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A planetary gear system is shown, where the gear teeth are omitted from the figure. Each of the three identical planet gears \(A, B,\) and \(C\) has a mass of \(0.8 \mathrm{kg},\) a radius \(r=50 \mathrm{mm},\) and a radius of gyration of \(30 \mathrm{mm}\) about its center. The spider \(E\) has a mass of \(1.2 \mathrm{kg}\) and a radius of gyration about \(O\) of \(60 \mathrm{mm} .\) The ring gear \(D\) has a radius \(R=150 \mathrm{mm}\) and is fixed. If a torque \(M=5 \mathrm{N} \cdot \mathrm{m}\) is applied to the shaft of the spider at \(O,\) determine the initial angular acceleration \(\alpha\) of the spider.

Short Answer

Expert verified
The initial angular acceleration of the spider is 400.64 rad/s².

Step by step solution

01

Identify Key Parameters

First, let's identify all the given parameters and convert any non-standard units.- Mass of each planet gear, \( m = 0.8 \text{ kg} \).- Radius of each planet gear, \( r = 0.05 \text{ m} \).- Radius of gyration of each planet gear, \( k_g = 0.03 \text{ m} \).- Mass of the spider, \( m_s = 1.2 \text{ kg} \).- Radius of gyration of the spider, \( k_s = 0.06 \text{ m} \).- Torque applied to the spider, \( M = 5 \text{ N} \cdot \text{m} \).
02

Calculate Moment of Inertia for Each Planet Gear

The moment of inertia \( I_g \) of each planet gear about its own axis is given by\[ I_g = m k_g^2 \].Substituting the given values:\[ I_g = 0.8 \times (0.03)^2 = 0.8 \times 0.0009 = 0.00072 \text{ kg} \cdot \text{m}^2 \].
03

Calculate Moment of Inertia for the Spider

The moment of inertia \( I_s \) of the spider about point \( O \) is given by\[ I_s = m_s k_s^2 \].Substituting the given values:\[ I_s = 1.2 \times (0.06)^2 = 1.2 \times 0.0036 = 0.00432 \text{ kg} \cdot \text{m}^2 \].
04

Consider Effective Moment of Inertia of the System

The total moment of inertia \( I_{total} \) of the system must account for both the spider and the planet gears. Since the planets also rotate around the center O, their effective moment of inertia is:\[ I_{planets} = 3 \times (I_g + m r^2) \].Substitute the values:\[ I_{planets} = 3 \times (0.00072 + 0.8 \times 0.05^2) = 3 \times (0.00072 + 0.8 \times 0.0025) = 3 \times 0.00272 = 0.00816 \text{ kg} \cdot \text{m}^2 \].
05

Calculate Total Effective Moment of Inertia

Now compute the total effective moment of inertia for the entire system:\[ I_{total} = I_s + I_{planets} \].Substituting in the values calculated:\[ I_{total} = 0.00432 + 0.00816 = 0.01248 \text{ kg} \cdot \text{m}^2 \].
06

Apply Torque Equation to Find Angular Acceleration

Using the relation between torque and angular acceleration:\[ M = I_{total} \alpha \].We solve for \( \alpha \):\[ \alpha = \frac{M}{I_{total}} = \frac{5}{0.01248} \].Calculating gives us:\[ \alpha = 400.64 \text{ rad/s}^2 \].

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Key Concepts

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

Angular Acceleration
In a planetary gear system, angular acceleration is a measure of how quickly a component within the system changes its rate of rotation. Specifically, it considers how the entire setup reacts to applied forces, like torque, which cause rotation or changes in rotation speed. When a torque is applied to the gear system, it introduces an angular acceleration \( \alpha \).

The relation between the applied torque \( M \) and the resulting angular acceleration \( \alpha \) can be visualized through its formula:
  • \( \alpha = \frac{M}{I_{total}} \)
Where \( I_{total} \) is the total moment of inertia of the system. This formula indicates that the greater the total moment of inertia, the lesser the angular acceleration for a given torque. Angular acceleration is measured in radians per second squared \( \text{rad/s}^2 \).

Understanding angular acceleration assists in predicting how a mechanism like the spider in our planetary system will react to forces and helps in designing systems that operate smoothly under specific requirements.
Moment of Inertia
Moment of inertia is a property of a body that determines its resistance to changes in rotational motion. It plays a crucial role in the gear dynamics of planetary systems.

Each component of a planetary gear system has its own moment of inertia, which depends on its mass distribution relative to the axis of rotation.

**For planet gears**:
The moment of inertia \( I_g \) of a planet gear about its axis is calculated as:
  • \( I_g = m k_g^2 \)
where \( m \) is the mass of the gear and \( k_g \) is the radius of gyration.

**For the spider**:
The spider's moment of inertia \( I_s \) about point O is:
  • \( I_s = m_s k_s^2 \)
The spider combines with the moving parts to create the system's total moment of inertia \( I_{total} \).

This collective moment of inertia impacts how the system responds to torque, making accurate calculations essential in predicting system behavior under different conditions.
Torque Equation
The torque equation is foundational in understanding rotational motion and dynamics within planetary gear systems. It is formally stated as \( M = I_{total} \alpha \), connecting the applied torque \( M \), the system's total moment of inertia \( I_{total} \), and the resultant angular acceleration \( \alpha \).

**Importance of Torque in Gear Dynamics**:
  • Torque is the force that causes objects to rotate. By applying torque to a system, you effectively set its components in motion.
  • This equation is crucial for solving problems where you need to predict how a planetary gear system like the spider will behave.
When you know the torque and total moment of inertia, you can easily find angular acceleration. It's a powerful tool that helps you design and analyze equipment performance, ensuring the efficiency and reliability of mechanical systems.
Gear Dynamics
Gear dynamics involve the study of motion and forces related to gears in a system like the planetary gear configuration. It takes into account various interactions and forces that act on different components as they transfer movement.

**Key Aspects of Gear Dynamics**:
  • **Motion Transfer:** Gears are designed to alter the transmission of motion, often increasing torque while reducing speed, or vice versa.
  • **Internal Forces:** Understanding how forces impact gear components, such as planet gears and the spider, allows for efficient system design.
In planetary gear systems, these dynamics are more complex due to multiple rotating gears interacting concurrently. Studying the forces and motion ensures that the system operates effectively under varying loads and conditions.

Gear dynamics is an integral part of mechanical design and plays a crucial role in ensuring that systems function smoothly, even under demanding mechanical stress.

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Most popular questions from this chapter

Small ball-bearing rollers mounted on the ends of the slender bar of mass \(m\) and length \(l\) constrain the motion of the bar in the horizontal \(x-y\) slots. If a couple \(M\) is applied to the bar initially at rest at \(\theta=\) \(45^{\circ},\) determine the forces exerted on the rollers at \(A\) and \(B\) as the bar starts to move.

The 6 -lb pendulum with mass center at \(G\) is pivoted at \(A\) to the fixed support \(C A .\) It has a radius of gyration of 17 in. about \(O-O\) and swings through an amplitude \(\theta=60^{\circ} .\) For the instant when the pendulum is in the extreme position, calculate the moments \(M_{x}, M_{y},\) and \(M_{z}\) applied by the base support to the column at \(C\)

The uniform semicircular ring of mass \(m=2.5 \mathrm{kg}\) and mean radius \(r=200 \mathrm{mm}\) is mounted on spokes of negligible mass and pivoted about a horizontal axis through \(O .\) If the ring is released from rest in the position \(\theta=30^{\circ},\) determine the force \(R\) supported by the bearing \(O\) just after release.

The wad of clay of mass \(m\) is initially moving with a horizontal velocity \(v_{1}\) when it strikes and sticks to the initially stationary uniform slender bar of mass \(M\) and length \(L\). Determine the final angular velocity of the combined body and the \(x\) -component of the linear impulse applied to the body by the pivot \(O\) during the impact.

The 20 -kg wheel has an eccentric mass which places the center of mass \(G\) a distance \(\bar{r}=70 \mathrm{mm}\) away from the geometric center \(0 .\) A constant couple \(M=6 \mathrm{N} \cdot \mathrm{m}\) is applied to the initially stationary wheel, which rolls without slipping along the horizontal surface and enters the curve of radius \(R=600 \mathrm{mm} .\) Determine the normal force under the wheel just before it exits the curve at \(C .\) The wheel has a rolling radius \(r=100 \mathrm{mm}\) and a radius of gyration \(k_{O}=65 \mathrm{mm}\)

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