Problem 78 - University Physics with Modern Physics

Question

The solid wood door of a gymnasium is 1.00 m wide and 2.00 m high, has total mass 35.0 kg, and is hinged along one side. The door is open and at rest when a stray basketball hits the center of the door head-on, applying an average force of 1500 N to the door for 8.00 ms. Find the angular speed of the door after the impact. [\(Hint:\) Integrating Eq. (10.29) yields \(\Delta L_z = \int_{t1}^{t2} (\Sigma\tau_z)dt = (\Sigma\tau_z)_av \Delta t\). The quantity \(\int_{t1}^{t2} (\Sigma\tau_z)dt\) is called the angular impulse.]

Step-by-Step Solution

Answer

The angular speed of the door after the impact is approximately 0.129 rad/s.

Step 1: Understanding the Problem

We need to find the angular speed of the door after it is hit by a basketball. We know the door's dimensions, mass, and the force applied by the basketball over a specified time.

Step 2: Key Equations and Concepts

The angular impulse applied to the door is given by the formula \( \Delta L_z = (\Sigma\tau_z)_{avg} \Delta t \), where \( (\Sigma\tau_z)_{avg} \) is the average torque. The relationship between torque, force, lever arm distance (d), and force is \( \Sigma\tau = F \cdot d \) where d is half of the door's width (distance from hinge to impact point).

Step 3: Calculate Torque

The force applies at the center of the door, which is 0.5 m from the hinge (half the width of the door). Therefore, the average torque \( (\Sigma\tau_z)_{avg} = 1500 \, \text{N} \times 0.5 \, \text{m} = 750 \, \text{Nm} \).

Step 4: Calculate Angular Impulse

Using the given time period, \( \Delta t = 8.00 \, \text{ms} = 8.00 \times 10^{-3} \, \text{s} \). Thus, the angular impulse \( \Delta L_z = 750 \, \text{Nm} \times 8.00 \times 10^{-3} \, \text{s} = 6.00 \, \text{Nms} \).

Step 5: Determine the Moment of Inertia

The door is a rectangular object hinged along one side. The moment of inertia \( I \) is \( \frac{1}{3} m h^2 \) based on its geometry. Here, \( m = 35.0 \, \text{kg} \), \( h = 2.00 \, \text{m} \). So, \( I = \frac{1}{3} \times 35.0 \, \text{kg} \times (2.00 \, \text{m})^2 = \frac{1}{3} \times 35.0 \times 4.00 = 46.67 \, \text{kg m}^2 \).

Step 6: Calculate Angular Speed

The change in angular momentum \( \Delta L_z \) equals the final angular momentum \( I \omega \) (assuming initial momentum is zero), where \( \omega \) is the angular speed. Thus, \( \omega = \frac{\Delta L_z}{I} = \frac{6.00 \, \text{Nms}}{46.67 \, \text{kg m}^2} = 0.129 \, \text{rad/s} \).

Key Concepts

Angular Impulse

Angular impulse is a crucial concept in physics that helps us understand how rotational motion changes over time. It is similar to the linear impulse, which dictates how linear motion changes. Angular impulse is defined as the change in angular momentum over a period of time.
In this problem, the stray basketball applies a force to the door, causing it to rotate. This force, applied over a certain time duration, gives the door its angular impulse. The mathematical expression for angular impulse is \[ \Delta L_z = \int_{t1}^{t2} (\Sigma\tau_z) dt = (\Sigma\tau_z)_{avg} \Delta t \] where \( \Delta L_z \) is the change in angular momentum and \((\Sigma\tau_z)_{avg}\) is the average torque applied. By applying the angular impulse, we can determine how quickly the door will start to rotate after being hit by the basketball.

Torque

Torque is the rotational equivalent of force. It represents the tendency of a force to rotate an object around an axis. When a force is applied at a distance from the rotational axis, it creates torque. The magnitude of torque depends on two factors: the size of the force applied and the distance from the pivot point to where the force is applied.
In this exercise, the basketball hits the door at its center, creating torque. The mathematical representation of torque is given by \[ \Sigma\tau = F \cdot d \] where \( F \) is the force and \( d \) is the lever arm distance from the center of rotation. For the rectangular door, the distance \( d \) is 0.5 meters, which is half the width of the door from the hinge to the impact point. With a force of 1500 N, this results in an average torque of 750 Nm, as calculated in the step-by-step solution.

Moment of Inertia

The moment of inertia is a measure of an object's resistance to changes in its rotational motion. Just like mass is a measure of an object's resistance to linear acceleration, moment of inertia plays a similar role in rotation. It depends on the mass distribution relative to the axis of rotation.
For the rectangular door hinged along one side, the moment of inertia \( I \) can be calculated using the formula \[ I = \frac{1}{3} m h^2 \] Here, \( m \) is the mass of the door, and \( h \) is its height. For the given door with a mass of 35.0 kg and a height of 2.00 meters, the moment of inertia is calculated to be 46.67 kg m². This value indicates how much torque is needed for a desired angular acceleration.

Rectangular Door

In this particular exercise, we are dealing with a rectangular door, which is a common real-world object with practical implications in physics. Understanding the dynamics of a rectangular door helps us grasp how forces and motions work in everyday situations.
The door has dimensions of 1.00 meters wide and 2.00 meters high and is hinged along one of its vertical sides. The motion of the door after being hit is analyzed in terms of its angular motion rather than linear motion. This approach is essential for objects that rotate about a fixed axis, like doors, gates, or pivots.

• Width: 1.00 m
• Height: 2.00 m
• Mass: 35.0 kg
• Acted on at the center

By understanding these characteristics, we can calculate the door's behavior when subjected to forces, such as angular impulse and torque.