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Chapter 10: Problem 4

(I) The blades in a blender rotate at a rate of 6500 \(\mathrm{rpm.}\) When the motor is turned off during operation, the blades slow to rest in 4.0 \(\mathrm{s}\) . What is the angular acceleration as the blades slow down?

Short Answer

Expert verified
The angular acceleration is -170.25 rad/s虏.

Step by step solution

01

Understand the Problem

We need to find the angular acceleration given that the blades rotate at an initial rate of 6500 revolutions per minute (rpm) and come to a stop (final rate is 0 rpm) in 4 seconds.
02

Convert Units

The initial angular velocity is given in rpm. Convert this to radians per second. 1 revolution = 2蟺 radians, and 1 minute = 60 seconds. Initial angular velocity, \( \omega_i = 6500 \, \text{rpm} = 6500 \times \frac{2\pi}{60} \, \text{rad/s} \). Calculate the initial angular velocity:\( \omega_i = 6500 \times \frac{2\pi}{60} = \frac{6500 \times 2\pi}{60} \approx 681.0 \, \text{rad/s} \).
03

Apply the Angular Motion Equation

Use the equation for angular acceleration, \( \alpha \), which is:\[ \alpha = \frac{\omega_f - \omega_i}{t} \]where \( \omega_f \) is the final angular velocity (0 rad/s), \( \omega_i \) is the initial angular velocity, and \( t \) is the time taken (4 seconds).
04

Substitute Values and Solve

Substitute the known values into the equation:\[ \alpha = \frac{0 - 681.0}{4} \]Calculate the angular acceleration: \[ \alpha = \frac{-681.0}{4} = -170.25 \, \text{rad/s}^2 \].
05

Interpret the Result

The negative sign indicates that the acceleration is in the opposite direction of the initial angular velocity, signaling a deceleration.

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Key Concepts

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

Angular Velocity
Angular velocity tells us how fast something is rotating. In the context of the exercise, it refers to the speed of the blender blades. It's similar to linear velocity, but instead of moving in a straight line, it describes motion along a circular path. Angular velocity is often denoted by the Greek letter \( \omega \).
In our exercise, the initial angular velocity is 6500 revolutions per minute (rpm). This is a common unit for measuring rotational speeds in daily devices, like engines or fans. However, physics often requires us to work in different units, particularly when dealing with equations. That's why we'll convert this to radians per second.
Unit Conversion
Unit conversion is essential when dealing with rotational motion. Different systems use different units, such as rpm for revolutions per minute or rad/s for radians per second. Understanding how to switch between them is crucial for solving equations effectively.
In our problem, we need to convert from rpm to rad/s to use common physics equations. To do this conversion, remember:
  • 1 revolution equals \( 2\pi \) radians.
  • 1 minute equals 60 seconds.
To convert 6500 rpm to rad/s, you鈥檒l multiply by \( \frac{2\pi}{60} \). This translates 6500 revolutions in a minute to approximately 681.0 rad/s. This new unit makes it easier to understand the speed in terms of standard physics measurements.
Angular Motion Equation
The angular motion equation helps us find the angular acceleration \( \alpha \), which measures how quickly the rotation speed changes. This is very similar to linear acceleration but in a rotational context.
The equation is:
  • \( \alpha = \frac{\omega_f - \omega_i}{t} \)
This means we find the change in angular velocity (from initial \( \omega_i \) to final \( \omega_f \)), and divide by the time \( t \) over which this change occurs.
For the blender exercise:
  • Final speed \( \omega_f = 0 \) rad/s, as the blades stop moving.
  • Initial speed \( \omega_i = 681.0 \) rad/s (from our conversion).
  • Time \( t = 4 \) seconds.
Plugging these into the equation, we get \( \alpha = \frac{0 - 681.0}{4} = -170.25 \) rad/s\(^2\). The negative sign tells us that the blades are slowing down, which is why we say the acceleration is negative.

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Most popular questions from this chapter

(III) A wheel with rotational inertia \(I=\frac{1}{2} M R^{2}\) about its hori- zontal central axle is set spinning with initial angular speed \(\omega_{0}\) . It is then lowered, and at the instant its edge touches the ground the speed of the axle (and CM) is zero. Initially the wheel slips when it touches the ground, but then begins to move forward and eventually rolls without slipping. (a) In what direction does friction act on the slipping wheel? \((b)\) How long does the wheel slip before it begins to roll without slipping? (c) What is the wheel's final translational speed, \(v_{\mathrm{cM}} ?[\)Hint. Use \(\Sigma \vec{\mathbf{F}}=m \vec{\mathbf{a}}, \Sigma \tau_{\mathrm{CM}}=I_{\mathrm{CM}} \alpha_{\mathrm{CM}},\) and recall that only when there is rolling without slipping is \(v_{\mathrm{CM}}=\omega R . ]\)

Suppose David puts a \(0.50-\mathrm{kg}\) rock into a sling of length \(1.5 \mathrm{~m}\) and begins whirling the rock in a nearly horizontal circle, accelerating it from rest to a rate of 85 rpm after \(5.0 \mathrm{~s}\). What is the torque required to achieve this feat, and where does the torque come from?

On a 12.0-cm-diameter audio compact disc (CD), digital bits of information are encoded sequentially along an outward spiraling path. The spiral starts at radius \(R_{1}=2.5 \mathrm{~cm}\) and winds its way out to radius \(R_{2}=5.8 \mathrm{~cm} .\) To read the digital information, a CD player rotates the CD so that the player's readout laser scans along the spiral's sequence of bits at a constant linear speed of \(1.25 \mathrm{~m} / \mathrm{s}\). Thus the player must accurately adjust the rotational frequency \(f\) of the \(\mathrm{CD}\) as the laser moves outward. Determine the values for \(f\) (in units of \(\mathrm{rpm}\) ) when the laser is located at \(R_{1}\) and when it is at \(R_{2}\).

(II) \(\mathrm{A} .72\) -m-diameter solid sphere can be rotated about an axis through its center by a torque of 10.8 \(\mathrm{m} \cdot \mathrm{N}\) which accelerates it uniformly from rest through a total of 180 revolutions in 15.0 \(\mathrm{s}\) s. What is the mass of the sphere?

A string passing over a pulley has a 3.80-kg mass hanging from one end and a \(3.15-\mathrm{kg}\) mass hanging from the other end. The pulley is a uniform solid cylinder of radius \(4.0 \mathrm{~cm}\) and mass \(0.80 \mathrm{~kg} .\) ( \(a\) ) If the bearings of the pulley were frictionless, what would be the acceleration of the two masses? \((b)\) In fact, it is found that if the heavier mass is given a downward speed of \(0.20 \mathrm{~m} / \mathrm{s},\) it comes to rest in \(6.2 \mathrm{~s}\). What is the average frictional torque acting on the pulley?
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