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Chapter 9: Problem 14

A circular saw blade 0.200 m in diameter starts from rest. In 6.00 s it accelerates with constant angular acceleration to an angular velocity of 140 rad/s. Find the angular acceleration and the angle through which the blade has turned.

Short Answer

Expert verified
Angular acceleration is 23.33 rad/s虏, and the blade turns 420 radians.

Step by step solution

01

Determine Initial Conditions

The saw blade starts from rest, so the initial angular velocity \( \omega_0 \) is 0 rad/s.
02

Identify Given Values

We have a final angular velocity \( \omega \) of 140 rad/s, an initial angular velocity \( \omega_0 \) of 0 rad/s, and a time interval \( t \) of 6.00 s.
03

Apply Angular Acceleration Formula

The formula for constant angular acceleration \( \alpha \) is \( \alpha = \frac{\omega - \omega_0}{t} \).\[ \alpha = \frac{140 - 0}{6.00} \]\[ \alpha = \frac{140}{6.00} = 23.33 \text{ rad/s}^2 \]
04

Calculate the Angle Turned

Use the angular displacement formula \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \). Since \( \omega_0 = 0 \), the equation simplifies to \( \theta = \frac{1}{2} \alpha t^2 \).\[ \theta = \frac{1}{2} \times 23.33 \times (6.00)^2 \]\[ \theta = \frac{1}{2} \times 23.33 \times 36 \]\[ \theta = 420 \text{ radians} \]

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Key Concepts

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

Angular Acceleration
Angular acceleration describes how quickly an object's angular velocity changes over time. It is similar to linear acceleration, but it applies to rotational motion. When a circular saw blade spins, it starts to rotate faster and faster, which results in an increase in angular velocity. Angular acceleration tells us how fast this increase happens.

To calculate angular acceleration, the formula used is:
  • \( \alpha = \frac{\omega - \omega_0}{t} \)
where:
  • \( \alpha \) is the angular acceleration in radians per second squared (rad/s虏)
  • \( \omega \) is the final angular velocity
  • \( \omega_0 \) is the initial angular velocity
  • \( t \) is the time taken to change the velocity
This measure shows us how much the angular velocity increases per second. In this exercise, the saw blade takes 6 seconds to reach an angular velocity of 140 rad/s from rest. So, its angular acceleration is calculated as \( 23.33 \text{ rad/s}^2 \).
Angular Velocity
Angular velocity is the speed of rotation of an object and is measured in radians per second (rad/s). While linear velocity looks at how fast something moves along a path, angular velocity measures how fast something spins or rotates.

When you consider the saw blade, imagine it as if you're watching a fan blade. You can tell how fast it spins, and that's what angular velocity describes.
  • It starts from rest, meaning its initial angular velocity, \( \omega_0 \), is 0 rad/s.
  • It reaches a final velocity of 140 rad/s.
Understanding how quickly the saw blade achieves this speed is crucial. We calculate angular acceleration with this information and confirm how intensely the blade has been set in motion. The 140 rad/s tells us exactly how fast the blade is spinning when it stops accelerating.
Angular Displacement
Angular displacement pertains to the angle through which an object rotates during its motion. It's similar to how linear displacement describes how far an object travels, but with rotations.

In the exercise, once we found the angular acceleration, we needed to determine how far, in terms of rotation, the saw blade has moved. To do this, we use:
  • \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \)
where:
  • \( \theta \) is the angular displacement, measured in radians
  • \( \alpha \) is the angular acceleration
  • \( t \) is the time
  • \( \omega_0 \) is the initial angular velocity
Since the blade starts from rest, the initial angular velocity \( \omega_0 \) is zero, simplifying the formula. Calculating gives us the total angle through which the blade has turned, which is 420 radians. This measure gives a clear picture of the blade鈥檚 overall motion during the acceleration phase.

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Most popular questions from this chapter

You are designing a rotating metal flywheel that will be used to store energy. The flywheel is to be a uniform disk with radius 25.0 cm. Starting from rest at \(t =\) 0, the flywheel rotates with constant angular acceleration 3.00 rad/s\(^2\) about an axis perpendicular to the flywheel at its center. If the flywheel has a density (mass per unit volume) of 8600 kg/m\(^3\), what thickness must it have to store 800 J of kinetic energy at \(t =\) 8.00 s?

Energy is to be stored in a 70.0-kg flywheel in the shape of a uniform solid disk with radius \(R =\) 1.20 m. To prevent structural failure of the flywheel, the maximum allowed radial acceleration of a point on its rim is 3500 m/s\(^2\). What is the maximum kinetic energy that can be stored in the flywheel?

A child is pushing a merry-go-round. The angle through which the merry-go- round has turned varies with time according to \(\theta{(t) = \gamma t + \beta t^3}\), where \(\gamma =\) 0.400 rad/s and \(\beta =\) 0.0120 rad/s\(^3\). (a) Calculate the angular velocity of the merry-go-round as a function of time. (b) What is the initial value of the angular velocity? (c) Calculate the instantaneous value of the angular velocity \(\omega$$_z\) at \(t =\) 5.00 s and the average angular velocity \(\omega_{av-z}\) for the time interval \(t =\) 0 to \(t =\) 5.00 s. Show that \(\omega_{av-z}\) is not equal to the average of the instantaneous angular velocities at \(t =\) 0 and \(t =\) 5.00 s, and explain.

A uniform 2.00-m ladder of mass 9.00 kg is leaning against a vertical wall while making an angle of 53.0\(^\circ\) with the floor. A worker pushes the ladder up against the wall until it is vertical. What is the increase in the gravitational potential energy of the ladder?

A uniform disk with radius \(R =\) 0.400 m and mass 30.0 kg rotates in a horizontal plane on a frictionless vertical axle that passes through the center of the disk. The angle through which the disk has turned varies with time according to \(\theta(t) =\) (1.10 rad/s)\(t +\) (6.30 rad/s\(^2)t^2\). What is the resultant linear acceleration of a point on the rim of the disk at the instant when the disk has turned through 0.100 rev?
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