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Chapter 11: Problem 43

Solid Cylinder-Thin Hoop (a) Show that the rotational inertia of a solid cylinder of mass \(M\) and radius \(R\) about its central axis is equal to the rotational inertia of a thin hoop of mass \(M\) and radius \(R / \sqrt{2}\) about its central axis. (b) Show that the rotational inertia \(I\) of any given body of mass \(M\) about any given axis is equal to the rotational inertia of an equivalent hoop about that axis, if the hoop has the same mass \(M\) and a radius \(k\) given by $$ k=\sqrt{\frac{I}{M}} . $$ The radius \(k\) of the equivalent hoop is called the radius of gyration of the given body.

Short Answer

Expert verified
The rotational inertia of a solid cylinder is equal to a thin hoop's inertia if their radius and the radius of gyration are correctly adjusted.

Step by step solution

01

Rotational Inertia of a Solid Cylinder

The rotational inertia of a solid cylinder about its central axis is given by the formula: \[ I_{\text{cylinder}} = \frac{1}{2} M R^2 \]
02

Rotational Inertia of a Thin Hoop

The rotational inertia of a thin hoop about its central axis is given by: \[ I_{\text{hoop}} = M R_{\text{hoop}}^2 \] Here, the radius of the thin hoop is given as \( R_{\text{hoop}} = \frac{R}{\frac{\text{sqrt}(2)}} \). Substituting, we get: \[ I_{\text{hoop}} = M \frac{R^2}{2} \]
03

Equivalence of Rotational Inertias

Comparing the results from step 1 and step 2, we see that: \[ I_{\text{cylinder}} = I_{\text{hoop}} = \frac{1}{2} M R^2 \] Therefore, the rotational inertias are equal.
04

Rotational Inertia of Any Body About Any Given Axis

The given body has rotational inertia \( I \), mass \( M \), and radius of gyration \( k \). According to the given relationship: \[ k = \text{sqrt}(\frac{I}{M}) \]
05

Rotational Inertia of Equivalent Hoop About Same Axis

A hoop equivalent to the given body will have the same mass \( M \) and radius \( k \). Therefore, the rotational inertia of the hoop will be: \[ I_{\text{hoop}} = M k^2 \] Substituting \( k = \text{sqrt}(\frac{I}{M}) \), we get: \[ I_{\text{hoop}} = M \times (\text{sqrt}(\frac{I}{M}))^2 \] Simplifying, we obtain: \[ I_{\text{hoop}} = M \times \frac{I}{M} = I \] Thus, \( I_{\text{hoop}} = I \), showing they are equal.

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Key Concepts

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

solid cylinder
In physics, we often analyze how objects rotate around different axes. A solid cylinder is a common object studied for its rotational properties. To calculate the rotational inertia (or moment of inertia) of a solid cylinder with mass \(M\) and radius \(R\) about its central axis, we use the formula:
I_{\text{cylinder}} = \frac{1}{2} M R^2.
This formula helps us understand how mass is distributed relative to the axis of rotation. The rotational inertia is crucial in determining how difficult it is to change the rotation of the cylinder.
thin hoop
A thin hoop is another object often analyzed in physics for its rotational properties. The rotational inertia of a thin hoop with mass \(M\) and radius \(R\) about its central axis is given by:
I_{\text{hoop}} = M R^2.
In some problems, the radius of the thin hoop is adjusted. For instance, if the radius is \(R / \text{sqrt}(2)\), the rotational inertia becomes: I_{\text{hoop}} = M \frac{R^2}{2}.

This adjustment shows that distribution of mass closer to the axis reduces rotational inertia, making it easier to rotate the object.
radius of gyration
The radius of gyration, \(k\), is a useful concept in rotational dynamics. It simplifies the complexity of mass distribution. For a body with mass \(M\) and rotational inertia \(I\), the radius of gyration is given by:
k = \text{sqrt}(\frac{I}{M}).
This radius is an imaginary distance from the axis where if the entire mass were concentrated, it would yield the same rotational inertia. It simplifies calculations especially when dealing with complex shapes, turning them into an equivalent thin hoop problem.
moment of inertia
Moment of inertia (also known as rotational inertia) is a fundamental concept in rotational dynamics. It measures how mass is distributed relative to a rotational axis and determines how resistant an object is to changes in its rotation. For various shapes, the moment of inertia varies. For example:
  • Solid cylinder: \(\frac{1}{2} M R^2\)
  • Thin hoop: \(M R^2\)


Calculating moments of inertia is critical in designing and understanding mechanical systems, ensuring their stability and performance.
physics problem-solving
Solving physics problems, especially in rotational dynamics, involves breaking down problems into manageable steps. Here's a general approach:
  • Understand the problem and the physical principles involved.
  • Identify and list the known and unknown variables.
  • Use appropriate formulas and relationships:
    • For rotational inertia, use the formulas for the shapes involved.
    • For radius of gyration, apply \(k = \text{sqrt}(\frac{I}{M})\).
  • Solve algebraically and simplify your expressions.
  • Review each step to ensure accuracy.


Staying methodical and organized helps in tackling complex problems effectively, understanding the underlying concepts better, and avoiding common mistakes.

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Most popular questions from this chapter

Four Identical Particles Four identical particles of mass \(0.50 \mathrm{~kg}\) each are placed at the vertices of a \(2.0 \mathrm{~m} \times 2.0 \mathrm{~m}\) square and held there by four massless rods, which form the sides of the square. What is the rotational inertia of this rigid body about an axis that (a) passes through the midpoints of opposite sides and lies in the plane of the square, (b) passes through the midpoint of one of the sides and is perpendicular to the plane of the square, and (c) lies in the plane of the square and passes through two diagonally opposite particles?

Thin Spherical Shell A thin spherical shell has a radius of \(1.90 \mathrm{~m}\). An applied torque of \(960 \mathrm{~N} \cdot \mathrm{m}\) gives the shell a rotational acceleration of \(6.20 \mathrm{rad} / \mathrm{s}^{2}\) about an axis through the center of the shell. What are (a) the rotational inertia of the shell about that axis and (b) the mass of the shell?

Fixed Axis An object rotates about a fixed axis, and the rotational position of a reference line on the object is given by \(\theta=\) \((0.40 \mathrm{rad}) e^{\left(2.0 \mathrm{~s}^{-1}\right) t}\). Consider a point on the object that is \(4.0 \mathrm{~cm}\) from the axis of rotation. At \(t=0\), what are the magnitudes of the point's (a) tangential component of acceleration and (b) radial component of acceleration?

Bicycle Pedal Arm The length of a bicycle pedal arm is \(0.152 \mathrm{~m}\), and a downward force of \(111 \mathrm{~N}\) is applied to the pedal by the rider's foot. What is the magnitude of the torque about the pedal arm's pivot point when the arm makes an angle of (a) \(30^{\circ}\), (b) \(90^{\circ}\). and (c) \(180^{\circ}\) with the vertical?

Flywheel The rotational position of a flywheel on a generator is given by \(\theta=(a \mathrm{rad} / \mathrm{s}) t+\left(b \mathrm{rad} / \mathrm{s}^{3}\right) t^{3}-\left(c \mathrm{rad} / \mathrm{s}^{4}\right) t^{4}\), where \(a, b\), and \(c\) are constants. Write expressions for the wheel's (a) rotational velocity and (b) rotational acceleration.
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