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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rotational Kinetic Energy
Rotational kinetic energy is the energy possessed by a rotating object. This form of kinetic energy is dependent on how fast an object rotates and how its mass is distributed. You can think of it as the spinning counterpart to the more familiar linear kinetic energy, which deals with movement in straight lines.
  • The formula for rotational kinetic energy (\( Er \)) is given by: \( Er = \frac{1}{2} I \omega^2 \)
  • Here, \( I \) represents the moment of inertia, showing how mass is spread out in relation to the axis of rotation.
  • \( \omega \), the angular velocity, describes how fast the object spins or rotates.

This formula reveals that both the mass distribution and rotational speed determine the energy an object carries due to its rotation.
Angular Momentum
Angular momentum, much like its linear cousin, is a measure of the rotational motion of an object. It describes how fast an object rotates and how its mass distribution impacts that rotation.
  • The angular momentum (\( L \)) is given by: \( L = I \omega \)
  • Again, \( I \) refers to the moment of inertia, giving insight into how the mass is allocated.
  • \( \omega \) is the angular velocity.

Angular momentum is conserved in isolated systems without external torques. This means that if no external forces act on a rotating object, its angular momentum remains constant, making it a critical concept in understanding rotational dynamics.
Moment of Inertia
Moment of inertia is an essential concept in rotational kinetics, representing how mass is distributed relative to an axis of rotation. It is sometimes called "rotational inertia," as it reflects how difficult it is to change an object's rotational motion.
- The greater the moment of inertia, the harder it is to accelerate or decelerate the rotation.- Different shapes and mass arrangements cause different moments of inertia.- It can be likened to mass in linear motion, where it indicates resistance to changes in motion.
The moment of inertia (\( I \)) plays a vital role in determining both rotational kinetic energy and angular momentum. Depending on the shape and rotation axis, calculating \( I \) can involve more complex mathematics, but its core function remains tied to mass distribution.
Angular Velocity
Angular velocity describes how fast an object rotates around an axis. It is a fundamental factor in both rotational kinetic energy and angular momentum.
  • Represented by the symbol \( \omega \), angular velocity is measured in radians per second.
  • Unlike linear velocity, which deals with straight-line motion, angular velocity deals with circular paths.
  • Having a higher angular velocity means the object spins faster.

Understanding angular velocity is key to predicting how systems behave when set into rotation, like the spinning blades of a fan or the wheels of a car. It's an intrinsic part of the equations that describe rotational dynamics, linking the rotational speed to energy and momentum.

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Most popular questions from this chapter

Two point masses of \(0.3 \mathrm{~kg}\) and \(0.7 \mathrm{~kg}\) are fixed at the ends of a rod of length \(1.4 \mathrm{~m}\) and of negligible mass. The rod is set rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum, is located at a distance of ..... \(\\{\mathrm{A}\\} 0.4 \mathrm{~m}\) from mass of \(0.3 \mathrm{~kg}\) \\{B \(0.98 \mathrm{~m}\) from mass of \(0.3 \mathrm{~kg}\) \\{C\\} \(0.7 \mathrm{~m}\) from mass of \(0.7 \mathrm{~kg}\) \\{D \(\\} 0.98 \mathrm{~m}\) from mass of \(0.7 \mathrm{~kg}\)

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 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 acceleration of the centre of mass of the disc is \(\\{\mathrm{A}\\} \mathrm{g} \sin \theta\) \(\\{B\\}[(2 g \sin \theta) / 3]\) \(\\{\mathrm{C}\\}[(\mathrm{g} \sin \theta) / 3]\) \(\\{\mathrm{D}\\}[(2 \mathrm{~g} \cos \theta) / 3]\)

Let I be the moment of inertia of a uniform square plate about an axis \(\mathrm{AB}\) that passes through its centre and is parallel to two of its sides \(\mathrm{CD}\) is a line in the plane of the plate that passes through the centre of the plate and makes an angle of \(\theta\) with \(\mathrm{AB}\). The moment of inertia of the plate about the axis \(\mathrm{CD}\) is then equal to.... \(\\{\mathrm{A}\\} \mathrm{I}\) \(\\{B\\} I \sin ^{2} \theta\) \(\\{C\\} I \cos ^{2} \theta\) \(\\{\mathrm{D}\\} I \cos ^{2}(\theta / 2)\)

From a circular disc of radius \(\mathrm{R}\) and mass \(9 \mathrm{M}\), a small disc of radius \(\mathrm{R} / 3\) is removed from the disc. The moment of inertia of the remaining portion about an axis perpendicular to the plane of the disc and passing through \(\mathrm{O}\) is.... \(\\{\mathrm{A}\\} 4 \mathrm{MR}^{2}\) \(\\{\mathrm{B}\\}(40 / 9) \mathrm{MR}^{2}\) \(\\{\mathrm{C}\\} 10 \mathrm{MR}^{2}\) \(\\{\mathrm{D}\\}(37 / 9) \mathrm{MR}^{2}\)

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