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Chapter 7: Problem 46

If the angular momentum of a rotating body is increased by \(200 \%\), then its kinetic energy of rotation will be increased by (a) \(400 \%\) (b) \(800 \%\) (c) \(200 \%\) (d) \(100 \%\)

Short Answer

Expert verified
The kinetic energy of rotation will be increased by \(800\%\). Thus, the correct option is (b) \(800\%\).

Step by step solution

01

Definition

The kinetic energy \(K\) of a rotating body is given by the relation \(K = \frac{1}{2} I {\omega}^2\), where \(I\) represents the moment of inertia and \(\omega\) the angular velocity. The angular momentum \(L\) of the object is given by the relation \(L = I \omega\).
02

Express angular velocity in terms of angular momentum

From \(L = I \omega\), we can express \(\omega\) in terms of \(L\) and \(I\). So, \(\omega = \frac{L}{I}\).
03

Substitute the expression of angular velocity in kinetic energy

Now using the relation of angular momentum, substitute the term \(\omega\) in the equation of kinetic energy. So, \(K = \frac{1}{2} I (\frac{L}{I})^2 = \frac{L^2}{2I}\). From this relation, we see that the kinetic energy \(K\) is proportional to the square of the angular momentum \(L\).
04

Find the increase in kinetic energy

Given that the angular momentum \(L\) increases by \(200\%\), it becomes \(3L\). The new kinetic energy \(K'\) = \frac{(3L)^2}{2I} = 9 \times \frac{L^2}{2I} = 9K. So, there is an increase of \(800\%\) in kinetic energy.

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Key Concepts

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

Angular Momentum
Angular momentum is a crucial concept in rotational motion. It describes the amount of rotational motion an object has and depends on two main factors: the object's moment of inertia (I) and its angular velocity (\omega). The formula for angular momentum (L) is given by \( L = I \omega \). This formula shows that angular momentum increases if either the moment of inertia or angular velocity increases.

When dealing with situations where the angular momentum changes, understanding how it affects other properties like kinetic energy is essential. For example, in the problem above, if the angular momentum increases by 200%, it impacts the rotational kinetic energy, as shown through calculations. This connection between angular momentum and kinetic energy is a key topic in studying rotational dynamics.
Kinetic Energy
Kinetic energy in rotational motion works similarly to linear motion but with a twist. It's the energy an object possesses due to its rotation and is calculated using the formula \( K = \frac{1}{2} I \omega^2 \). Here, the moment of inertia (I) and angular velocity (\omega) both play a role.

Understanding the relationship between kinetic energy and angular momentum is essential. As we've seen, the kinetic energy can be expressed in terms of angular momentum: \( K = \frac{L^2}{2I} \). This shows that kinetic energy depends on the square of the angular momentum, meaning any change in angular momentum significantly affects the kinetic energy. For instance, if angular momentum increases by 200%, the kinetic energy increases by 800%, demonstrating the powerful impact of these variables on each other in rotational motion scenarios.
Moment of Inertia
The moment of inertia is akin to mass in linear motion, representing an object's resistance to change in its rotational motion. It depends on both the distribution of an object's mass and the axis around which it rotates. A larger moment of inertia means the object is harder to spin or stop spinning.

In the relation \( L = I \omega \), the moment of inertia (I) determines how changes in angular velocity affect angular momentum. If you keep mass constant but change the shape or the axis, you can alter the moment of inertia, directly impacting rotational dynamics. Comprehending this concept helps in analyzing how angular momentum and angular velocity contribute to the overall rotational kinetic energy.
Angular Velocity
Angular velocity refers to how fast something is rotating. It's the rate of change of angular displacement and is usually measured in radians per second. Understanding angular velocity is essential for describing any rotational motion.

In the formulas \( L = I \omega \) and \( K = \frac{1}{2} I \omega^2 \), angular velocity (\omega) is a key component. Since it's directly involved in these expressions, changes in angular velocity can lead to alterations in both angular momentum and kinetic energy. Learning how angular velocity interacts with moment of inertia and angular momentum provides a comprehensive view of an object's rotational characteristics, helping to solve complex problems involving rotation.

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Most popular questions from this chapter

A ring, a solid sphere and a thin disc of different masses rotate with the same kinetic energy. Equal torques are applied to stop them. Which will make the least number of rotations before coming to rest (a) Disc (b) Ring (c) Solid sphere (d) All will make same number of rotations

Three point masses each of mass \(m\) are placed at the corners of an equilateral triangle of side \(a\). Then the moment of inertia of this system about an axis passing along one side of the triangle is [AIIMS 1995. (a) \(m a^{2}\) (b) \(3 m a^{2}\) (c) \(\frac{3}{4} m a^{2}\) (d) \(\frac{2}{3} m a^{2}\)

The wheel of a car is rotating at the rate of 1200 revolutions per minute. On pressing the accelerator for 10 sec it starts rotating at 4500 revolutions per minute. The angular acceleration of the wheel is [MP PET 2001] (a) 30 radians \(/ \mathrm{sec}^{2}\) (b) 188 o degrees \(/ \mathrm{sec}^{2}\) (c) 40 radians \(/ \mathrm{sec}^{2}\) (d) 1980

An automobile engine develops \(100 \mathrm{~kW}\) when rotating at a speed of \(1800 \mathrm{rev} / \mathrm{min}\). What torque does it deliver (a) \(350 \mathrm{~N}-\mathrm{m}\) (b) \(440 \mathrm{~N}-\mathrm{m}\) (c) \(531 \mathrm{~N}-\mathrm{m}\) (d) \(628 \mathrm{~N}-\mathrm{m}\)

A horizontal heavy uniform bar of weight \(W\) is supported at its ends by two men. At the instant, one of the men lets go off his end of the rod, the other feels the force on his hand changed to (a) \(W\) (b) \(\frac{W}{2}\) (c) \(\frac{3 W}{4}\) (d) \(\frac{W}{4}\)Solution: (d) Let the mass of the rod is \(M \quad \therefore\) Weight \((W)=M g\) Initially for the equilibrium \(F+F=M g \Rightarrow F=M g / 2\) When one man withdraws, the torque on the rod \(\tau=I \alpha=M g \frac{l}{2}\) \(\Rightarrow \frac{M l^{2}}{3} \alpha=M g \frac{l}{2} \quad\left[\right.\) As \(\left.I=M l^{2} / 3\right]\) \(\Rightarrow\) Angular acceleration \(\alpha=\frac{3}{2} \frac{g}{l}\) and linear acceleration \(a=\frac{l}{2} \alpha=\frac{3 g}{4}\) Now if the new normal force at \(A\) is \(F\) then \(M g-F=M a\) \(\Rightarrow F=M g-M a=M g-\frac{3 M g}{4}=\frac{M g}{4}=\frac{W}{4} .\)
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