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Chapter 10: Problem 20

A man stands on a frictionless platform that is rotating with an angular speed of \(1.22 \mathrm{rev} / \mathrm{s} ;\) his arms are outstretched and he holds a weight in each hand. With his hands in this position the total rotational inertia of the man, the weights, and the platform is \(6.13 \mathrm{~kg} \cdot \mathrm{m}^{2}\). If by moving the weights the man decreases the rotational inertia to \(1.97 \mathrm{~kg} \cdot \mathrm{m}^{2}\), what is the resulting angular speed of the platform?

Short Answer

Expert verified
After calculating the above expression, the resulting angular speed of the platform, when the man moves the weights and decreases the rotational inertia, is approximately 19.22 rad/s.

Step by step solution

01

Establish the angular momentum conservation equation

According to the conservation of angular momentum, the initial angular momentum of the system equals its final momentum, represented as \(L_{i}=L_{f}\). Angular momentum is calculated as the product of rotational inertia and angular speed, then this equation can be written as \[I_{i} \cdot \omega_{i} = I_{f} \cdot \omega_{f}\], whereby \(\omega_{i}\) represents the initial angular speed, \(I_{i}\) represents the initial rotational inertia, \(\omega_{f}\) represents the final angular speed and \(I_{f}\) represents the final rotational inertia.
02

Substitute the given values

Substitute the provided values: \(I_{i}=6.13~kg.m^2\), \(\omega_{i}=1.22~rev/s\), \(I_{f}=1.97~kg.m^2\). The rotational speed was given in revolutions per second, but this needs to be converted to radians per second because the standard units in physics for angular speed are radians per second. To do so, recall that \(1~rev = 2\pi~rad\). Thus, \(\omega_{i}=1.22~rev/s = 1.22\cdot2\pi~rad/s\). Substituting the given values into the angular momentum conservation equation gives \[6.13~kg.m^2\cdot1.22\cdot2\pi~rad/s = 1.97~kg.m^2\cdot\omega_{f}\].
03

Solve the equation for \(\omega_{f}\)

Solving for \(\omega_{f}\) gives \(\omega_{f} = \frac{6.13~kg.m^2\cdot1.22\cdot2\pi~rad/s}{1.97~kg.m^2}\) .

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Key Concepts

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

Rotational Inertia
Rotational inertia, also known as the moment of inertia, is a physical quantity that determines how much torque is required for a desired angular acceleration about an axis of rotation. This concept is pivotal in understanding how objects rotate and how difficult it is to change their rotational motion.
In the context of the problem provided, rotational inertia plays a crucial role. Initially, the man, along with the weights and platform, has a higher rotational inertia of \(6.13 \text{ kg} \cdot \text{m}^2\). By pulling the weights closer, the man reduces this value to \(1.97 \text{ kg} \cdot \text{m}^2\).
Key factors that affect rotational inertia include the distribution of mass around the axis of rotation and the size and shape of the object. The further away the mass is distributed from the axis, the higher the rotational inertia. This explains why bringing the weights closer decreased the rotational inertia. The change in distribution of mass closer to the axis of rotation significantly alters the rotational inertia and subsequently affects how easily something spins.
Angular Speed
Angular speed refers to how fast an object rotates or revolves relative to another point, often the center of a circle. It is usually measured in radians per second in the International System of Units (SI).
In the given problem, the initial angular speed of the system is \(1.22 \text{ rev/s}\), which is equivalent to \(1.22 \cdot 2\pi \text{ rad/s}\). By altering the distribution of the mass, and thus the rotational inertia, the system's angular speed changes to preserve angular momentum.
Essentially, when the man pulls the weights inwards, the overall rotational inertia decreases, causing the angular speed to increase in order to maintain the same angular momentum. This showcases the concept of conservation of angular momentum: when rotational inertia decreases, angular speed increases, if no external torque acts on the system.
Physics Problem Solving
Physics problem solving often requires a step-by-step approach to inferentially make sense of the underlying principles at work. This methodical strategy is exemplified by applying the conservation of angular momentum, which asserts that in a closed system with no external forces, the total angular momentum remains constant.
To tackle complex problems, follow these general steps:
  • Identify and define the physical quantities involved, such as rotational inertia and angular speed in our problem.
  • Apply appropriate physics principles, like the conservation of angular momentum equation.
  • Substitute known values and solve for any unknowns, while keeping units consistent. Remember that consistency in units (conversion from revolutions per second to radians per second) is crucial.
  • Recheck your answers and consider if the results are logically satisfying the conditions of the problem.
This approach provides a systematic way to analyze and solve physics problems efficiently, while deeply understanding the laws governing motion and forces.

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Most popular questions from this chapter

Let \(\overrightarrow{\mathbf{r}}_{\mathrm{cm}}\) be the position vector of the center of mass \(C\) of a system of particles with respect to the origin \(O\) of an inertial reference frame, and let \(\overrightarrow{\mathbf{r}}_{i}^{\prime}\) be the position vector of the \(i\) th particle, of mass \(m_{i}\), with respect to the center of mass \(C\). Hence \(\overrightarrow{\mathbf{r}}_{i}=\overrightarrow{\mathbf{r}}_{\mathrm{cm}}+\overrightarrow{\mathbf{r}}_{i}^{\prime}\) (see Fig. \(\left.10-22\right)\). Now define the total angular momentum of the system of particles relative to the center of mass \(C\) to be \(\overrightarrow{\mathbf{L}}^{\prime}=\Sigma\left(\overrightarrow{\mathbf{r}}_{i}^{\prime} \times \overrightarrow{\mathbf{p}}_{i}^{\prime}\right), \quad\) where \(\overrightarrow{\mathbf{p}}_{i}^{\prime}=m_{i} d \overrightarrow{\mathbf{r}}_{i}^{\prime} / d t\) (a) Show that \(\overrightarrow{\mathbf{p}}_{i}^{\prime}=m_{i} d \overrightarrow{\mathbf{r}}_{i} / d t-\) \(m_{i} d \overrightarrow{\mathbf{r}}_{\mathrm{cm}} / d t=\overrightarrow{\mathbf{p}}_{i}-m_{i} \overrightarrow{\mathbf{v}}_{\mathrm{cm}}\) (b) Show next that \(d \overrightarrow{\mathbf{L}}^{\prime} / d t=\) \(\Sigma\left(\overrightarrow{\mathbf{r}}_{i}^{\prime} \times d \overrightarrow{\mathbf{p}}_{i}^{\prime} / d t\right) .(c)\) Combine the results of \((a)\) and \((b)\) and, using the definition of center of mass and Newton's third law, show that \(\vec{\tau}_{\mathrm{ext}}^{\prime}=d \overrightarrow{\mathbf{L}}^{\prime} / d t\), where \(\vec{\tau}_{\mathrm{ext}}^{\prime}\) is the sum of all the \(\mathrm{ex}\) - ternal torques acting on the system about its center of mass.

Astronomical observations show that from 1870 to 1900 the length of the day increased by about \(6.0 \times 10^{-3}\) s. (a) What corresponding fractional change in the Earth's angular velocity resulted? \((b)\) Suppose that the cause of this change was a shift of molten material in the Earth's core. What resulting fractional change in the Earth's rotational inertia could account for the answer to part \((a) ?\)

In a playground there is a small merry-go-round of radius \(1.22 \mathrm{~m}\) and mass \(176 \mathrm{~kg} .\) The radius of gyration (see Exercise \(9-20\) ) is \(91.6 \mathrm{~cm}\). A child of mass \(44.3 \mathrm{~kg}\) runs at a speed of \(2.92 \mathrm{~m} / \mathrm{s}\) tangent to the rim of the merry-go-round when it is at rest and then jumps on. Neglect friction between the bearings and the shaft of the merry-go-round and find the angular speed of the merry-go-round and child.

A wheel with rotational inertia \(1.27 \mathrm{~kg} \cdot \mathrm{m}^{2}\) is rotating with an angular speed of 824 rev/min on a shaft whose rotational inertia is negligible. A second wheel, initially at rest and with rotational inertia \(4.85 \mathrm{~kg} \cdot \mathrm{m}^{2}\), is suddenly coupled to the same shaft. What is the angular speed of the resultant combination of the shaft and two wheels?

If we are given \(r, p\), and \(\theta\), we can calculate the angular momentum of a particle from Eq. 10-2. Sometimes, however, we are given the components \((x, y, z)\) of \(\overrightarrow{\mathbf{r}}\) and \(\left(v_{x}, v_{y}, v_{z}\right)\) of \(\overrightarrow{\mathbf{v}}\) instead. ( \(a\) ) Show that the components of \(\overrightarrow{\mathbf{1}}\) along the \(x, y\), and \(z\) axes are then given by $$ \begin{aligned} &l_{x}=m\left(y v_{z}-z v_{y}\right) \\ &l_{y}=m\left(z v_{x}-x v_{z}\right) \\ &l_{z}=m\left(x v_{y}-y v_{x}\right) \end{aligned} $$ (b) Show that if the particle moves only in the \(x y\) plane, the resultant angular momentum vector has only a \(z\) component.
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