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A circular disc \(\mathrm{x}\) of radius \(\mathrm{R}\) is made from an iron plate of thickness \(t\). and another disc \(Y\) of radius \(4 R\) is made from an iron plate of thickness \(t / 4\) then the rotation between the moment of inertia \(\mathrm{I}_{\mathrm{x}}\) and \(\mathrm{I}_{\mathrm{y}}\) is \(\\{\mathrm{A}\\} \mathrm{I}_{\mathrm{y}}=64 \mathrm{I}_{\mathrm{x}}\) \(\\{B\\} I_{y}=32 I_{x}\) \(\\{\mathrm{C}\\} \mathrm{I}_{\mathrm{y}}=16 \mathrm{I}_{\mathrm{x}}\) \(\\{\mathrm{D}\\} \mathrm{I}_{\mathrm{y}}=\mathrm{I}_{\mathrm{x}}\)

Short Answer

Expert verified
The relationship between the moments of inertia of the discs is \(I_y = 64 I_x\), which corresponds to option A.

Step by step solution

01

Moment of Inertia for a Uniform Circular Disc

The moment of inertia for a thin uniform circular disc rotating about an axis perpendicular to the plane of the disc and passing through its center is given by: \[ I = \frac{1}{2} M R^{2} \] Where I is the moment of inertia, M is the mass of the disc, and R is the radius of the disc.
02

Express Mass in terms of Density, Thickness and Area

We can write the mass of the disc in terms of its density ρ, thickness t, and area A, where A is the area of the disc given by πR². So, \(M = \rho A t = \rho(\pi R^{2})t \). Now, let's substitute the mass into the formula for the moment of inertia of our discs.
03

Moment of Inertia In terms of Density, Thickness and Radius

Substitute the mass expression into the moment of inertia formula: \[ I = \frac{1}{2}(\rho \pi R^{2} t) R^{2} \]
04

Calculate the Moment of Inertia for Disc X and Disc Y

Now, we'll plug in the values for disc X and disc Y and find their moments of inertia. For disc X: Radius RX = R Thickness tX = t \[I_{x} = \frac{1}{2}(\rho \pi R^{2} t) R^{2} \] For disc Y: Radius RY = 4R Thickness tY = t/4 \[I_{y} = \frac{1}{2}(\rho \pi (4R)^{2} \frac{t}{4}) (4R)^{2} \]
05

Find the Relationship between Ix and Iy

Now, let's divide Iy by Ix to find the relationship between them. \[\frac{I_{y}}{I_{x}} = \frac{\frac{1}{2}(\rho \pi (4R)^{2} \frac{t}{4}) (4R)^{2}}{\frac{1}{2}(\rho \pi R^{2} t) R^{2}} \] Simplify the equation: \[\frac{I_{y}}{I_{x}} = \frac{(4R)^{2} (4R)^{2} \frac{t}{4}}{R^{2} R^{2} t} \] \[\frac{I_{y}}{I_{x}} = \frac{256 R^{4} t}{4 R^{4} t} \] \[\frac{I_{y}}{I_{x}} = 64 \] From the given options, the relationship between the moments of inertia is Iy = 64 Ix, which corresponds to option A.

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Key Concepts

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

Circular Disc
A circular disc is a common geometrical figure that can be visualized as a perfectly round, flat, and thin sheet. It is a 2-dimensional shape characterized by its radius, which is the distance from its center to the outer edge. The disc has a circular boundary called the circumference. It is important to note that a circular disc is an idealized model commonly used in physics and engineering.

In the context of rotational dynamics, the properties of a circular disc, such as its radius and mass distribution, are critical to understanding its behavior under rotation. When we analyze the moment of inertia of a disc, we're essentially looking at how mass is distributed relative to an axis of rotation, which affects how the disc accelerates when a torque is applied. Understanding these properties is fundamental for solving rotational motion problems.

For example, if you consider a compact disc (like a CD), it rotates around its center when played, and its moment of inertia determines the energy required to spin it at a certain angular speed.
Density
Density is a key property in physics that describes how much mass is contained in a given volume of a material. It is expressed mathematically as \( \rho = \frac{m}{V} \), where \( \rho \) is density, \( m \) is mass, and \( V \) is volume.

In the context of a circular disc, density allows us to determine the mass of the disc if we know its thickness and surface area. This is crucial because the mass of the disc contributes to its moment of inertia, which is calculated differently depending on how mass is distributed in relation to the axis of rotation. For example, in the given exercise, the mass is derived from density, thickness, and area as \( M = \rho \times A \times t \), where \( A \) is the area \((\pi R^2)\) and \( t \) is thickness.

Additionally, different materials have different densities. For instance, iron, the material for the discs in the problem, is denser than materials like aluminum or plastic. This difference in density impacts the moment of inertia, which is why it is important to consider when conducting calculations or experiments involving rotational dynamics.
Axis of Rotation
An axis of rotation is an imaginary line around which an object rotates. In 3-dimensional space, an object can have multiple axes of rotation, but for a circular disc in many physics problems, the standard axis is usually vertical and passes through the disc's center.

Understanding the axis of rotation is crucial because it directly affects the calculation of moment of inertia. For a circular disc, the moment of inertia about an axis perpendicular to its plane and through its center is a measure of how difficult it is to change the disc's rotational motion. The formula \( I = \frac{1}{2} M R^{2} \) indicates that moment of inertia is reliant on radius squared, meaning a larger radius significantly increases the inertia value.

If you change the axis of rotation from the center to the edge, a different formula is applied. This property exemplifies why identifying the correct axis of rotation is pivotal in solving physics problems, as it influences both the formulas used and the interpretation of dynamics.
Uniform Circular Disc
A uniform circular disc is a type of circular disc where the mass is evenly distributed across its entire surface. This uniformity implies that any segment of the disc has the same thickness and density as any other segment.

This assumption simplifies calculations, especially when calculating the moment of inertia, because it allows us to use formulas that presume a consistent distribution of mass. For the given exercise, the discs are treated as uniform, meaning the mass calculated through density and area can be directly used without considering any variations throughout the disc.

Uniformity plays a crucial role in theoretical physics as it provides an ideal scenario to study fundamental concepts, like moment of inertia, rotational kinetics, and dynamics of bodies. While most real-world objects are not perfectly uniform, analyzing uniform discs helps in building the foundational understanding required to tackle more complex, non-uniform objects.

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Most popular questions from this chapter

A spherical ball rolls on a table without slipping, then the fraction of its total energy associated with rotation is \(\\{\mathrm{A}\\} 2 / 5\) \\{B \\} \(3 / 5\) \(\\{\mathrm{C}\\} 2 / 7\) \(\\{\mathrm{D}\\} 3 / 7\)

A player caught a cricket ball of mass \(150 \mathrm{gm}\) moving at a rate of \(20 \mathrm{~m} / \mathrm{s}\) If the catching process is Completed in \(0.1\) sec the force of the flow exerted by the ball on the hand of the player ..... N \(\\{\mathrm{A}\\} 3\) \(\\{B\\} 30\) \(\\{\mathrm{C}\\} 150\) \(\\{\mathrm{D}\\} 300\)

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From a uniform circular disc of radius \(\mathrm{R}\), a circular disc of radius \(\mathrm{R} / 6\) and having centre at a distance \(+\mathrm{R} / 2\) from the centre of the disc is removed. Determine the centre of mass of remaining portion of the disc. \(\\{\mathrm{A}\\}[(-\mathrm{R}) / 70]\) \(\\{\mathrm{B}\\}[(+\mathrm{R}) / 70]\) \(\\{\mathrm{C}\\}[(-\mathrm{R}) / 7]\) \(\\{\mathrm{D}\\}[(+\mathrm{R}) / 7]\)

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