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Chapter 11: Problem 137

The peripheral speed of the tooth of a 10 -in-diameter circular saw blade is \(150 \mathrm{ft} / \mathrm{s}\) when the power to the saw is turned off. The speed of the tooth decreases at a constant rate, and the blade comes to rest in \(9 \mathrm{s}\). Determine the time at which the total acceleration of the tooth is \(130 \mathrm{ft} / \mathrm{s}^{2}\).

Short Answer

Expert verified
At approximately 3.6 seconds.

Step by step solution

01

Identify the Problem

We are looking to find the time at which the total acceleration of the tooth is \(130 \text{ ft/s}^2\), given the saw blade's initial conditions and the time it takes to stop.
02

Determine Angular Deceleration

First, calculate the angular deceleration. The initial angular speed, \( \omega_0 \), can be calculated from the peripheral speed:\[ \omega_0 = \frac{v_0}{r} = \frac{150 \text{ ft/s}}{5/12 \text{ ft}} \] The radius \( r \) is half of the diameter, converted to feet. Then, use the equation \( \omega = \omega_0 + \alpha t \) and the condition that the blade stops after 9 seconds (\( \omega = 0 \)) to find \( \alpha \):\[ 0 = \omega_0 + \alpha \times 9 \] Solving for \( \alpha \) gives \( \alpha = -\frac{\omega_0}{9} \).
03

Angular Speed Function

Calculate the angular speed as a function of time:\[ \omega(t) = \omega_0 + \alpha t \] Substitute \( \alpha = -\frac{\omega_0}{9} \) to get:\[ \omega(t) = \omega_0 (1 - \frac{t}{9}) \].
04

Total Acceleration Condition

The total acceleration is the vector sum of radial and tangential accelerations. Radial acceleration \( a_r \) is given by:\[ a_r = r\omega^2 \] Tangential acceleration \( a_t \) is:\[ a_t = r \alpha = -r \frac{\omega_0}{9} \]. The magnitude of total acceleration \( a\) is:\[ a = \sqrt{a_t^2 + a_r^2} = 130 \text{ ft/s}^2 \].
05

Solve for Time

We solve for \( t \) using the relationship we established. Replace \( a_t \) and \( a_r \) with the expressions in terms of \( t \) and solve the quadratic equation for the given total acceleration \( a = 130 \text{ ft/s}^2 \). The radial acceleration term becomes:\[ a_r = r\left( \omega_0 (1 - \frac{t}{9}) \right)^2 \]. Insert both into the total acceleration equation and solve for \( t \):\[ \sqrt{(-r \frac{\omega_0}{9})^2 + r^2\left( \omega_0 (1 - \frac{t}{9}) \right)^4} = 130 \].

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Key Concepts

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

Peripheral Speed
The peripheral speed is a key concept in understanding the motion of rotating objects, like a saw blade. Peripheral speed refers to the linear speed of a point on the edge of a rotating object, such as a circular saw. It is the rate at which a point on the outer rim of the object moves along its circular path. Given in the exercise, the saw blade has a diameter of 10 inches. Therefore, the peripheral speed of the blade's tooth is 150 ft/s when the power is initially cut off. To find the angular speed, you first recognize that peripheral speed relates to angular speed via the formula:- \[ v_0 = r \cdot \omega_0 \] - where:
  • \( v_0 \) is the peripheral speed (150 ft/s).
  • \( r \) is the radius of the saw blade (5/12 ft after conversion from inches).
  • \( \omega_0 \) is the initial angular speed.
By rearranging the equation, you can calculate \( \omega_0 \). Peripheral speed is important because it provides an understanding of how fast the outer edge of an object is moving. It's directly related to the angular speed of the entire object.
Total Acceleration
Total acceleration of a point on a rotating object is a combination of different types of acceleration. In our scenario, total acceleration refers to the combined effect of radial (centripetal) and tangential accelerations. The total acceleration \( a \) can be expressed as:- \[ a = \sqrt{a_t^2 + a_r^2} \]- The problem specifies a total acceleration of 130 ft/s虏 at a certain time, and solving for when this condition is met involves understanding both components:
  • **Recalculating components:** Radial, \( a_r \), shows how the position along the circular path changes, while tangential, \( a_t \), changes the speed along this path.
  • The radial acceleration \( a_r \) is dependent on the speed at which a point moves along the circular path, given by \( a_r = r\omega^2 \).
  • The tangential acceleration is primarily driven by changes in speed, calculated via \( a_t = r \alpha \).
Understanding total acceleration not only helps in finding when the saw blade reaches a certain speed but also contributes to predicting how quickly it can stop or reach a certain condition.
Radial and Tangential Acceleration
Radial and tangential accelerations are components that work in different directions within a circular motion. Knowing them helps to understand the motion of an object going around a circle. **Radial Acceleration**When an object moves in a circular path, it changes direction constantly. This change in direction is handled by radial acceleration, also known as centripetal acceleration. Given by:- \[ a_r = r \omega^2 \]- It points toward the center of the circle, ensuring the object stays along its circular path.**Tangential Acceleration**On the other hand, tangential acceleration is responsible for changes in the speed of the object along the path. It can be calculated as:- \[ a_t = r \alpha \]- When power to the saw blade is turned off, the tangential acceleration embodies the rate of deceleration as the blade coasts to a stop. These two accelerations combine to produce the total acceleration, making them crucial in understanding the dynamics of rotating objects. Knowing about radial and tangential accelerations gives insight into both the speed and direction changes for an object in circular motion.

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