Chapter 26: Problem 3
Find the moment of inertia (in \(\mathrm{g} \cdot \mathrm{cm}^{2}\) ) and the radius of gyration (in \(\mathrm{cm}\) ) with respect to the origin of each of the given arrays of masses located at the given points on the \(x\) -axis. $$4.2 \mathrm{g} \text { at }(1.7,0), 3.2 \mathrm{g} \text { at }(3.5,0)$$
Short Answer
Step by step solution
Understand the Moment of Inertia
Calculate Moment of Inertia for 4.2 g Mass
Calculate Moment of Inertia for 3.2 g Mass
Sum the Moments of Inertia
Calculate the Radius of Gyration
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Radius of Gyration
The formula to calculate the radius of gyration is given by:
- \( k = \sqrt{\frac{I}{M}} \)
This formula allows us to conceptualize the moment of inertia as if all the mass were concentrated at a single distance \( k \) from the axis. Such simplification is quite valuable in both mathematical computations and practical applications, like engineering and physics.
Calculating Moment of Inertia
- \( I = \sum m_i \cdot r_i^2 \)
For instance, when calculating the moment of inertia for a 4.2 g mass located at 1.7 cm from the origin, this simply becomes \( I_1 = 4.2 \cdot (1.7)^2 \), which is evaluated to \( 12.138 \; \text{g} \cdot \text{cm}^2 \).
Repeating this process for other masses within the system allows us to sum these individual contributions to find the system's total moment of inertia. Remember, this is crucial for knowing how much torque is needed to produce a desired angular acceleration.
Point Mass
Consider a mass of 3.2 g situated at a point on the x-axis, like in our problem. Despite real objects having dimensions, treating them as point masses reduces the problem to one that can be handled easily with basic mathematical formulas.
- This idealization helps us focus on key quantities like mass and distance without worrying about an object's shape or volume.
- It also forms the foundation for more complex models where the distribution of mass in space becomes relevant.
Mathematics Education
By using step-by-step solutions as shown in this exercise, students can understand how each part of a formula interacts. This reinforces their ability to deconstruct various parts of complex equations and apply them in new contexts.
- Using practical examples helps students see the usefulness of mathematical concepts in real-world situations.
- Teachers can facilitate learning by breaking down complicated topics into simpler parts, building confidence in students.