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Chapter 5: Problem 658

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 \\} [ \(\left[\mathrm{ML}^{2} / 12\right]\) \(\\{\mathrm{C}\\}\left[\left(\mathrm{ML}^{2} / 9\right]\right.\) \(\\{\mathrm{D}\\}\left[\left(\mathrm{ML}^{2} / 3\right]\right.\)

Short Answer

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

Step by step solution

01

Recall the parallel axis theorem

The parallel axis theorem states that the moment of inertia I about any axis can be found using the moment of inertia Ic about a parallel axis passing through the center of mass and the mass M of the object: \[ I = I_{c} + Md^2\] where d is the distance between the two parallel axes.
02

Calculate the moment of inertia about the center of mass

First, we need to find the moment of inertia of the thin rod about a parallel axis passing through the center of mass. For a thin uniform rod rotating about an axis perpendicular to the rod at its midpoint, the moment of inertia is: \[I_c = \frac{1}{12}ML^2\] We will use this equation in the parallel axis theorem to find the moment of inertia about the given axis.
03

Calculate the distance between the axes

The problem states that the axis passes through a point at distance L/4 from one end. Since the axis is perpendicular to the rod, and the axis through the center of mass is also perpendicular, we can say that the two axes are parallel to each other. The distance d between the axes is the distance from the center of mass to the given axis, which is: \[d = \frac{L}{4}\]
04

Use the parallel axis theorem to find the moment of inertia

Now, we use the parallel axis theorem to find the moment of inertia I about the given axis: \[ I = I_c + Md^2 = \frac{1}{12} ML^2 + M\left(\frac{L}{4}\right)^2\] \[ I = \frac{1}{12} ML^2 + \frac{1}{16}ML^2\] \[ I = \left(\frac{1}{12}+\frac{1}{16}\right)ML^2 = \frac{7}{48}ML^2\]
05

Compare the result with the given options

We found that the moment of inertia of the thin rod about the given axis is (7/48)ML^2. Comparing this result with the given options, we can see that it matches option A. So the correct answer is: \(\\{A\\}\left[\left(7 \mathrm{ML}^{2}\right) / 48\right]\)

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Most popular questions from this chapter

A wheel of mass \(10 \mathrm{~kg}\) has a moment of inertia of $160 \mathrm{~kg} \mathrm{~m}$ radius of gyration will be $\begin{array}{llll}\\{\mathrm{A}\\} 10 & \\{\mathrm{~B}\\} 8 & \\{\mathrm{C}\\} 6 & \\{\mathrm{D}\\} 4\end{array}$

A thin circular ring of mass \(\mathrm{M}\) and radius \(\mathrm{r}\) is rotating about its axis with a constant angular velocity \(\mathrm{w}\). Two objects each of mass \(\mathrm{m}\) are attached gently to the opposite ends of a diameter of the ring. The ring will now rotate with an angular velocity.... $\\{\mathrm{A}\\}[\\{\omega(\mathrm{M}-2 \mathrm{~m})\\} /\\{\mathrm{M}+2 \mathrm{~m}\\}]$ \(\\{\mathrm{B}\\}[\\{\omega \mathrm{M}\\} /\\{\mathrm{M}+2 \mathrm{~m}\\}]\) \(\\{C\\}[\\{\omega M)\\} /\\{M+m\\}]\) \(\\{\mathrm{D}\\}[\\{\omega(\mathrm{M}+2 \mathrm{~m})\\} / \mathrm{M}]\)

A uniform disc of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) rolls without slipping down a plane inclined at an angle \(\theta\) with the horizontal. The frictional force on the disc is \(\\{\mathrm{A}\\}[(\mathrm{Mg} \sin \theta) / 3]\) \(\\{\mathrm{B}\\}[(2 \mathrm{Mg} \sin \theta) / 3]\) \(\\{\mathrm{C}\\} \mathrm{Mg} \sin \theta\) \(\\{\mathrm{D}\\}\) None

A body of mass \(\mathrm{m}\) is tied to one end of spring and whirled round in a horizontal plane with a constant angular velocity. The elongation in the spring is one centimetre. If the angular velocity is doubted, the elongation in the spring is \(5 \mathrm{~cm}\). The original length of spring is... \(\\{\mathrm{A}\\} 16 \mathrm{~cm}\) \(\\{B\\} 15 \mathrm{~cm}\) \(\\{\mathrm{C}\\} 14 \mathrm{~cm}\) \(\\{\mathrm{D}\\} 13 \mathrm{~cm}\)

Identify the correct statement for the rotational motion of a rigid body \(\\{A\\}\) Individual particles of the body do not undergo accelerated motion \\{B \\} The centre of mass of the body remains unchanged. \\{C\\} The centre of mass of the body moves uniformly in a circular path \\{D\\} Individual particle and centre of mass of the body undergo an accelerated motion.
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