Problem 95 - University Physics with Modern Physics

Question

A \(500.0-\mathrm{g}\) bird is flying horizontally at 2.25 \(\mathrm{m} / \mathrm{s}\) not paying much attention, when it suddenly flies into a stationary vertical bar, hitting it 25.0 \(\mathrm{cm}\) below the top (Fig. P10.95). The bar is uniform, 0.750 \(\mathrm{m}\) long, has a mass of \(1.50 \mathrm{kg},\) and is hinged at its base. The collision stuns the bird so that it just drops to the ground afterward (but soon recovers to fly happily away). What is the angular velocity of the bar (a) just after it is hit by the bird and (b) just as it reaches the ground?

Step-by-Step Solution

Answer

(a) 2.0 rad/s; (b) 5.59 rad/s.

Step 1: Analyze the System Before Collision

We need to determine the system's state before the collision. The bird is flying horizontally with a velocity of 2.25 m/s. The mass of the bird is 0.5 kg (converted from 500 g). The vertical bar is stationary. We are interested in how this initial state contributes to the angular momentum around the hinge point at the base of the bar.

Step 2: Calculate Initial Angular Momentum of Bird

Using the formula for angular momentum \( L = mvr \), where \( m \) is mass, \( v \) is velocity, and \( r \) is the distance from the hinge point to where the bird hits (0.25 m from the top or 0.5 m from the bottom). Thus, the initial angular momentum of the bird about the hinge is \( L = 0.5 \times 2.25 \times 0.50 = 0.5625 \ \text{kg}\cdot\text{m}^2/\text{s} \).

Step 3: Calculate Moment of Inertia of the Bar

For a uniform bar hinged at the end, the moment of inertia \( I \) is given by \( I = \frac{1}{3}ML^2 \). Here, \( M = 1.5 \ \text{kg} \) and \( L = 0.75 \ \text{m} \), so \( I = \frac{1}{3} \times 1.5 \times (0.75)^2 = 0.28125 \ \text{kg}\cdot\text{m}^2 \).

Step 4: Apply Conservation of Angular Momentum

Prior to the collision, only the bird has angular momentum. After the collision, the total system's angular momentum is conserved. Set the initial angular momentum equal to the final angular momentum: \( L_{\text{initial}} = I \cdot \omega_{\text{final}} \). Solve for \( \omega_{\text{final}} \): \( 0.5625 = 0.28125 \cdot \omega_{\text{final}} \). Therefore, \( \omega_{\text{final}} = 2.0 \ \text{rad/s} \).

Step 5: Calculate Potential Energy at Top and Kinetic Energy at Bottom

As the bar rotates down, potential energy due to gravity converts to kinetic energy. At the highest point, potential energy is \( U = Mgh \), where \( h = \frac{L}{2} = 0.375 \ \text{m} \). So, \( U = 1.5 \times 9.8 \times 0.375 = 5.5125 \ \text{J} \). This will convert to rotational kinetic energy at the bottom: \( K = \frac{1}{2}I \omega^2 \).

Step 6: Solve for Final Angular Velocity as Bar Hits Ground

Set the initial potential energy equal to the final kinetic energy: \( 5.5125 = \frac{1}{2} \times 0.28125 \times \omega^2 \). Solving \( \omega = \sqrt{\frac{5.5125 \times 2}{0.28125}} \), we find \( \omega \approx 5.59 \ \text{rad/s} \).

Key Concepts

Moment of Inertia

Moment of inertia is a fundamental concept in rotational dynamics. Think of it as the rotational equivalent of mass in linear motion. Just as a massive object is harder to move than a lighter one, an object with a large moment of inertia is harder to rotate. For our exercise, the moment of inertia, denoted as \( I \), depends on how the mass of the object is distributed relative to the axis of rotation.
For a uniform bar of length \( L \) and mass \( M \) that is hinged at one end, the formula for calculating the moment of inertia is \( I = \frac{1}{3} ML^2 \). This formula accounts for the mass being distributed along the length of the bar. Using this equation allows us to calculate how resistant the bar is to changes in its rotational state.
In our specific problem, we found \( I = 0.28125 \ \text{kg} \cdot \text{m}^2 \). This value is crucial as it is used to determine the angular velocity of the bar both after collision and when it reaches the ground.

Conservation of Angular Momentum

The principle of conservation of angular momentum is vital in the analysis of rotational motion, similar to conservation of momentum in linear motion. Essentially, it states that the total angular momentum of a system remains constant if no external torques act on it. This principle helps us understand how the initial angular momentum of the system (just the bird, in this case) translates into angular motion of the bar.
When the bird strikes the bar, the system includes both the bird and the bar. Angular momentum is initially present only due to the bird's motion, given by the formula \( L = mvr \), where \( m \) is mass, \( v \) is velocity, and \( r \) is the distance from the pivot to the point of impact.
After the collision, this initial angular momentum is transferred to the bar, allowing us to solve for the bar's angular velocity just after the collision. This illustrates the conservation of momentum principle: the bird's angular momentum is distributed to the bar, yielding a post-collision angular velocity of \( 2.0 \ \text{rad/s} \).

Rotational Kinetic Energy

When dealing with rotating objects, understanding rotational kinetic energy is essential. It's the energy due to the rotation of an object and is given by the formula \( K = \frac{1}{2}I \omega^2 \), where \( \omega \) is the angular velocity. This concept is analogous to kinetic energy in linear motion, but applied to spinning objects.
In our context, after the bar is struck and starts rotating, it gains rotational kinetic energy. At the base of its downward path, just as it hits the ground, this kinetic energy determines its angular velocity. The rotational kinetic energy originally comes from the potential energy the bar had at the top of its swing.
By calculating the kinetic energy when the bar reaches the ground, and setting this equal to the potential energy it had initially, we use the conversion of energy principle to find the final angular velocity of \( 5.59 \ \text{rad/s} \). This demonstrates how energy changes form but remains constant in a closed system.

Potential Energy Conversion

Potential energy conversion is an area that connects gravitational forces to motion. For an object at height, potential energy is the energy it possesses due to its position in a gravitational field. The formula used is \( U = Mgh \), where \( h \) is the height above a reference point. This energy can be converted into other forms, like kinetic energy, as the object falls.
In the problem, the bar starts with a certain amount of potential energy at the top of its path. This potential energy, calculated to be \( 5.5125 \ \text{J} \), converts into kinetic energy as the bar swings downwards.
Understanding potential energy conversion allows us to see how energy moves from one form to another without being lost, demonstrating crucial principles of physics like energy conservation. By calculating the potential energy and equating it to the kinetic energy just before the bar hits the ground, we found the final angular velocity, linking these stages of energy transformation.