Problem 92 - University Physics with Modern Physics

Question

The solid wood door of a gymnasium is 1.00 \(\mathrm{m}\) wide and 2.00 \(\mathrm{m}\) high, has total mass \(35.0 \mathrm{kg},\) 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 \(\mathrm{N}\) to the door for 8.00 \(\mathrm{ms} .\) Find the angular speed of the door after the impact. [Hint: Integrating Eq. \((10.29)\) yields \(\Delta L_{z}=\int_{t_{1}}^{t_{2}}\left(\Sigma \tau_{z}\right) d t=\left(\sum \tau_{z}\right)_{\mathrm{av}} \Delta t .\) The quantity \(\int_{t_{1}}^{t_{2}}\left(\Sigma \tau_{z}\right) d t\) is called the angular impulse.]

Step-by-Step Solution

Answer

The angular speed of the door after the impact is 0.514 rad/s.

Step 1: Understand the problem

The problem involves calculating the angular speed of a door after a basketball impacts it. Knowing the dimensions, mass of the door, the applied force, and the duration of impact allows us to use angular impulse to find the angular speed.

Step 2: Calculate the moment of inertia

The door can be modeled as a thin rectangle rotating about the vertical edge. The formula for a rectangular plate's moment of inertia about an edge is \( I = \frac{1}{3} M L^2 \), where \( M = 35.0\, \mathrm{kg} \) and \( L = 1.00\, \mathrm{m} \) represent the mass and width of the door respectively. Substituting these values in, we get \( I = \frac{1}{3} \times 35.0 \times (1.00)^2 = 11.67\, \mathrm{kg}\cdot\mathrm{m}^2 \).

Step 3: Calculate the torque applied during impact

The torque \( \tau \) is calculated by \( \tau = F \times r \), where \( F = 1500\, \mathrm{N} \) is the force applied and \( r = 0.5 \times 1.00 = 0.5\, \mathrm{m} \) (the force is applied in the center of the door). Thus, \( \tau = 1500 \times 0.5 = 750\, \mathrm{N}\cdot\mathrm{m} \).

Step 4: Determine the angular impulse

The angular impulse, given by the formula \( \Delta L_z = \tau \times \Delta t \), represents the change in angular momentum. With \( \Delta t = 8.00\, \mathrm{ms} = 8.00 \times 10^{-3} \mathrm{\, s} \), the angular impulse is \( \Delta L_z = 750 \times 8.00 \times 10^{-3} = 6.00\, \mathrm{N}\cdot\mathrm{m}\cdot\mathrm{s} \).

Step 5: Calculate the angular speed

Using the principle of conservation of angular momentum, where the angular impulse equals the change in angular momentum \( \Delta L_z = I \times \omega \). Solve for angular speed \( \omega \): \( \omega = \frac{\Delta L_z}{I} = \frac{6.00}{11.67} = 0.514\, \mathrm{rad/s} \).

Key Concepts

Moment of Inertia

The moment of inertia is a crucial concept in rotational dynamics. It is often compared to mass in linear motion. Simply put, it measures how much an object resists rotational acceleration around an axis.
For a gymnasium door, which essentially forms a rectangular shape rotating around a vertical edge, the moment of inertia is derived using the formula: \[I = \frac{1}{3} M L^2,\] where \(M\) is the door's mass and \(L\) is the width of the door.

This formula suggests that the larger the mass and size of the object, the more force it will require to achieve the same angular acceleration as an object with a smaller moment of inertia.

• In our problem, the door was given a mass of 35.0 kg and a width of 1.00 m. When we plug these values into the formula, we find the moment of inertia equals 11.67 kg·m².

This value tells us how "difficult" it is to set this door into rotational motion from a standstill.

Torque

Torque is the rotational equivalent of force and is essential in making objects rotate. It determines how effectively a force can cause an object to spin about an axis. Torque is calculated by multiplying the force applied to the object by the lever arm distance (the perpendicular distance from the axis of rotation to the line of action of the force): \[ \tau = F \times r.\]
In the given exercise, the door experienced a force due to the basketball impact. If this force is 1500 N and it hits at the door's midpoint (0.5 m from the hinge), we calculate: \[ \tau = 1500 \times 0.5 = 750 \text{ N}\cdot\text{m}. \]

• This torque indicates how many Newton meters of force are effectively working to rotate the door around its hinges during the basketball's brief contact.

Torque is a key factor in understanding how rotational motion is initiated and modified.

Angular Impulse

Angular impulse is a concept that connects the ideas of torque and time to changes in angular momentum, much like linear impulse relates force and time to changes in linear momentum. This relationship is expressed as: \[ \Delta L_z = \tau \times \Delta t. \]
The term on the left, \(\Delta L_z\), represents the change in angular momentum. In the door scenario, it is the product of the torque applied during the impact and the duration of that impact.

Given the impact lasted for 8.00 ms or 8.00 x 10⁻³ s, the angular impulse is: \[ \Delta L_z = 750 \times 8.00 \times 10^{-3} = 6.00 \text{ N}\cdot\text{m}\cdot\text{s}. \]

• This angular impulse causes a change in the door's angular momentum, leading it from rest to a specific angular speed.

By connecting these dots, we see how a brief force application can significantly change the rotational state of an object.