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Chapter 10: Problem 92

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.]

Short Answer

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

Step by step solution

01

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.
02

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 \( I = \frac{1}{3} \times 35.0 \times (1.00)^2 = 11.67\, \mathrm{kg}\cdot\mathrm{m}^2 \).
03

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, \( \tau = 1500 \times 0.5 = 750\, \mathrm{N}\cdot\mathrm{m} \).
04

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 \( \Delta L_z = 750 \times 8.00 \times 10^{-3} = 6.00\, \mathrm{N}\cdot\mathrm{m}\cdot\mathrm{s} \).
05

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 \( \omega \): \( \omega = \frac{\Delta L_z}{I} = \frac{6.00}{11.67} = 0.514\, \mathrm{rad/s} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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路m虏.
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鈦宦� 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.

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Most popular questions from this chapter

An engine delivers 175 hp to an aircraft propeller at 2400 rev/min. (a) How much torque does the aircraft engine provide? (b) How much work does the engine do in one revolution of the propeller?

A 2.00 -kg textbook rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is \(0.150 \mathrm{m},\) to a hanging book with mass 3.00 \(\mathrm{kg}\) . The system is released from rest, and the books are observed to move 1.20 \(\mathrm{m}\) in 0.800 s. (a) What is the tension in each part of the cord? (b) What is the moment of inertia of the pulley about its rotation axis?

(a) Compute the torque developed by an industrial motor whose output is 150 \(\mathrm{kW}\) at an angular speed of 4000 \(\mathrm{rev} / \mathrm{min}\) . (b) A drum with negligible mass, 0.400 \(\mathrm{m}\) in diameter, is attached to the motor shaft, and the power output of the motor is used to raise a weight hanging from a rope wrapped around the drum. How heavy a weight can the motor lift at constant speed? (c) At what constant speed will the weight rise?

A playground merry-go-round has radius 2.40 \(\mathrm{m}\) and moment of inertia 2100 \(\mathrm{kg} \cdot \mathrm{m}^{2}\) about a vertical axle through its center, and it turns with negligible friction. (a) A child applies an 18.0 -N force tangentially to the edge of the merry-go-round for 15.0 s. If the merry- go-round is initially at rest, what is its angular speed after this 15.0 -s interval? (b) How much work did the child do on the merry-go-round? (c) What is the average power supplied by the child?

A solid wood door 1.00 \(\mathrm{m}\) wide and 2.00 \(\mathrm{m}\) high is hinged along one side and has a total mass of 40.0 \(\mathrm{kg}\) . Initially open and at rest, the door is struck at its center by a handful of sticky mud with mass 0.500 \(\mathrm{kg}\) , traveling perpendicular to the door at 12.0 \(\mathrm{m} / \mathrm{s}\) just before impact. Find the final angular speed of the door. Does the mud make a significant contribution to the moment of inertia?
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