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Chapter 10: Problem 46

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?

Short Answer

Expert verified
Calculations based on the given values will yield the final angular speed of the door. Using similar reasoning, you'll find that the mud does make a measurable, but not significant, contribution to the moment of inertia.

Step by step solution

01

Calculate moment of inertia of the door

The formula to calculate the moment of inertia for a rod rotating about its end is \( I_{door}=\frac{1}{3}m_{door}L_{door}^{2} \), where \( m_{door} \) is the mass of the door and \( L_{door} \) is its length. So, plug the given values into the formula to find \( I_{door}=\frac{1}{3}*40.0 \mathrm{~kg}*(2.00 \mathrm{~m})^2 \) and calculate.
02

Calculate the initial angular momentum of the mud

The initial angular momentum of the mud can be calculated using \( L_{mud}=m_{mud} * v_{mud} * r \), where \( m_{mud} \) is the mass of the mud, \( v_{mud} \) is the mud's velocity, and \( r \) is the distance from the rotation axis to where the force is applied, which is half the width of the door. So, plug the given values into the formula to find \( L_{mud}=0.500 \mathrm{~kg}*(12.0 \mathrm{~m/s})*(1.00 \mathrm{~m}/2) \) and calculate.
03

Calculate the final angular speed of the door

Given that the total angular momentum does not change after the collision (angular momentum conservation), you can equate the initial angular momentum of the mud with the final angular momentum of the door and solve for the final angular speed w of the door using \( L_{mud}= I_{door} * w \). Then calculate the w.
04

Check the contribution of the mud to the moment of inertia

After the mud sticks to the door, the mud鈥檚 moment of inertia is given by \( I_{mud}=m_{mud} * r_{mud}^2 \). Then you can compare the moment of inertias of the mud and the door to check if the mud makes a significant contribution.

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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 an essential concept in physics, especially when dealing with rotational motion. It determines how much torque is needed to rotate an object about an axis. For our door problem, the moment of inertia helps measure the door's resistance to changes in its rotational motion.

In this specific exercise, we calculate the moment of inertia of a solid door using the formula for a rod rotating about its end: \[ I_{door} = \frac{1}{3} m_{door} L_{door}^{2} \]
  • \( m_{door} \) is the mass of the door, which is 40.0 kg.
  • \( L_{door} \) is the length of the door, equal to 2.00 m.
By plugging these values into the formula, you will be able to determine how hard it is to alter the door's rotation around the hinge due to its mass distribution. This provides a foundation to solve the rest of the exercise with respect to how the mud affects the door's motion.
Angular Velocity
Angular velocity refers to how fast an object rotates or revolves relative to another point, usually the center of a circle or a hinge. It is a crucial part of our exercise since the problem ultimately asks for the final angular speed of the door after being hit by the mud.

In the original problem, the goal is to find the door's angular velocity after the collision using the fact that angular momentum is conserved. This means that the initial angular momentum of the mud is transferred to the door such that:\[ L_{mud} = I_{door} \cdot \omega \]where:
  • \( L_{mud} \) represents the initial angular momentum of the mud.
  • \( I_{door} \) is the moment of inertia of the door we calculated previously.
  • \( \omega \) is the angular velocity, which we solve for.
By solving this equation, you can determine how quickly the door will spin as a result of the impact.
Conservation of Angular Momentum
The conservation of angular momentum is a fundamental principle in physics whereby the total angular momentum of a closed system remains constant if no external torques act upon it. This principle is particularly useful when analyzing rotational systems and collisions, such as the one presented in our exercise involving the door and the mud.

In our problem, the system is considered closed once the mud sticks to the door, meaning that all angular momentum generated when the mud hits the door remains within the system. Before the collision, the mud's travel provides it with a linear momentum, which, upon impact, converts into angular momentum. The equation used for this is:\[ m_{mud} \cdot v_{mud} \cdot r = I_{door} \cdot \omega \]Here, the term on the left represents the initial angular momentum of the mud, and the term on the right is the resulting angular momentum of the door post-collision.

This principle allows us to link these two states, giving us a straightforward way to solve for the post-collision angular velocity and analyze any contributions, like that of the mud, to the door's motion dynamics.

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Most popular questions from this chapter

A solid uniform sphere and a thin-walled, hollow sphere have the same mass \(M\) and radius \(R .\) If they roll without slipping up a ramp that is inclined at an angle \(\beta\) above the horizontal and if both have the same \(v_{\mathrm{cm}}\) before they start up the incline, calculate the maximum height above their starting point reached by each object. Which object reaches the greater height, or do both of them reach the same height?

The moment of inertia of the empty turntable is \(1.5 \mathrm{~kg} \cdot \mathrm{m}^{2}\). With a constant torque of \(2.5 \mathrm{~N} \cdot \mathrm{m},\) the turntable-person system takes \(3.0 \mathrm{~s}\) to spin from rest to an angular speed of \(1.0 \mathrm{rad} / \mathrm{s} .\) What is the person's moment of inertia about an axis through her center of mass? Ignore friction in the turntable axle. (a) \(2.5 \mathrm{~kg} \cdot \mathrm{m}^{2}\) (b) \(6.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\) (c) \(7.5 \mathrm{~kg} \cdot \mathrm{m}^{2} ;\) (d) \(9.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\).

A thin-walled, hollow spherical shell of mass \(m\) and radius \(r\) starts from rest and rolls without slipping down a track (Fig. \(\mathbf{P 1 0 . 7 2}\) ). Points \(A\) and \(B\) are on a circular part of the track having radius \(R\). The diameter of the shell is very small compared to \(h_{0}\) and \(R,\) and the work done by rolling friction is negligible. (a) What is the minimum height \(h_{0}\) for which this shell will make a complete loop-the-loop on the circular part of the track? (b) How hard does the track push on the shell at point \(B,\) which is at the same level as the center of the circle? (c) Suppose that the track had no friction and the shell was released from the same height \(h_{0}\) you found in part (a). Would it make a complete loop-theloop? How do you know? (d) In part (c), how hard does the track push on the shell at point \(A,\) the top of the circle? How hard did it push on the shell in part (a)?

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?

The flywheel of an engine has moment of inertia \(1.60 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about its rotation axis. What constant torque is required to bring it up to an angular speed of 400 rev \(/ \min\) in 8.00 s, starting from rest?
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