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Chapter 5: Problem 566

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: \[{B}\\ \, 3:1 \]

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Most popular questions from this chapter

A solid cylinder of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) rolls down an inclined plane of height \(\mathrm{h}\). The angular velocity of the cylinder when it reaches the bottom of the plane will be. \(\\{\mathrm{A}\\}(2 / \mathrm{R}) \sqrt{(\mathrm{gh})}\) \(\\{\mathrm{B}\\}(2 / \mathrm{R}) \sqrt{(\mathrm{gh} / 2)}\) \(\\{\mathrm{C}\\}(2 / \mathrm{R}) \sqrt{(\mathrm{gh} / 3)}\) \(\\{\mathrm{D}\\}(1 / 2 \mathrm{R}) \sqrt{(\mathrm{gh})}\)

Two identical hollow spheres of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) are joined together and the combination is rotated about an axis tangential to one sphere and perpendicular to the line connecting their centres. The moment of inertia of the combination is \(\\{\mathrm{A}\\} 10 \mathrm{MR}^{2}\) \(\\{\mathrm{B}\\}(4 / 3) \mathrm{MR}^{2}\) \(\\{\mathrm{C}\\}(32 / 3) \mathrm{MR}^{2}\) \(\\{\mathrm{D}\\}(34 / 3) \mathrm{MR}^{2}\)

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 \(\\{\mathrm{A}\\}(1 / 4) \mathrm{mv}^{2}\) \(\\{\mathrm{B}\\}(1 / 2) \mathrm{mv}^{2}\) \(\\{\mathrm{C}\\}(3 / 4) \mathrm{mv}^{2}\) \(\\{\mathrm{D}\\} \mathrm{mv}^{2}\)

The moment of inertia of a meter scale of mass \(0.6 \mathrm{~kg}\) about an axis perpendicular to the scale and passing through \(30 \mathrm{~cm}\) position on the scale is given by (Breath of scale is negligible). \(\\{\mathrm{A}\\} 0.104 \mathrm{kgm}^{2}\) \\{B \(\\} 0.208 \mathrm{kgm}^{2}\) \(\\{\mathrm{C}\\} 0.074 \mathrm{kgm}^{2}\) \(\\{\mathrm{D}\\} 0.148 \mathrm{kgm}^{2}\)

A constant torque of \(1500 \mathrm{Nm}\) turns a wheel of moment of inertia \(300 \mathrm{~kg} \mathrm{~m}^{2}\) about an axis passing through its centre the angular velocity of the wheel after 3 sec will be.......... $\mathrm{rad} / \mathrm{sec}$ \(\\{\mathrm{A}\\} 5\) \\{B \\} 10 \(\\{C\\} 15\) \(\\{\mathrm{D}\\} 20\)
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