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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 solid cylinder of mass \(\mathrm{M}\) and \(\mathrm{R}\) is mounted on a frictionless horizontal axle so that it can freely rotate about this axis. A string of negligible mass is wrapped round the cylinder and a body of mass \(\mathrm{m}\) is hung from the string as shown in figure the mass is released from rest then The angular speed of cylinder is proportional to \(\mathrm{h}^{\mathrm{n}}\), where \(\mathrm{h}\) is the height through which mass falls, Then the value of \(n\) is \(\\{\mathrm{A}\\}\) zero \(\\{\mathrm{B}\\} 1\) \(\\{\mathrm{C}\\}(1 / 2)\) \([\mathrm{D}] 2\)

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 car is moving at a speed of \(72 \mathrm{~km} / \mathrm{hr}\) the radius of its wheel is \(0.25 \mathrm{~m}\). If the wheels are stopped in 20 rotations after applying breaks then angular retardation produced by the breaks is \(\ldots .\) \(\\{\mathrm{A}\\}-25.5 \mathrm{rad} / \mathrm{s}^{2}\) \(\\{\mathrm{B}\\}-29.52 \mathrm{rad} / \mathrm{s}^{2}\) \(\\{\mathrm{C}\\}-33.52 \mathrm{rad} / \mathrm{s}^{2}\) \(\\{\mathrm{D}\\}-45.52 \mathrm{rad} / \mathrm{s}^{2}\)

In HCl molecule the separation between the nuclei of the two atoms is about \(1.27 \mathrm{~A}\left(1 \mathrm{~A}=10^{-10}\right)\). The approximate location of the centre of mass of the molecule is \(-\mathrm{A}\) i \(\wedge\) with respect of Hydrogen atom (mass of CL is \(35.5\) times of mass of Hydrogen \()\) \(\\{\mathrm{A}\\} 1 \mathrm{i}\) \\{B \\} \(2.5 \mathrm{i}\) \\{C \(\\} 1.24 \mathrm{i}\) \\{D \(1.5 \mathrm{i}\)

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 \(\\} 0.208 \mathrm{kgm}^{2}\) \(\\{\mathrm{C}\\} 0.074 \mathrm{kgm}^{2}\) \(\\{\mathrm{D}\\} 0.148 \mathrm{kgm}^{2}\)

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