/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 64 A ballerina begins a tour jet茅 ... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 64

A ballerina begins a tour jet茅 (Fig. \(11-19 a\) ) with angular speed \(\omega_{i}\) and a rotational inertia consisting of two parts: \(I_{\text {leg }}=1.44 \mathrm{~kg} \cdot \mathrm{m}^{2}\) for her leg extended outward at angle \(\theta=90.0^{\circ}\) to her body and \(I_{\text {trunk }}=0.660 \mathrm{~kg} \cdot \mathrm{m}^{2}\) for the rest of her body (primarily her trunk). Near her maximum height she holds both legs at angle \(\theta=30.0^{\circ}\) to her body and has angular speed \(\omega_{f}\) (Fig. \(11-19 b\) ). Assuming that \(I_{\text {trunk }}\) has not changed, what is the ratio \(\omega_{f} / \omega_{i}\) ?

Short Answer

Expert verified
The ratio \(\omega_f / \omega_i\) is approximately 1.20.

Step by step solution

01

Understand the conservation of angular momentum

In this problem, angular momentum is conserved because no external torques are acting on the ballerina. This means that the initial angular momentum \(L_i\) is equal to the final angular momentum \(L_f\).
02

Write the expression for initial angular momentum

The initial angular momentum \(L_i\) is given by the product of the initial angular velocity \(\omega_i\) and the initial total moment of inertia \(I_i\). So, \(L_i = (I_{\text{leg}} + I_{\text{trunk}}) \cdot \omega_i\). With the given values, \(L_i = (1.44 + 0.660)\, \text{kg} \cdot \text{m}^2 \cdot \omega_i\).
03

Calculate the adjusted moment of inertia after the leg position changes

When both legs are positioned at \(\theta = 30.0^{\circ}\), the moment of inertia for the leg changes. The moment of inertia for each leg (assuming symmetry) can be written as \(I_{\text{leg}} \cdot \cos^2(\theta)\). Substituting \(\theta = 30.0^{\circ}\), \(I_{\text{leg, new}} = 2 \cdot (1.44\ \cos^2(30^{\circ}))\). Here, the factor of 2 accounts for both legs.
04

Calculate the total final moment of inertia

The final moment of inertia \(I_f\) will be the sum of the adjusted leg moment of inertia for both legs and the unchanged trunk moment of inertia: \(I_f = 2 \cdot (1.44\ \cos^2(30^{\circ})) + 0.660\).
05

Write the expression for final angular momentum

The final angular momentum \(L_f\) is given by the product of the final angular velocity \(\omega_f\) and the final moment of inertia \(I_f\). Thus, \(L_f = I_f \cdot \omega_f\).
06

Equate initial and final angular momentum using conservation law

Since \(L_i = L_f\), we have \[(I_{\text{leg}} + I_{\text{trunk}}) \cdot \omega_i = I_f \cdot \omega_f.\]
07

Solve for the ratio \(\omega_f / \omega_i\)

Rearrange the expression from Step 6 to find \(\omega_f / \omega_i\): \[\frac{\omega_f}{\omega_i} = \frac{I_{\text{leg}} + I_{\text{trunk}}}{I_f}.\] Substitute the values \[(1.44 + 0.660) / (2 \cdot 1.44 \cdot \cos^2(30^{\circ}) + 0.660).\] After calculating, we find the ratio.

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Key Concepts

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

Conservation of Angular Momentum
The conservation of angular momentum is a fundamental principle in physics. This law states that if no external torque acts on a system, the total angular momentum of the system remains constant. In simpler terms, how fast or slow something spins can only change if something from outside pushes or pulls on it.
For example, consider the ballerina in the exercise. As she performs her dance, there are no external forces like someone grabbing her to slow her down or speed her up. Hence, her angular momentum stays the same.
  • Initial angular momentum is represented as \(L_i = I_i \cdot \omega_i\), where \(I_i\) is the initial moment of inertia and \(\omega_i\) is the initial angular velocity.
  • Final angular momentum is represented as \(L_f = I_f \cdot \omega_f\), where \(I_f\) is the final moment of inertia and \(\omega_f\) is the final angular velocity.
By setting \(L_i = L_f\) due to conservation, we can solve for unknown quantities like the final angular velocity.
Moment of Inertia
The moment of inertia is like the rotational equivalent of mass in linear motion. It's a measure of how hard it is to change the rotational speed of an object. The moment of inertia depends on how the mass is distributed relative to the axis of rotation.
In our ballerina example, her moment of inertia changes when she changes the position of her legs. Initially, with the leg extended at \(90^\circ\), the inertia is higher because the mass is further from the rotation axis.
  • Moment of inertia is symbolized as \(I\).
  • Changes in posture affect the value of \(I\).
  • The new position at \(30^\circ\) reduces the moment of inertia, calculated using \(I_{\text{leg, new}} = 2 \cdot (1.44\ \cos^2(30^{\circ}))\).
Understanding this helps us figure out how movements like tucking in arms or legs lead to spinning faster.
Rotational Kinematics
Rotational kinematics is the study of motion around an axis. It helps us describe how objects spin or rotate. Just like linear kinematics deals with speed and acceleration, rotational kinematics deals with angular speed and angular acceleration.
In the problem, the ballerina's angular speed changes due to her changing body position, a clear application of rotational kinematics. Here's what you need to know:
  • Angular speed (\(\omega\)) tells us how fast something spins.
  • Adjustment of the moment of inertia leads to changes in angular speed while conserving angular momentum.
  • The ratio \(\omega_f / \omega_i\) helps understand how speed changes when posture or configuration changes.
In practical terms, mastering these concepts can help us understand movements in sports, engineering applications, and much more.

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Most popular questions from this chapter

A uniform solid ball rolls smoothly along a floor, then up a ramp inclined at \(15.0^{\circ} .\) It momentarily stops when it has rolled \(1.50 \mathrm{~m}\) along the ramp. What was its initial speed?

In unit-vector notation, what is the torque about the origin on a particle located at coordinates \((0,-4.0 \mathrm{~m}, 3.0 \mathrm{~m})\) if that torque is due to (a) force \(\vec{F}_{1}\) with components \(F_{1 x}=2.0 \mathrm{~N}, F_{1 y}=F_{1 z}=0\), and (b) force \(\vec{F}_{2}\) with components \(F_{2 x}=0, F_{2 y}=2.0 \mathrm{~N}, F_{2 z}=4.0 \mathrm{~N} ?\)

a wheel is rotating freely at angular speed 800 rev/min on a shaft whose rotational inertia is negligible. A second wheel, initially at rest and with twice the rotational inertia of the first, is suddenly coupled to the same shaft. (a) What is the angular speed of the resultant combination of the shaft and two wheels? (b) What fraction of the original rotational kinetic energy is lost?

A uniform block of granite in the shape of a book has face dimensions of \(20 \mathrm{~cm}\) and \(15 \mathrm{~cm}\) and a thickness of \(1.2 \mathrm{~cm}\). The density (mass per unit volume) of granite is \(2.64 \mathrm{~g} / \mathrm{cm}^{3}\). The block rotates around an axis that is perpendicular to its face and halfway between its center and a corner. Its angular momentum about that axis is \(0.104 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\). What is its rotational kinetic energy about that axis?

Figure 11-51 shows an overhead view of a ring that can rotate about its center like a merrygo-round. Its outer radius \(R_{2}\) is \(0.800 \mathrm{~m}\), its inner radius \(R_{1}\) is \(R_{2} / 2.00\), its mass \(M\) is \(8.00 \mathrm{~kg}\), and the mass of the crossbars at its center is negligible. It initially rotates at an angular speed of \(8.00 \mathrm{rad} / \mathrm{s}\) with a cat of mass \(m=M / 4.00\) on its outer edge, at radius \(R_{2}\). By how much does the cat increase the kinetic energy of the cat-ring system if the cat crawls to the inner edge, at radius \(R_{1}\) ?
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