Problem 3 - Basic Technical Mathematics with Calculus

Question

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. $$4.2 \mathrm{g} \text { at }(1.7,0), 3.2 \mathrm{g} \text { at }(3.5,0)$$

Step-by-Step Solution

Answer

I = 51.338 g·cm², radius of gyration = 2.633 cm.

Step 1: Understand the Moment of Inertia

The moment of inertia about the origin for point masses is given by the formula \( I = \sum m_i \cdot r_i^2 \), where \( m_i \) is the mass and \( r_i \) is the distance of the mass from the origin. We'll calculate this for each given mass.

Step 2: Calculate Moment of Inertia for 4.2 g Mass

For the mass 4.2 g located at (1.7,0), calculate \( I_1 = 4.2 \, \text{g} \times (1.7 \, \text{cm})^2 \). This becomes \( I_1 = 4.2 \times 2.89 = 12.138 \text{ g} \cdot \text{cm}^2 \).

Step 3: Calculate Moment of Inertia for 3.2 g Mass

For the mass 3.2 g located at (3.5,0), calculate \( I_2 = 3.2 \, \text{g} \times (3.5 \, \text{cm})^2 \). This becomes \( I_2 = 3.2 \times 12.25 = 39.2 \text{ g} \cdot \text{cm}^2 \).

Step 4: Sum the Moments of Inertia

Add the individual moments of inertia to find the total moment of inertia: \( I = I_1 + I_2 = 12.138 + 39.2 = 51.338 \text{ g} \cdot \text{cm}^2 \).

Step 5: Calculate the Radius of Gyration

The radius of gyration \( k \) is found using the formula \( k = \sqrt{\frac{I}{M}} \), where \( M \) is the total mass. Calculate \( M = 4.2 + 3.2 = 7.4 \, \text{g} \). Then \( k = \sqrt{\frac{51.338}{7.4}} \approx \sqrt{6.936}\approx 2.633 \, \text{cm} \).

Key Concepts

Radius of Gyration

The radius of gyration, denoted by \( k \), is a very interesting concept that helps us understand how mass is distributed around an axis. It is particularly useful because it provides a single value to describe the distribution of a system's mass, involving a complex calculation.
The formula to calculate the radius of gyration is given by:

• \( k = \sqrt{\frac{I}{M}} \)

where \( I \) is the total moment of inertia, and \( M \) is the total mass of all point masses combined.
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

Calculating the moment of inertia is fundamental in understanding how different bodies resist rotational motion. The general formula for a system of discrete point masses is:

• \( I = \sum m_i \cdot r_i^2 \)

Here, \( m_i \) represents each individual mass and \( r_i \) is the distance of the mass from the axis of rotation, in our case, the origin.
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

Understanding point masses is a stepping stone for grasping more complex concepts in physics and engineering. A point mass is an idealization where all the mass of an object is assumed to be concentrated at a single point. This simplification is particularly useful in theoretical calculations, such as determining the moment of inertia discussed earlier.
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.

Thus, point masses are essential for simplifying real-world physics into comprehensible problems that can be solved mathematically.

Mathematics Education

Mathematics education thrives on building intuition as well as problem-solving skills. Concepts like moment of inertia play a significant role in developing a student's ability to approach problems from different angles.
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.

Effective mathematics education not only prepares students to solve complex problems but also stimulates their curiosity and interest in the world around them. Each new concept, like the radius of gyration or point masses, builds upon this foundation, driving a deeper understanding of both physics and mathematics.