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Chapter 5: Problem 624

According to the theorem of parallel axis \(\mathrm{I}=\mathrm{I}_{\mathrm{cm}}+\mathrm{md}^{2}\) the graph between \(\mathrm{I} \rightarrow \mathrm{d}\) will be

Short Answer

Expert verified
The graph between the moment of inertia (I) and distance (d) based on the parallel axis theorem will be a parabolic curve that opens upwards, showing a positive and quadratic relation between I and d. This is due to the equation \(I = I_{cm} + md^2\), where Icm and m are fixed values for a specific object.

Step by step solution

01

Understanding The Variables and Formula

The equation for parallel axis theorem is given by: \(I = I_{cm} + md^{2}\) where: - I: Moment of inertia about an axis parallel to the axis passing through the center of mass - Icm: Moment of inertia about the axis passing through the center of mass - m: Mass of the object - d: Distance between the two parallel axes The equation represents the relationship between the moment of inertia around an axis parallel to the one passing through the center of mass and distance between both axes, mass and moment of inertia around the center of mass.
02

Fixing the Known Values

In order to create a graph between I and d, we need to fix the values of mass (m) and moment of inertia about the center of mass (Icm). These values will remain constant so that we can understand how the moment of inertia (I) varies with distance (d). For example, let's choose m = 5 kg and Icm = 10 kg m^2. (Feel free to choose any other values for clarification)
03

Writing the Equation with Fixed Values

Now that we have our fixed values for mass and moment of inertia about the center of mass, let's substitute them into the formula: \(I = 10 + 5 * d^{2}\)
04

Creating a Table of Values

To create a graph between I and d, we'll need to create a table with a variety of values for distance (d) and use the equation to calculate the corresponding moment of inertia (I). The table may look like this: d | I --------|------- 0 | 10 1 | 15 2 | 30 3 | 55 4 | 90
05

Plotting the Graph

With the table of values for I and d, we can now plot the graph. On the horizontal x-axis, represent the distance (d), and on the vertical y-axis, represent the moment of inertia (I). Plot the points from the table and connect them with a smooth curve. Considering the equation has a quadratic form (\(I = 10 + 5d^2\)), the graph will be a parabolic curve that opens upwards. #Conclusion# The graph between the moment of inertia (I) and distance (d) will be a parabolic curve that opens upwards, based on the theorem of the parallel axis, with the fixed values for mass (m) and moment of inertia about the center of mass (Icm). The graph will show a positive and quadratic relation between I and d.

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Most popular questions from this chapter

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 \(\\{\mathrm{A}\\} 1: 81\) \(\\{\mathrm{B}\\} 1: 27\) \(\\{C\\} 1: 9\) \(\\{\mathrm{D}\\} 1: 3\)

A gramophone record of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) is rotating with angular speed \(\mathrm{W}\). If two pieces of wax each of mass \(\mathrm{M}\) are kept on it at a distance of \(R / 2\) from the centre on opposite side then the new angular velocity will be..... \(\\{\mathrm{A}\\}(\omega / 2)\) \(\\{B\\}[m \omega /(M+m)\) \(\\{C\\}[M \omega /(M+m)]\) \(\\{\mathrm{D}\\}[(\mathrm{M}+\mathrm{m}) / \mathrm{M} \omega]\)

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 \(\ldots .\) \(\\{\mathrm{A}\\}-25.5 \mathrm{rad} / \mathrm{s}^{2}\) \(\\{\mathrm{B}\\}-29.52 \mathrm{rad} / \mathrm{s}^{2}\) \(\\{\mathrm{C}\\}-33.52 \mathrm{rad} / \mathrm{s}^{2}\) \(\\{\mathrm{D}\\}-45.52 \mathrm{rad} / \mathrm{s}^{2}\)

Particles of \(1 \mathrm{gm}, 1 \mathrm{gm}, 2 \mathrm{gm}, 2 \mathrm{gm}\) are placed at the corners \(A, B, C, D\), respectively of a square of side $6 \mathrm{~cm}$ as shown in figure. Find the distance of centre of mass of the system from geometrical centre of square. (A) \(1 \mathrm{~cm}\) (B) \(2 \mathrm{~cm}\) (C) \(3 \mathrm{~cm}\) (D) \(4 \mathrm{~cm}\)

A particle performing uniform circular motion has angular momentum \(L\)., its angular frequency is doubled and its \(K . E\). halved, then the new angular momentum is \(\\{\mathrm{A}\\} 1 / 2\) \\{B \(\\} 1 / 4\) \(\\{\mathrm{C}\\} 2 \mathrm{~L}\) \(\\{\mathrm{D}\\} 4 \mathrm{~L}\)
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