/*! 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 30 \(\bullet\) A twirler's baton is... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

\(\bullet\) 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 \( \left(\frac{1}{12} M + \frac{1}{2} m \right) L^2 \).

Step by step solution

01

Understand the setup

First, visualize the baton as a slender cylinder with a mass \(M\) and length \(L\). Each end of the baton has a rubber cap with mass \(m\). The baton is being twirled about an axis perpendicular to its length and through its center. The moment of inertia of the entire baton about this axis needs to be determined.
02

Moment of inertia of the cylinder

The moment of inertia \(I_{cyl}\) for a slender cylinder about an axis through its center and perpendicular to its length is given by the formula: \[ I_{cyl} = \frac{1}{12} ML^2 \] where \(M\) is the mass of the cylinder and \(L\) is its length.
03

Moment of inertia of the rubber caps

Each rubber cap is considered a point mass located at a distance \(\frac{L}{2}\) from the axis of rotation. The moment of inertia \(I_{cap}\) for a point mass is calculated using: \[ I_{cap} = m \left(\frac{L}{2}\right)^2 = m \frac{L^2}{4} \] Since there are two caps, the total moment of inertia for both caps is: \[ 2 \times I_{cap} = 2 \left(m \frac{L^2}{4}\right) = m \frac{L^2}{2} \]
04

Total moment of inertia

Add the moments of inertia of the cylinder and the two caps to find the total moment of inertia \(I_{total}\) of the baton: \[ I_{total} = I_{cyl} + 2 \times I_{cap} = \frac{1}{12} ML^2 + m \frac{L^2}{2} \] Simplifying, \(I_{total}\) is given by: \[ I_{total} = \frac{1}{12} ML^2 + \frac{1}{2} mL^2 = \left(\frac{1}{12} M + \frac{1}{2} m \right)L^2 \]
05

Conclusion

The formula \(I_{total} = \left(\frac{1}{12} M + \frac{1}{2} m \right)L^2 \) expresses the total moment of inertia of the baton, considering the contributions from both the metal cylinder and the two rubber caps.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Rotational Dynamics and Its Role
Rotational dynamics is all about how objects spin or rotate, and why. It's the rotational equivalent of linear dynamics, which covers motion in a straight line. For our twirler's baton, rotational dynamics help us determine how it will behave as it spins. This involves understanding its moment of inertia, which is a measure of how difficult it is to change the baton's rotation rate.
The moment of inertia is dependent on the mass distribution of the object. For the baton, this takes into account both the slender metal cylinder and the rubber caps at each end. When analyzing an object in rotational dynamics, you must consider how far each part of the object is from the axis of rotation. The further the mass is from the axis, the greater its contribution to the moment of inertia. This is why the rubber caps, being at the ends, significantly affect the baton's total moment of inertia.
In work and sports, understanding rotational dynamics is key for anything involving spins and rotations, from car engines to ice skaters' turns. This field gives insight into the forces and torques necessary to cause various rotational motions.
Tackling Physics Problem Solving
Solving physics problems effectively requires a structured approach. First, it's important to thoroughly understand what the problem is asking before diving into calculations. With our baton, we started by recognizing it was a combination of a cylinder and point masses (rubber caps). Each part contributes to the moment of inertia of the entire baton.
Next, identify the known values and the relationships between them, such as equations or constants that apply. Here, the formulas for the moment of inertia of a cylinder and point masses guided us. By knowing these, you can systematically find the moment of inertia for each component and sum them up for the complete baton.
Consistent practice and an organized method can transform seemingly complex physics problems into simple, manageable tasks. Once solved, review your solution to ensure it makes sense in the context of the problem. This will build both confidence and competence in tackling physics challenges.
Understanding Rigid Body Rotation
Rigid body rotation refers to objects that rotate around an axis, where all points in the object move in circles around the axis with a common angular velocity. For the baton, the handle is the main axis of rotation, with both the metal shaft and rubber caps moving with it.
A key concept in rigid body rotation is that all parts of the body remain in fixed positions relative to each other, no bending or flexing occurs. This assumption simplifies calculations and makes it easier to analyze rotational dynamics.
When dealing with rigid bodies, one must consider the axis of rotation. For instance, the axis of our baton is through its center perpendicular to its length. The moment of inertia captures how the mass is distributed in relation to this axis, which affects how the baton will respond to a given torque. Only by understanding and calculating these aspects can you gain comprehensive insights into the motion and forces acting on the baton during twirling.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

You need to design an industrial turntable that is 60.0 cm in diameter and has a kinetic energy of 0.250 J when turning at 45.0 rpm v (a) What must be the moment of inertia of the turntable about the rotation axis? (b) If your workshop makes this turntable in the shape of a uniform solid disk, what must be its mass?

A solid uniform marble and a block of ice, each with the same mass, start from rest at the same height \(H\) above the bottom of a hill and move down it. The marble rolls without slipping, but the ice slides without friction. (a) Find the speed of each of these objects when it reaches the bottom of the hill. (b) Which object is moving faster at the bottom, the ice or the marble? (c) Which object has more kinetic energy at the bottom, the ice or the marble?

3\. An apparatus for launching a small boat consists of a 150.0 kg cart that rides down a set of tracks on four solid steel wheels, each with radius 20.0 \(\mathrm{cm}\) and mass 45.0 \(\mathrm{kg}\) . The tracks slope at an angle of \(7.50^{\circ}\) to the horizontal, and the boat's mass is 750.0 kg. If the boat is released from rest a distance of 16.0 \(\mathrm{m}\) from the water (measured along the slope), how fast will it be mov- ing when it reaches the water? Assume the wheels roll without slipping, and that there is no energy loss due to friction.

\(\bullet\) What fraction of the total kinetic energy is rotational for the following objects rolling without slipping on a horizontal surface? (a) a uniform solid cylinder; (b) a uniform sphere; (c) a thin-walled, hollow sphere; (d) a hollow cylinder with outer radius \(R\) and inner radius \(R / 2\) .

Emilie's potter's wheel rotates with a constant 2.25 \(\mathrm{rad} / \mathrm{s}^{2}\) angular acceleration. After 4.00 \(\mathrm{s}\) , the wheel has rotated through an angle of 60.0 rad. What was the angular velocity of the wheel at the beginning of the 4.00 s interval?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.