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Chapter 5: Problem 655

A cylinder of mass \(\mathrm{M}\) has length \(\mathrm{L}\) that is 3 times its radius what is the ratio of its moment of inertia about its own axis and that about an axis passing through its centre and perpendicular to its axis? \(\\{\mathrm{A}\\} 1\) \(\\{\mathrm{B}\\}(1 / \sqrt{3})\) \(\\{\mathrm{C}\\} \sqrt{3}\) \(\\{\mathrm{D}\\}(\sqrt{3} / 2)\)

Short Answer

Expert verified
The short answer based on the step-by-step solution is: None of the given options match the correct ratio, which is \(\frac{3}{7}\).

Step by step solution

01

Moment of Inertia about its own axis

The moment of inertia, \(I_A\), of a cylinder about its own axis, assuming it is a solid cylinder, can be found using the following formula: \[ I_A = \frac{1}{2}MR^2, \] where \(M\) is the mass of the cylinder and \(R\) is its radius. In this case, \(R = r\). So, the moment of inertia about its own axis can be calculated as: \[ I_A = \frac{1}{2}Mr^2. \]
02

Moment of Inertia about the perpendicular axis

The moment of inertia of a cylinder about an axis passing through its center and perpendicular to its axis, \(I_B\), can be found using this formula: \[ I_B = \frac{1}{12}M(3d^2 + h^2), \] where d is the diameter of the cylinder which is \(2r\) (2 times the radius) and h is the height of the cylinder which is \(3r\), given the length is 3 times its radius. So, the moment of inertia about the perpendicular axis can be calculated as: \[ I_B = \frac{1}{12}M(3(2r)^2 + (3r)^2). \] Now, we have both moments of inertia \(I_A\) and \(I_B\). To find the ratio of these moments of inertia, we just need to divide them.
03

Finding the ratio

We will now find the ratio as follows: \[ \text{ratio} = \frac{I_A}{I_B} = \frac{\frac{1}{2}Mr^2}{\frac{1}{12}M(3(2r)^2 + (3r)^2)}. \] We can simplify the fraction, canceling out \(M\) and \(r^2\) from both the numerator and denominator and simplifying the expression: \[ \text{ratio}=\frac{\frac{1}{2}}{\frac{1}{12}(3(4) + 9)} = \frac{\frac{1}{2}}{\frac{1}{12}(12+9)} = \frac{\frac{1}{2}}{\frac{1}{12}(21)} = \frac{1}{2}\times\frac{12}{21}. \] Further simplifying the expression: \[ \text{ratio} = \frac{1}{2}\times\frac{6}{7} = \frac{6}{14} = \frac{3}{7}. \] Hence, none of the given options matches the correct ratio.

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Most popular questions from this chapter

A thin uniform rod \(A B\) of mass \(M\) and length \(L\) is hinged at one end \(\mathrm{A}\) to the horizontal floor. Initially it stands vertically. It is allowed to fall freely on the floor in the vertical plane. The angular velocity of the rod when its end \(B\) strikes the floor is \(\\{\mathrm{A}\\} \sqrt{(\mathrm{g} / \mathrm{L})}\) \(\\{\mathrm{B}\\} \sqrt{(2 \mathrm{~g} / \mathrm{L})}\) \(\\{C\\} \sqrt{(3 g / L)}\) \(\\{\mathrm{D}\\} 2 \sqrt{(\mathrm{g} / \mathrm{L})}\)

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 : \(\\{\mathrm{A}\\} 30 \mathrm{~m} / \mathrm{s}\) \(\\{\mathrm{B}\\} 20 \mathrm{~m} / \mathrm{s}\) \(\\{\mathrm{C}\\} 10 \mathrm{~m} / \mathrm{s}\) \(\\{\mathrm{D}\\} 5 \mathrm{~m} / \mathrm{s}\)

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\\} \(1: 9\) \(\\{\mathrm{D}\\} 9: 1\)

The moment of inertia of a thin rod of mass \(\mathrm{M}\) and length \(\mathrm{L}\) about an axis passing through the point at a distance $\mathrm{L} / 4$ from one of its ends and perpendicular to the rod is \(\\{\mathrm{A}\\}\left[\left(7 \mathrm{ML}^{2}\right) / 48\right]\) \\{B \\} [ \(\left[\mathrm{ML}^{2} / 12\right]\) \(\\{\mathrm{C}\\}\left[\left(\mathrm{ML}^{2} / 9\right]\right.\) \(\\{\mathrm{D}\\}\left[\left(\mathrm{ML}^{2} / 3\right]\right.\)

Two discs of the same material and thickness have radii \(0.2 \mathrm{~m}\) and \(0.6 \mathrm{~m}\) their moment of inertia about their axes will be in the ratio \(\\{\mathrm{A}\\} 1: 81\) \(\\{\mathrm{B}\\} 1: 27\) \(\\{C\\} 1: 9\) \(\\{\mathrm{D}\\} 1: 3\)
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