{1} = 3:1\] So the answer to the problem is: \[{B}\\ \, 3:1 \]

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 Vaia!

Key Concepts

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

Disc Density
Density is a key property of materials that tells us how much mass is packed into a given volume. For a disc, the density \( \rho \) is defined as the mass \( M \) divided by the volume \( V \). This relationship is expressed by:
  • \( \rho = \frac{M}{V} \)
For circular discs, volume can be calculated as \( V = \pi R^2 h \), where \( R \) is the radius and \( h \) is the thickness. This means that even if two discs have the same mass, differences in density will affect their size, specifically radius if thickness is constant.
Understanding density helps us see how different materials can occupy space differently, leading to different physical properties and behaviors.
Mass and Volume Relationship
The relationship between mass and volume is crucial in physics. When we talk about discs made of different materials but having the same mass, we're really exploring how changes in volume reflect the properties of those materials. Mass is constant here, leading to the formula:
  • \( M = \rho V \)
For discs with equal thicknesses, the volume focuses on the radius. Hence, if one disc is made from a material three times as dense as the other, the denser material's disc must have a smaller radius to maintain the same mass. This gives rise to a direct way to compare properties based on radius and volume.
Ratio of Densities
In this problem, the two materials have densities in a ratio of \( 1:3 \). This tells us that one material is three times as dense as the other. By setting mass equal for both discs, we can deduce:
  • \( \frac{\rho_1}{\rho_2} = \frac{3}{1} \)
With density ratio known and mass same, this allows us to derive a relationship between their radii:
  • \( \frac{R_1^2}{R_2^2} = \frac{3}{1} \)
This step simplifies solving the problem, as finding the ratios of any property linked to these will inherently depend on the density ratio.
Circular Disc Properties
Circular discs are defined by properties like radius, thickness, and mass. These features determine calculations like the moment of inertia, which tells us how difficult it is to change the disc's rotational motion. The formula for the moment of inertia \( I \) of a disc is:
  • \( I = \frac{1}{2}MR^2 \)
Where \( I \) depends directly on both mass \( M \) and the square of the radius \( R \).
When comparing two discs with equal mass but different radii, the difference in their radius squared directly influences their moments of inertia. As demonstrated, if one disc has a radius that scales in a particular ratio due to density differences, this ratio directly influences the moment of inertia, providing a clear understanding of rotational dynamics.

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

Match list I with list II and select the correct answer $$ \begin{aligned} &\begin{array}{|l|l|} \hline \text { List-I } & \begin{array}{l} \text { List - II } \\ \text { System } \end{array} & \text { Moment of inertia } \\ \hline \text { (x) A ring about it axis } & \text { (1) }\left(\mathrm{MR}^{2} / 2\right) \\ \hline \text { (y) A uniform circular disc about it axis } & \text { (2) }(2 / 5) \mathrm{MR}^{2} \\ \hline \text { (z) A solid sphere about any diameter } & \text { (3) }(7 / 5) \mathrm{MR}^{2} \\ \hline \text { (w) A solid sphere about any tangent } & \text { (4) } \mathrm{MR}^{2} \\ \cline { 2 } & \text { (5) }(9 / 5) \mathrm{MR}^{2} \\ \hline \end{array}\\\ &\text { Select correct option }\\\ &\begin{array}{|l|l|l|l|l|} \hline \text { Option? } & \mathrm{X} & \mathrm{Y} & \mathrm{Z} & \mathrm{W} \\\ \hline\\{\mathrm{A}\\} & 2 & 1 & 3 & 4 \\ \hline\\{\mathrm{B}\\} & 4 & 3 & 2 & 5 \\ \hline\\{\mathrm{C}\\} & 1 & 5 & 4 & 3 \\ \hline\\{\mathrm{D}\\} & 4 & 1 & 2 & 3 \\ \hline \end{array} \end{aligned} $$

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 small object of uniform density rolls up a curved surface with initial velocity 'u'. It reaches up to maximum height of \(3 \mathrm{v}^{2} / 4 \mathrm{~g}\) with respect to initial position then the object is \(\\{\mathrm{A}\\}\) ring \(\\{B\\}\) solid sphere \(\\{\mathrm{C}\\}\) disc \\{D\\} hollow sphere

A circular disc of radius \(\mathrm{R}\) and thickness \(\mathrm{R} / 6\) has moment of inertia I about an axis passing through its centre and perpendicular to its plane. It is melted and re-casted in to a solid sphere. The moment of inertia of the sphere about its diameter as axis of rotation is \(\ldots\) \(\\{\mathrm{A}\\} \mathrm{I}\) \(\\{\mathrm{B}\\}(2 \mathrm{I} / 8)\) \(\\{\mathrm{C}\\}(\mathrm{I} / 5)\) \(\\{\mathrm{D}\\}(\mathrm{I} / 10)\)

A circular disc of mass \(\mathrm{m}\) and radius \(\mathrm{r}\) is rolling on a smooth horizontal surface with a constant speed v. Its kinetic energy is \(\\{\mathrm{A}\\}(1 / 4) \mathrm{mv}^{2}\) \(\\{\mathrm{B}\\}(1 / 2) \mathrm{mv}^{2}\) \(\\{\mathrm{C}\\}(3 / 4) \mathrm{mv}^{2}\) \(\\{\mathrm{D}\\} \mathrm{mv}^{2}\)

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.

Sign-up for free