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Chapter 10: Problem 3

A thin cylindrical hoop having a diameter of \(1.0 \mathrm{~m}\) and a mass of \(400 \mathrm{~g}\), rolls down the street. What is the hoop's moment of inertia about its central axis of rotation? It follows from the first part of Fig. \(10-1\) that for a hoop $$ I=M R^{2}=(0.400 \mathrm{~kg})(0.50 \mathrm{~m})^{2}=0.10 \mathrm{~kg} \cdot \mathrm{m}^{2} $$

Short Answer

Expert verified
The hoop's moment of inertia is 0.1 kg路m虏.

Step by step solution

01

Identify Given Information

We are given a thin cylindrical hoop with a diameter of 1.0 m and a mass of 400 g. First, convert the mass from grams to kilograms for consistency in SI units.
02

Convert Units

Convert the mass from 400 g to kilograms. \[ M = 400\, \text{g} = 0.4\, \text{kg} \]
03

Determine Radius

The diameter of the hoop is 1.0 m, so calculate the radius by dividing the diameter by 2.\[ R = \frac{1.0\, \text{m}}{2} = 0.5\, \text{m} \]
04

Apply Formula for Moment of Inertia

Use the formula for the moment of inertia of a cylindrical hoop about its central axis: \[ I = M R^{2} \]Substitute the mass (M) and radius (R) into the formula.
05

Calculate Moment of Inertia

Substitute the values into the formula and calculate:\[ I = (0.4\, \text{kg}) \times (0.5\, \text{m})^{2} = 0.1\, \text{kg} \cdot \text{m}^{2} \]

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Key Concepts

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

Cylindrical Hoop
A cylindrical hoop, often referred to in physics as a ring or thin hoop, is a simple geometrical object that is quite common in many problems due to its unique shape. Picture a wheel or a bracelet made out of lightweight material. In these types of problems, we usually assume that the hoop's thickness is negligible compared to its diameter, meaning it is very thin. This simplification helps focus on the motion of the hoop around its axis without getting bogged down in complex geometry. For our problem, the hoop has a given diameter of 1.0 m. This is fundamental as it allows us to calculate the radius. The radius is crucial because it is part of the moment of inertia formula, which we will delve into more as we explore the central axis.
Central Axis
In physics, especially regarding rotational dynamics, the central axis is a line through an object around which the object rotates. Imagine a merry-go-round spinning around its pole; that pole acts as its central axis. For our cylindrical hoop, its central axis passes through the center of the circle and is perpendicular to the hoop's plane. This central axis is critical for calculating the moment of inertia because it determines how the rotational mass is distributed around it. When calculating the moment of inertia for the hoop, we use the formula: \[ I = M R^2 \] Here, \(I\) represents the moment of inertia, \(M\) is the mass of the hoop, and \(R\) is the radius. Understanding the orientation of the central axis helps us correctly apply this formula to find the hoop's resistance to changes in its rotational motion.
SI Units
The SI unit system is a standard way of measuring physical quantities. It simplifies exchanging and comparing physical data worldwide. When solving physics problems, it's essential to ensure all measurements are in SI units to maintain consistency and accuracy. Typically, mass is measured in kilograms (kg), distance in meters (m), and time in seconds (s). In our problem, we start with the mass of the hoop given in grams (g), which we have converted to kilograms (kg) as the SI unit for mass. Likewise, the diameter of the hoop is provided directly in meters (m), which is convenient since that is already the SI unit for length. Using these standardized units allows you to apply mathematical formulas correctly without errors caused by unit conversion mistakes.
Physics Problem Solving
Solving physics problems involves a structured approach of identifying, converting, and applying formulas. It starts with reading the problem to note given information, such as measurements and quantities. Next, ensuring all units are consistent, preferably in SI units, sets a solid foundation for calculations.When calculating the moment of inertia, proceed to determine any necessary variables, like the radius of the hoop from the provided diameter. Following this, identify the correct physics formula, often influenced by the object's shape or the type of motion involved. Here it's the moment of inertia for a hoop: \[ I = M R^2 \] Substitute known values into this formula, calculate, and interpret the solution based on the context of the problem. This step-by-step approach ensures organized working, minimizing mistakes and increasing your understanding and confidence in solving similar problems.

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Most popular questions from this chapter

A large horizontal disk is rotating on a vertical axis through its center. Its moment of inertia is \(I=4000 \mathrm{~kg} \cdot \mathrm{m}^{2}\). The disk is revolving freely at a rate of \(0.150\) rev/s when a \(90.0\) -kg person drops straight down onto it from an overhanging tree limb. The person lands and remains at a distance of \(3.00 \mathrm{~m}\) from the axis of rotation. What will be the rate of rotation after the person has landed?

An airplane propeller has a mass of \(70 \mathrm{~kg}\) and a radius of gyration of \(75 \mathrm{~cm}\). Find its moment of inertia. How large a torque is needed to give it an angular acceleration of \(4.0 \mathrm{rev} / \mathrm{s}^{2}\) ? $$I=M k^{2}=(70 \mathrm{~kg})(0.75 \mathrm{~m})^{2}=39 \mathrm{~kg} \cdot \mathrm{m}^{2}$$ To be able to use \(\tau=I \alpha\), we must have \(\alpha\) in \(\mathrm{rad} / \mathrm{s}^{2}\) :

A 500 -g uniform sphere of \(7.0\) -cm radius spins frictionlessly at 30 rev/s on an axis through its center. Find its ( \(a\) ) \(\mathrm{KE}_{r}\), (b) angular momentum, and ( \(c\) ) radius of gyration. We need the moment of inertia of a uniform sphere about an axis through its center. From \(\underline{\text { Fig. } 10-1}\), $$I=\frac{2}{5} M r^{2}=(0.40)(0.50 \mathrm{~kg})(0.070 \mathrm{~m})^{2}=0.00098 \mathrm{~kg} \cdot \mathrm{m}^{2}$$ (a) Knowing that \(\omega=30 \mathrm{rev} / \mathrm{s}=188 \mathrm{rad} / \mathrm{s}\), we have $$\mathrm{KE}_{r}=\frac{1}{2} I \omega^{2}=\frac{1}{2}\left(0.00098 \mathrm{~kg} \cdot \mathrm{m}^{2}\right)(188 \mathrm{rad} / \mathrm{s})^{2}=0.017 \mathrm{~kJ}$$ Notice that \(\omega\) must be in \(\mathrm{rad} / \mathrm{s}\). (b) Its angular momentum is $$L=I \omega=\left(0.00098 \mathrm{~kg} \cdot \mathrm{m}^{2}\right)(188 \mathrm{rad} / \mathrm{s})=0.18 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}$$ (c) For any object, \(I=M k^{2}\), where \(k\) is the radius of gyration. Therefore, $$k=\sqrt{\frac{I}{M}}=\sqrt{\frac{0.00098 \mathrm{~kg} \cdot \mathrm{m}^{2}}{0.50 \mathrm{~kg}}}=0.044 \mathrm{~m}=4.4 \mathrm{~cm}$$ Notice that this is a reasonable value in view of the fact that the radius of the sphere is \(7.0 \mathrm{~cm}\).

Starting from rest, a hoop with a 20-cm radius rolls down a hill to a place \(5.0 \mathrm{~m}\) below its starting point. How fast is it rotating as it rolls through that point? The hoop descends \(5.0 \mathrm{~m}\), whereupon an amount of gravitational PE is converted into KE: $$\begin{array}{l} \mathrm{PE}_{\mathrm{G}} \text { at start }=\left(\mathrm{KE}_{r}+\mathrm{KE}_{t}\right) \text { at end } \\ M g h=\frac{1}{2} I \omega^{2}+\frac{1}{2} M v^{2} \end{array}$$ Here \(I=M r^{2}\) for a hoop and \(v=\omega r\). The above equation becomes $$\begin{array}{l} \text { from which }\\\ \begin{aligned} M g h &=\frac{1}{2} M \omega^{2} r^{2}+\frac{1}{2} M \omega^{2} r^{2} \\ \quad \omega &=\sqrt{\frac{g h}{r^{2}}}=\sqrt{\frac{\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(5.0 \mathrm{~m})}{(0.20 \mathrm{~m})^{2}}}=35 \mathrm{rad} / \mathrm{s} \end{aligned} \end{array}$$

A student stands on a freely rotating platform, as shown in Fig. 10 - 10 . With his arms extended, his rotational frequency is \(0.25\) rev/s. But when he draws his arm in, that frequency becomes \(0.80\) rev/s. Find the ratio of his moment of inertia in the first case to that in the second. Because there is no external torque on the system (why?), the law of conservation of angular momentum tells us that Angular momentum before \(=\) Angular momentum after $$I_{i} \omega_{i}=I_{f} \omega_{f}$$ Or, since we require \(I_{i} / I_{f}\) $$\frac{I_{i}}{I_{f}}=\frac{\omega_{f}}{\omega_{i}}=\frac{0.80 \mathrm{rev} / \mathrm{s}}{0.25 \mathrm{rev} / \mathrm{s}}=3.2$$
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