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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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