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Chapter 7: Problem 13

(a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of \(40 \mathrm{rev} / \mathrm{min} .\) How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to \(2 / 5\) times the initial value ? Assume that the turntable rotates without friction. (b) Show that the child's new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?

Short Answer

Expert verified
(a) The angular speed is 100 rev/min. (b) The child's new kinetic energy is greater due to work done by internal forces.

Step by step solution

01

Understand the Conservation of Angular Momentum

The principle of conservation of angular momentum states that if no external torque acts on a system, the total angular momentum of the system remains constant. This means:\[ L_i = L_f \]where \( L_i \) is the initial angular momentum and \( L_f \) is the final angular momentum.
02

Determine Angular Momentum Formula

Angular momentum \( L \) is given by \( L = I \omega \), where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. Let's apply this formula for initial and final states:\[ I_i \omega_i = I_f \omega_f \]
03

Find Final Angular Speed

Assume the initial moment of inertia is \( I_i \) and initial angular velocity is \( \omega_i = 40 \mathrm{rev/min} \). Given that \( I_f = \frac{2}{5} I_i \), substitute into the angular momentum equation:\[ I_i \times 40 = \frac{2}{5} I_i \times \omega_f \]Solving for \( \omega_f \), we find:\[ \omega_f = \frac{5}{2} \times 40 = 100 \mathrm{rev/min} \]
04

Calculate Initial and Final Kinetic Energy

The rotational kinetic energy \( KE \) is given by:\[ KE = \frac{1}{2} I \omega^2 \]First, calculate the initial kinetic energy \( KE_i \):\[ KE_i = \frac{1}{2} I_i (40 \times \frac{2\pi}{60})^2 \]Next, calculate the final kinetic energy \( KE_f \):\[ KE_f = \frac{1}{2} \times \frac{2}{5} I_i (100 \times \frac{2\pi}{60})^2 \]
05

Compare Initial and Final Kinetic Energy

Simplify the expressions:- Initial: \[ KE_i = \frac{1}{2} I_i \times \left(\frac{4\pi}{3}\right)^2 \]- Final: \[ KE_f = \frac{1}{2} \times \frac{2}{5} I_i \times \left(\frac{10\pi}{3}\right)^2 \]Calculate the ratio \( \frac{KE_f}{KE_i} \) to show \( KE_f > KE_i \). Thus:\[ \frac{KE_f}{KE_i} = \frac{1}{5} \times 2 \times \left(\frac{10}{4}\right)^2 = rac{1}{5} \times 2 \times \frac{100}{16} > 1 \].Therefore, \( KE_f > KE_i \).
06

Explain Increase in Kinetic Energy

The increase in kinetic energy can be attributed to the work done on the system when the child folds his arms. By pulling his arms inward, the child adds energy to the system through internal forces, thus increasing the rotational speed and kinetic energy.

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Key Concepts

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

Moment of Inertia
The moment of inertia is a property of a rotating body that determines how much resistance it offers to changes in its angular motion. Think of it as the rotational equivalent of mass for linear motion. The more spread out the mass is from the axis of rotation, the higher the moment of inertia. This means that a figure skating child with arms extended has a higher moment of inertia, requiring more effort to spin around. However, when the child pulls their arms in, they effectively reduce their moment of inertia.
  • The moment of inertia depends not just on how much mass an object has, but also on how that mass is distributed regarding the axis around which it rotates.
  • Mathematically, it's expressed as: \[ I = extstyle rac{1}{ extstyle 2} m r^2 \], where \( I \) is the moment of inertia, \( m \) is the mass of the object, and \( r \) is the distance of the mass from the axis of rotation.
  • In this exercise, the moment of inertia decreased when the child tucked their arms, taking it down to two-fifths of the initial value.
Angular Velocity
Angular velocity tells you how fast something is rotating. It is the measure of how many turns or revolutions the object makes in a unit time, usually expressed in revolutions per minute (rpm) or radians per second (rad/s). In the exercise, the child's angular velocity changes as a result of pulling in their arms, enhancing the conservation of angular momentum.
  • Initially, the child and turntable rotated together at 40 rpm. By folding the arms, the moment of inertia is reduced, which increases the angular velocity to 100 rpm due to the conservation of angular momentum.
  • This relationship is due to the fact that angular momentum \( L \) \[ L = I \omega \] where \( I \) is the moment of inertia and \( \omega \) is the angular velocity, must be conserved in the absence of external torques.
  • A vital point to remember is that decreasing the moment of inertia results in an increase in angular velocity, given no external force acts upon the system.
Rotational Kinetic Energy
Rotational kinetic energy is the part of the total kinetic energy in a rotating system. It arises from the object's rotation around an axis and is directly linked to both its moment of inertia and angular velocity. This exercise demonstrates how the change in distribution of mass affects the kinetic energy of a rotating system.
  • The formula for rotational kinetic energy is \[ KE = \frac{1}{2} I \omega^2 \], where \( KE \) is the kinetic energy, \( I \) is the moment of inertia, and \( \omega \) is the angular velocity.
  • Initially, the child has a certain kinetic energy with extended arms. However, when the arms are folded, both the moment of inertia and kinetic energy change, leading to an increase in kinetic energy.
  • This increase occurs because, while the moment of inertia decreases, the angular velocity increases significantly, which compensates and results in higher kinetic energy than initially.
  • This concept illustrates how internal changes, like altering an object's shape and mass distribution, can impact energy outcomes in rotational dynamics.

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