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Chapter 5: Problem 627

Let \(\mathrm{Er}\) is the rotational kinetic energy and \(\mathrm{L}\) is angular momentum then the graph between \(\log \mathrm{e}^{\mathrm{Er}}\) and $\log \mathrm{e}^{\mathrm{L}}$ can be

Short Answer

Expert verified
The graph between \( \log e^{Er} \) and \( \log e^{L} \) is a straight line with a slope of 2.

Step by step solution

01

Recall the formulas for rotational kinetic energy and angular momentum

Rotational kinetic energy (Er) is given by the formula: \[ Er = \frac{1}{2}Iω^2 \] where \(I\) is the moment of inertia and ω is the angular velocity. Angular momentum (L) is given by the formula: \[ L = Iω \]
02

Derive the relationship between Er and L

Using the formulas for Er and L, we can observe that \(Er = \frac{1}{2}(L^2/I)\). Now let's take the natural logarithm (ln) of both sides: \[ \ln Er = \ln\left(\frac{1}{2}\frac{L^2}{I}\right) \] Using the properties of logarithms, we can break this up further: \[ \ln Er = \ln\left(\frac{1}{2}\right) + \ln\left(\frac{L^2}{I}\right) \] \[ \ln Er = \ln\left(\frac{1}{2}\right) + 2\ln L - \ln I \]
03

Express the relationship by using log_e

We can express the natural logarithm (ln) using base e as: \( \ln x = \log_ e^x \). Applying this to the relationship we derived in step 2, we have: \[ \log_e e^{Er} = \log_e e^{\ln\left(\frac{1}{2}\right) + 2\ln L - \ln I} \]
04

Recognize the form of the graph

Now, let \(x = \log_e e^{L}\) and \(y = \log_e e^{Er}\). Also let \(a = \ln\left(\frac{1}{2}\right)\) and \(b = -\ln I\). Then we have the following relationship: \[y = a + 2x + b\] This relationship is the equation for a straight line y = mx + c, where m = 2, \(x = \log_e e^{L}\), and y = \(\log_e e^{Er}\). The graph between \(\log_e e^{Er}\) and \(\log_e e^{L}\) is a straight line with a slope of 2.

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

A rod of length L rotate about an axis passing through its centre and normal to its length with an angular velocity \(\omega\). If A is the cross-section and \(D\) is the density of material of rod. Find its rotational $\mathrm{K} . \mathrm{E}$. \(\\{\mathrm{A}\\}(1 / 2) \mathrm{AL}^{3} \mathrm{D} \omega^{2}\) \\{B \(\\}(1 / 6) \mathrm{AL}^{3} \mathrm{D} \omega^{2}\) \(\\{C\\}(1 / 24) A L^{3} D \omega^{2}\) \(\\{\mathrm{D}\\}(1 / 12) \mathrm{AL}^{3} \mathrm{D} \omega^{2}\)

A cylinder of mass \(5 \mathrm{~kg}\) and radius \(30 \mathrm{~cm}\), and free to rotate about its axis, receives an angular impulse of $3 \mathrm{~kg} \mathrm{M}^{2} \mathrm{~S}^{-1}$ initially followed by a similar impulse after every \(4 \mathrm{sec}\). what is the angular speed of the cylinder 30 sec after initial impulse? The cylinder is at rest initially. \(\\{\mathrm{A}\\} 106.7 \mathrm{rad} \mathrm{S}^{-1}\) \\{B\\} \(206.7 \mathrm{rad} \mathrm{S}^{-1}\) \\{C\\} \(107.6 \mathrm{rad} \mathrm{S}^{-1}\) \\{D \(\\} 207.6 \mathrm{rad} \mathrm{S}^{-1}\)

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
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