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Chapter 5: Problem 594

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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Most popular questions from this chapter

The moment of inertia of a uniform circular disc of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) about any of its diameter is \((1 / 4) \mathrm{MR}^{2}\), what is the moment of inertia of the disc about an axis passing through its centre and normal to the disc? \(\\{\mathrm{A}\\} \mathrm{MR}^{2}\) \\{B \(\\}(1 / 2) \mathrm{MR}^{2}\) \(\\{\mathrm{C}\\}(3 / 2) \mathrm{MR}^{2}\) \(\\{\mathrm{D}\\} 2 \mathrm{MR}^{2}\)

The M.I. of a body about the given axis is \(1.2 \mathrm{kgm}^{2}\) initially the body is at rest. In order to produce a rotational kinetic energy of $1500 \mathrm{~J}\(. an angular acceleration of \)25 \mathrm{rad} \mathrm{sec}^{2}$ must be applied about that axis for duration of \(\ldots\) \(\\{\mathrm{A}\\} 4 \mathrm{sec}\) \\{B \(\\} 2 \mathrm{sec}\) \(\\{C\\} 8 \mathrm{sec}\) \(\\{\mathrm{D}\\} 10 \mathrm{sec}\)

A straight rod of length \(L\) has one of its ends at the origin and the other end at \(\mathrm{x}=\mathrm{L}\) If the mass per unit length of rod is given by Ax where \(A\) is constant where is its centre of mass. \(\\{\mathrm{A}\\} \mathrm{L} / 3\) \(\\{\mathrm{B}\\} \mathrm{L} / 2\) \(\\{\mathrm{C}\\} 2 \mathrm{~L} / 3\) \(\\{\mathrm{D}\\} 3 \mathrm{~L} / 4\)

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In a bicycle the radius of rear wheel is twice the radius of front wheel. If \(\mathrm{r}_{\mathrm{F}}\) and \(\mathrm{r}_{\mathrm{r}}\) are the radius, \(\mathrm{v}_{\mathrm{F}}\) and \(\mathrm{v}_{\mathrm{r}}\) are speed of top most points of wheel respectively then... \(\\{\mathrm{A}\\} \mathrm{v}_{\mathrm{r}}=2 \mathrm{v}_{\mathrm{F}}\) $\\{\mathrm{B}\\} \mathrm{v}_{\mathrm{F}}=2 \mathrm{v}_{\mathrm{r}} \quad\\{\mathrm{C}\\} \mathrm{v}_{\mathrm{F}}=\mathrm{v}_{\mathrm{r}}$ \(\\{\mathrm{D}\\} \mathrm{v}_{\mathrm{F}}>\mathrm{v}_{\mathrm{r}}\)
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