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Chapter 9: Problem 3

The angular velocity of a fywheel obeys the equation \(\omega_{z}(t)=A+B t^{2},\) where \(t\) is in seconds and \(A\) and \(B\) are constants having numerical values 2.75\((\text { for } A)\) and 1.50\((\text { for } B)\) . (a) What are the units of \(A\) and \(B\) if \(\omega\) is in rad/s? \((b)\) What is the angular acceleration of the wheel at (i) \(t=0.00\) and (ii) \(t=5.00 \mathrm{s} ?\) (c) Through what angle does the flywheel turn during the first 2.00 \(\mathrm{s} ?(\text { Hint: See Section } 2.6 .)\)

Short Answer

Expert verified
(a) Units of \(A\) are rad/s, \(B\) are rad/s鲁. (b) \(0\text{ rad/s}^2\) at \(t=0.00\), \(15.0\text{ rad/s}^2\) at \(t=5.00\). (c) \(11.5\text{ rad}\) over first 2 seconds.

Step by step solution

01

Determine Units for A and B

The angular velocity \( \omega_z(t) \) is given in radians per second (rad/s). For \( \omega_z(t) = A + Bt^2 \), both \( A \) and \( Bt^2 \) must have units of rad/s. \( A \) itself must be in rad/s. \( Bt^2 \) implies that \( B \) must have units such that \( \text{units of } B \times \text{seconds}^2 = \text{rad/s} \). Therefore, \( B \) must have units of rad/s鲁.
02

Calculate Angular Acceleration Function

Angular acceleration \( \alpha(t) \) is the derivative of angular velocity \( \omega(t) \) with respect to time \( t \). Based on the equation, \( \omega(t) = A + Bt^2 \), we find \( \alpha(t) = \frac{d\omega}{dt} = \frac{d}{dt}(A + Bt^2) = 2Bt \).
03

Find Angular Acceleration at t=0.00s

Using the angular acceleration function \( \alpha(t) = 2Bt \), substitute \( t = 0.00 \). This yields \( \alpha(0) = 2B(0) = 0 \text{ rad/s}^2 \).
04

Find Angular Acceleration at t=5.00s

Substitute \( t = 5.00 \) into the angular acceleration function: \( \alpha(5) = 2B \times 5 = 2(1.50) \times 5 = 15.0 \text{ rad/s}^2 \).
05

Determine Rotation Angle over First 2 Seconds

The angle \( \theta(t) \) is the integral of angular velocity \( \omega(t) \): \( \theta(t) = \int_0^t (A + Bt^2) \, dt \). Evaluating from \( 0 \) to \( 2.00 \) sec gives \( \theta = \int_0^2 (2.75 + 1.50t^2) \, dt = [2.75t + 0.5(1.50)t^3]_0^2 = [2.75(2) + 0.75(2)^3] - [2.75(0) + 0.75(0)^3] = 5.5 + 6 = 11.5 \text{ rad} \).

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Key Concepts

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

Angular Velocity
Angular velocity describes how fast something is rotating. It tells us the rate at which an object, such as a flywheel, spins around its axis. In the equation given in the exercise, the angular velocity \( \omega_z(t) = A + Bt^2 \) combines a constant part \( A \) and a time-dependent part \( Bt^2 \).
- The constant \( A \) represents an initial angular velocity in radians per second (rad/s). This is the angular speed at time \( t = 0 \).- The term \( Bt^2 \) adds a time-dependent component, meaning the velocity can change with time \( t \). Here, \( B \) must have units of rad/s鲁 to ensure \( Bt^2 \) maintains the unit of rad/s.
Understanding angular velocity is vital because it informs us not just about the speed, but also the direction of rotation, as direction is integral in rotational motion.
Angular Acceleration
Angular acceleration measures how quickly the angular velocity of an object changes with time. It is the rate of change of angular velocity and is symbolized by \( \alpha(t) \). In the exercise, the angular acceleration formula is derived from the angular velocity equation by taking its derivative with respect to time: \( \alpha(t) = \frac{d}{dt}(A + Bt^2) = 2Bt \).
- At \( t = 0.00 \) seconds, the angular acceleration is \( \alpha(0) = 0 \) rad/s虏, meaning there's no change in rotation at the start.- At \( t = 5.00 \) seconds, substituting in the equation gives \( \alpha(5) = 15.0 \) rad/s虏, indicating the rotation is speeding up significantly at this time point.
Angular acceleration is crucial for understanding how forces or conditions (like motor power) change the spinning motion over time.
Rotation Angle
Rotation angle \( \theta \) indicates how far an object has rotated over a given time period. It's essentially the angular equivalent of linear distance and is measured in radians. To find the rotation angle over a specific timeframe, you need to integrate the angular velocity over time. For the exercise, integrating \( \omega(t) = A + Bt^2 \) between \( t = 0 \) and \( t = 2 \) seconds gives:\[ \theta(t) = \int_0^t (A + Bt^2) \, dt = [2.75t + 0.75t^3]_0^2, \] which evaluates to \( 11.5 \) radians.
- This calculation yields the total angle through which the flywheel turns.- It shows how integral calculus applies to real-world scenarios by summing up infinitesimal rotations.
Understanding rotation angle helps in scenarios like gear rotation or tracking how fast a mechanism moves from one point to another in circular paths.

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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 \(a_{\mathrm{rad}}=\omega^{2} r\) and \((\mathrm{b})\) from the relationship \(a_{\mathrm{red}}=v^{2} / r\)

A passenger bus in Zurich, Switzerland, derived its motive power from the energy stored in a large flywheel. The wheel was brought up to speed periodically, when the bus stopped at a station, by an electric motor, which could then be attached to the electric power lines. The flywheel was a solid cylinder with mass 1000 \(\mathrm{kg}\) and diameter \(1.80 \mathrm{m} ;\) its top angular speed was 3000 \(\mathrm{rev} / \mathrm{min}\) . (a) At this angular speed, what is the kinetic energy of the flywheel? (b) If the average power required to operate the bus is \(1.86 \times 10^{4} \mathrm{W},\) how long could it operate between stops?

At \(t=0\) the current to a de electric motor is reversed, resulting in an angular displacement of the motor shaft given by \(\theta(t)=(250 \mathrm{rad} / \mathrm{s}) t-\left(20.0 \mathrm{rad} / \mathrm{s}^{2}\right) t^{2}-\left(1.50 \mathrm{rad} / \mathrm{s}^{3}\right) t^{3} .(\mathrm{a}) \mathrm{At}\) 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? ( c) Calculate the average angular velocity for the time period from \(t=0\) to the time calculated in part (a).

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?

A safety device brings the blade of a power mower from an initial angular speed of \(\omega_{1}\) to rest in 1.00 revolution. At the same constant acceleration, how many revolutions would it take the blade to come to rest from an initial angular speed \(\omega_{3}\) that was three times as great, \(\omega_{3}=3 \omega_{1} ?\)
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