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Chapter 9: Problem 8

A wheel is rotating about an axis that is in the \(z\) -direction. The angular velocity \(\omega_{z}\) is \(-6.00\) rad/s at \(t=0\) , increases lin early with time, and is \(+8.00 \mathrm{rad} / \mathrm{s}\) at \(t=7.00 \mathrm{s}\) . We have taken counterclockwise rotation to be positive. (a) Is the angular acceleration during this time interval positive or negative? (b) During what time interval is the speed of the wheel increasing? Decreasing? (c) What is the angular displacement of the wheel at \(t=7.00\) s?

Short Answer

Expert verified
(a) Positive angular acceleration. (b) Speed increases [0,3] s; decreases [3,7] s. (c) Angular displacement is 7.00 rad.

Step by step solution

01

Determine Angular Acceleration

The angular acceleration \( \alpha \) can be found using the formula for angular velocity change: \( \alpha = \frac{\Delta \omega}{\Delta t} \). Here, the final angular velocity \( \omega_f = 8.00 \) rad/s, the initial angular velocity \( \omega_i = -6.00 \) rad/s, and the time interval \( \Delta t = 7.00 \) s. Substitute these values into the formula to find \( \alpha = \frac{8.00 - (-6.00)}{7.00} = \frac{14.00}{7.00} = 2.00 \) rad/s虏. Since \( \alpha \) is positive, the angular acceleration is positive.
02

Determine Speed Increase and Decrease Interval

The wheel speeds up when the angular velocity and angular acceleration have the same sign. Here, \( \alpha = 2.00 \) rad/s虏 is positive, so the wheel speeds up when \( \omega \) is negative (becoming less negative) until \( \omega = 0 \), then continues in the positive direction. To find when \( \omega = 0 \), use \( \omega_f = \omega_i + \alpha t \), where \( 0 = -6.00 + 2.00t \). Solving gives \( t = 3.00 \) s. Thus, speed increases: \([0, 3] \) s and decreases when \( \omega \) and \( \alpha \) are opposite: \([3, 7] \) s (when \( \omega \) is positive).
03

Calculate Angular Displacement at t=7.00 s

Angular displacement \( \theta \) can be calculated using \( \theta = \omega_i t + \frac{1}{2} \alpha t^2 \). With \( \omega_i = -6.00 \) rad/s, \( \alpha = 2.00 \) rad/s虏, and \( t = 7.00 \) s: \( \theta = (-6.00)(7.00) + \frac{1}{2}(2.00)(7.00)^2 \). Simplify: \( \theta = -42.00 + 49.00 = 7.00 \) rad.

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Key Concepts

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

Angular Acceleration
Angular acceleration refers to how quickly the angular velocity of an object changes over time. To visualize this, think about a wheel that starts to spin faster as time goes on. In our example, we dealt with a wheel whose angular velocity changed from -6.00 rad/s to 8.00 rad/s within 7 seconds. The formula to determine angular acceleration (\(\alpha\)) is:
  • \(\alpha = \frac{\Delta \omega}{\Delta t}\)
where \(\Delta \omega\) represents the change in angular velocity, and \(\Delta t\) is the change in time.
By plugging in the values from the problem, \(\alpha\) comes out to be 2.00 rad/s虏, a positive number. This positive sign indicates that the wheel's angular velocity is increasing in the counterclockwise direction, just like positive linear acceleration means a vehicle is speeding up as it moves forward.
Angular Velocity
Angular velocity describes how fast an object rotates or revolves relative to another point, in this case, around the wheel's axis. Imagine tracking a point on the spinning wheel. How fast this point goes around is dictated by the angular velocity \(\omega\).
In the stated problem, the wheel's initial angular velocity \(\omega_i\) was -6.00 rad/s, meaning it began rotating in the clockwise direction. Over time, because of the positive angular acceleration, it sped up, transitioning to an angular velocity of 8.00 rad/s counterclockwise at \(t = 7.00\) seconds. Here's an interesting fact: Angular velocity is analogous to linear velocity but in a rotational sense and uses radian (rad) as its unit of measurement instead of meters per second (m/s).
It's crucial to understand that the sign of \(\omega\) provides direction. In this problem, negative velocity indicates clockwise rotation, while positive velocity points to counterclockwise motion.
Angular Displacement
Angular displacement measures the angle through which an object has rotated or moved. It's similar to how we think of distance when discussing linear motion. If a wheel turns, the angular displacement tells us by how many radians it's twisted from its initial position.
Using the formula:
  • \(\theta = \omega_i t + \frac{1}{2} \alpha t^2\)
we can find out how far the wheel has spun in terms of angular displacement. Plugging in the initial velocity \(\omega_i = -6.00\) rad/s, angular acceleration \(\alpha = 2.00\) rad/s虏, and time \(t = 7.00\) seconds, we compute:
  • \(\theta = (-6.00 \times 7.00) + \frac{1}{2}(2.00)(7.00)^2\)
  • \(\theta = -42.00 + 49.00 = 7.00\) rad
So, by the time 7 seconds have passed, the wheel executes an angular displacement of 7.00 radians, describing a net motion in the counterclockwise direction. This formula demonstrates how angular displacement connects to both angular velocity and acceleration, paving a complete picture of rotational motion.

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Most popular questions from this chapter

A wheel of diameter 40.0 \(\mathrm{cm}\) starts from rest and rotates with a constant angular acceleration of 3.00 \(\mathrm{rad} / \mathrm{s}^{2} .\) At the instant the wheel has computed its second revolution, compute the radial acceleration of a point on the rim in two ways: (a) using the relationship \(a_{\text { rad }}=\omega^{2} r\) and \((b)\) from the relationship \(a_{\text { rad }}=v^{2} / r\) .

A uniform, solid disk with mass \(m\) and radius \(R\) is pivoted about a horizontal axis through its center. A small object of the same mass \(m\) is glued to the rim of the disk. If the disk is released from rest with the small object at the end of a horizontal radius, find the angular speed when the small object is directly below the axis.

At \(t=0\) the current to a dc electric motor is reversed, resulting in an angular displacement of the motor shaft given by \(\theta(t)=(250 \mathrm{rad} / \mathrm{s}) t-\left(20.0 \mathrm{rad} / \mathrm{s}^{2}\right) t^{2}-\left(1.50 \mathrm{rad} / \mathrm{s}^{3}\right) t^{3} .\) (a) \(\mathrm{At}\) what time is the angular velocity of the motor shaft zero? (b) Calculate the angular acceleration at the instant that the motor shaft has zero angular velocity. (c) How many revolutions does the motor shaft turn through between the time when the current is reversed and the instant when the angular velocity is zero? (d) How fast was the motor shaft rotating at \(t=0,\) when the current was reversed? (e) Calculate the average angular velocity for the time period from \(t=0\) to the time calculated in part (a).

A thin, light wire is wrapped around the rim of a wheel, as shown in Fig. E9.49. The wheel rotates about a stationary horizontal axle that passes through the center of the wheel. The wheel has radius 0.180 \(\mathrm{m}\) and moment of inertia for rotation about the axle of \(I=0.480 \mathrm{kg} \cdot \mathrm{m}^{2}\) . A small block with mass 0.340 \(\mathrm{kg}\) is suspended from the free end of the wire. When the system is released from rest, the block descends with constant acceleration. The bearings in the wheel at the axle are rusty, so friction there does \(-6.00 \mathrm{J}\) of work as the block descends 3.00 \(\mathrm{m}\) . What is the magnitude of the angular velocity of the wheel after the block has descended 3.00 \(\mathrm{m} ?\)

While redesigning a rocket engine, you want to reduce its weight by replacing a solid spherical part with a hollow spherical shell of the same size. The parts rotate about an axis through their center. You need to make sure that the new part always has the same rotational kinetic energy as the original part had at any given rate of rotation. If the original part had mass \(M,\) what must be the mass of the new part?
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