Problem 82 - Fundamentals of Physics

Question

A uniform rod rotates in a horizontal plane about a vertical axis through one end. The rod is \(6.00 \mathrm{~m}\) long, weighs \(10.0 \mathrm{~N}\), and rotates at \(240 \mathrm{rev} / \mathrm{min} .\) Calculate (a) its rotational inertia about the axis of rotation and (b) the magnitude of its angular momentum about that axis.

Step-by-Step Solution

Answer

Rotational inertia is approximately 12.24 kg m²; angular momentum is approximately 307.37 kg m²/s.

Step 1: Convert Rotational Speed to Radians per Second

To begin, we need to convert the given rotational speed from revolutions per minute to radians per second. 1 revolution equals \(2\pi\) radians. Therefore, \(240\) revolutions per minute can be converted as follows:\[\omega = 240 \frac{\text{rev}}{\text{min}} \times \frac{2\pi \text{ rad}}{1 \text{rev}} \times \frac{1 \text{ min}}{60 \text{ seconds}}\]Calculating this results in:\[\omega \approx 25.13 \text{ rad/s}\]

Step 2: Calculate Rotational Inertia

The formula for the rotational inertia (\(I\)) of a uniform rod of length \(L\) and mass \(m\) rotating about an axis through one end perpendicular to its length is:\[I = \frac{1}{3} m L^2\]First, we calculate the mass \(m\) from the weight \(10.0 \mathrm{~N}\) using the relationship \(W = mg\), where \(g = 9.8 \mathrm{~m/s}^2\):\[m = \frac{10.0 \mathrm{~N}}{9.8 \mathrm{~m/s}^2} \approx 1.02 \mathrm{~kg}\]Substitute \(m\) and \(L = 6.00 \mathrm{~m}\) into the formula for \(I\):\[I = \frac{1}{3} (1.02 \mathrm{~kg}) (6.00 \mathrm{~m})^2 \approx 12.24 \mathrm{~kg \, m^2}\]

Step 3: Calculate Angular Momentum

The angular momentum (\(L\)) about the axis of rotation is given by the product of the rotational inertia (\(I\)) and angular speed (\(\omega\)):\[L = I \omega\]Substituting the known values:\[L = (12.24 \mathrm{~kg \, m^2}) \times (25.13 \text{ rad/s}) \approx 307.37 \mathrm{~kg \, m^2/s}\]

Step 4: Summarize the Results

The calculations show that the rotational inertia of the rod about the axis of rotation is approximately \(12.24 \mathrm{~kg \, m^2}\). The magnitude of its angular momentum around that axis is approximately \(307.37 \mathrm{~kg \, m^2/s}\).

Key Concepts

Understanding Angular Momentum in Rotational Motion

Angular momentum is a key concept when discussing objects in rotational motion. It is the rotational equivalent of linear momentum. Think of it as a measure of how difficult it is to stop a spinning object. This concept can be defined as the product of an object's rotational inertia and its angular velocity. Angular momentum, often represented as \(L\), is calculated by the equation:

• \(L = I \omega\)

where \(I\) is the rotational inertia and \(\omega\) is the angular velocity in radians per second.
In our exercise, the angular momentum of the rotating rod is calculated using its inertia and angular velocity. The result tells us how much effort would be needed to bring the rod to a stop. The calculated angular momentum for our uniform rod was approximately \(307.37 \, \text{kg m}^2/\text{s}\).
Understanding angular momentum is crucial for students studying physics as it applies to various scenarios, such as understanding how the Earth's rotation is influenced by external forces or how athletes can perform spins and flips in gymnastics.

Rotational Inertia: The Moment of Inertia Explained

Rotational inertia, sometimes called the moment of inertia, describes how difficult it is to change the rotational motion of an object. It is similar to mass in linear motion. Rotational inertia depends on two main factors: the mass of the object and how that mass is distributed relative to the axis of rotation.
In the exercise above, calculation of rotational inertia \(I\) of the uniform rod was crucial. We used the formula:

• \(I = \frac{1}{3} m L^2\)

Here, \(m\) represents the mass, and \(L\) is the length of the rod. The factor of \(\frac{1}{3}\) arises due to the different mass distribution along the rod compared to a point mass. The calculated rotational inertia for our rod was \(12.24 \, \text{kg}\,\text{m}^2\).
This measure essentially tells us how the mass distribution in a body affects its ability to rotate. More spread out mass leads to higher rotational inertia, making it harder to spin.

Characteristics of a Uniform Rod in Rotational Motion

A uniform rod is a type of rigid body with mass evenly distributed along its length. In terms of physics, this uniform distribution gives us distinct characteristics for calculations.
For a uniform rod rotating about an end, like in our exercise, the uniform distribution means we can apply straightforward formulas to calculate inertia and angular momentum, as these formulas account for the distribution.A uniform rod, rotating about its end, uses the formula \(I = \frac{1}{3}m L^2\) for its moment of inertia because the entire mass is spread continuously over its length. The calculations considered both the length (6 m) and the weight (10 N) to determine these properties, leading to accurate results without needing complex integration or approximation.
Understanding that the mass is uniformly spread makes predictions about the rod's motion valid and teaches students about simplifying assumptions in physics that help tackle real-world problems efficiently.