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A twirler's baton is made of a slender metal cylinder of mass \(M\) and length \(L\) . Each end has a rubber cap of mass \(m,\) and you can accurately treat each cap as a particle in this problem. Find the total moment of inertia of the baton about the usual twirling axis (perpendicular to the baton through its center).

Short Answer

Expert verified
The total moment of inertia is \(\frac{ML^2 + 6mL^2}{12}\).

Step by step solution

01

Understand the Components

The baton consists of a slender metal cylinder of mass \(M\) and length \(L\), with rubber caps at each end, each having mass \(m\). The entire baton twirls about an axis perpendicular to its length and through its center.
02

Moment of Inertia of the Cylinder

The moment of inertia for a slender rod or cylinder about an axis perpendicular to its length through its center is given by the formula: \(I_{cylinder} = \frac{1}{12}ML^2\). This result comes from standard formulas for rotational dynamics.
03

Moment of Inertia of the Rubber Caps

Each rubber cap can be treated as a point mass located at a distance of \(\frac{L}{2}\) from the axis of rotation. The moment of inertia for a point mass is \(I = mr^2\), where \(r\) is the distance from the axis. Therefore, the moment of inertia for each cap is \(I_{cap} = m(\frac{L}{2})^2 = \frac{mL^2}{4}\).
04

Total Moment of Inertia

The total moment of inertia of the baton is the sum of the moment of inertia of the cylinder and the moments of inertia of the two caps. Thus, the total moment of inertia can be expressed as follows: \[I_{total} = I_{cylinder} + 2I_{cap} = \frac{1}{12}ML^2 + 2\left(\frac{mL^2}{4}\right)\].
05

Simplify the Total Moment of Inertia

Simplify the expression for the total moment of inertia: \[I_{total} = \frac{1}{12}ML^2 + \frac{2mL^2}{4} = \frac{1}{12}ML^2 + \frac{mL^2}{2}\]. Combine these terms over a common denominator to further simplify: \[I_{total} = \frac{ML^2}{12} + \frac{6mL^2}{12} = \frac{ML^2 + 6mL^2}{12}\].

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Key Concepts

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

Rotational Dynamics
Rotational dynamics is a branch of physics that deals with the motion of objects rotating about an axis. Just as linear dynamics concerns itself with objects moving in a straight line, rotational dynamics focuses on objects that spin or twist. The basic principle is similar—forces cause acceleration. However, in rotational dynamics, these forces are known as torques.

Here are key components in rotational dynamics:
  • Torque: This is a measure of how much a force acting on an object causes it to rotate. It's the rotational equivalent of force in linear motion.
  • Angular Momentum: This is a conserved quantity, similar to linear momentum, but it applies to rotation. Coffee mugs, footballs, and even the Earth demonstrate rotational dynamics because they spin around their axes.
  • Moment of Inertia: A key concept in understanding rotational dynamics and the topic of our exercise.
Expanding our knowledge of rotational dynamics makes it easier to grasp how various parts of a rotating body contribute to its overall spin, as we see with our baton example.
Point Mass
A point mass approach treats an object as if all of its mass is concentrated at a single point. This simplification is commonly used in physics to make complex problems more manageable.

When considering rotational dynamics involving point masses, these approximations help simplify calculations, especially when finding moments of inertia. For example:
  • If you have a pair of point masses located at either end of a baton's length, these masses can be considered singular points of mass distribution contributing to the overall moment of inertia.
  • The moment of inertia for a point mass is defined as: \(I = mr^2\), where \(r\) is the distance of the point mass from the axis of rotation.
This approach is particularly helpful when dealing with objects like the rubber caps on our baton's ends. Treating them as point masses allows us to focus on their distances from the axis, making the calculations simpler and more precise.
Cylinder
A cylinder, in the context of physics, often refers to an object that has a long, symmetrical shape. This makes it easier to calculate its moment of inertia since it can be treated as a uniform, continuous mass distribution along its length.

For our exercise, the baton is described as a slender metal cylinder. Calculating the moment of inertia of such a structure involves a straightforward formula due to its symmetry:
  • The moment of inertia of a cylinder (or slender rod) about an axis through its center and perpendicular to its length is \(I_{cylinder} = \frac{1}{12}ML^2\).
This formula takes into account the entire length and mass of the cylinder, allowing for an accurate calculation when combined with other forms of inertia, such as from point masses like the rubber caps.

Understanding this concept is crucial as it provides the foundation for solving more complex rotational motion problems involving cylinders and similar shapes.
Axis of Rotation
The axis of rotation is an imaginary line around which an object rotates. It plays a critical role in determining the dynamics of the rotating object, including its moment of inertia.

Several factors influence how an object behaves when rotating about an axis, such as:
  • The distance of the mass elements from the axis: The further the mass is distributed from this axis, the higher the moment of inertia.
  • The orientation of the axis: Different orientations can change the rotational dynamics of an object.
When it comes to our baton, the axis of rotation is perpendicular to its length and passes through its center. This setup ensures that the calculations for the moment of inertia encompass all aspects of the baton's mass distribution.

In physics, recognizing the axis of rotation is like identifying a map's compass—it guides the way to understanding how forces lead to rotational motion.

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Most popular questions from this chapter

A compound disk of outside diameter 140.0 \(\mathrm{cm}\) is made up of a uniform solid disk of radius 50.0 \(\mathrm{cm}\) and area density 3.00 \(\mathrm{g} / \mathrm{cm}^{2}\) surrounded by a concentric ring of inner radius 50.0 \(\mathrm{cm}\) , outer radius \(70.0 \mathrm{cm},\) and area density 2.00 \(\mathrm{g} / \mathrm{cm}^{2} .\) Find the moment of inertia of this object about an axis perpendicular to the plane of the object and passing through its center.

A roller in a printing press turns through an angle \(\theta(t)\) given by \(\theta(t)=\gamma t^{2}-\beta t^{3},\) where \(\gamma=3.20 \operatorname{rad} / \mathrm{s}^{2}\) and \(\beta=\) 0.500 \(\mathrm{rad} / \mathrm{s}^{3}\) , (a) Calculate the angular velocity of the roller as a function of time. (b) Calculate the angular acceleration of the roller as a function of time. (c) What is the maximum positive angular velocity, and at what value of \(t\) does it occur?

A pendulum is made of a uniform solid sphere with mass \(M\) and radius \(R\) suspended from the end of a light rod. The distance from the pivot at the upper end of the rod to the center of the sphere is \(L\) . The pendulum's moment of inertia I for rotation about the pivot is usually approximated as \(M L^{2} .\) (a) Use the parallel-axis theorem to show that if \(R\) is 5\(\%\) of \(L\) and the mass of the rod is ignored, \(I_{p}\) is only 0.1\(\%\) greater than \(M L^{2} .\) (b) If the mass of the rod is 1\(\%\) of \(M\) and \(R\) is much less than \(L,\) what is the ratio of \(I_{\text { rod }}\) for an axis at the pivot to \(M L^{2} ?\)

At \(t=0\) the current to a de electric motor is reversed, resulting in an angular displacement of the motor shaft given by \(\theta(t)=(250 \mathrm{rad} / \mathrm{s}) t-\left(20.0 \mathrm{rad} / \mathrm{s}^{2}\right) t^{2}-\left(1.50 \mathrm{rad} / \mathrm{s}^{3}\right) t^{3} .(\mathrm{a}) \mathrm{At}\) what time is the angular velocity of the motor shaft zero? (b) Calculate the angular acceleration at the instant that the motor shaft has zero angular velocity. (c) How many revolutions does the motor shaft turn through between the time when the current is reversed and the instant when the angular velocity is zero? (d) How fast was the motor shaft rotating at \(t=0,\) when the current was reversed? ( c) Calculate the average angular velocity for the time period from \(t=0\) to the time calculated in part (a).

(a) What angle in radians is subtended by an arc 1.50 \(\mathrm{m}\) long on the circumference of a circle of radius 2.50 \(\mathrm{m} ?\) What is this angle in degrees? (b) An are 14.0 \(\mathrm{cm}\) long on the circumference of a circle subtends an angle of \(128^{\circ} .\) What is the radius of the circle? (c) The angle between two radii of a circle with radius 1.50 \(\mathrm{m}\) is 0.700 rad. What length of arc is intercepted on the circumference of the circle by the two radii?

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