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A turntable rotates with a constant 2.25 rad/s\(^2\) angular acceleration. After 4.00 s it has rotated through an angle of 30.0 rad. What was the angular velocity of the wheel at the beginning of the 4.00-s interval?

Short Answer

Expert verified
The initial angular velocity is 3.0 rad/s.

Step by step solution

01

Identify the Known Variables

We are given:- Angular acceleration, \( \alpha = 2.25 \, \text{rad/s}^2 \)- Time interval, \( t = 4.00 \, \text{s} \)- Angular displacement, \( \theta = 30.0 \, \text{rad} \)We need to find the initial angular velocity \( \omega_0 \) of the wheel.
02

Use the Angular Kinematics Equation

The angular kinematics equation that relates initial angular velocity (\( \omega_0 \)), angular acceleration (\( \alpha \)), time (\( t \)), and angular displacement (\( \theta \)) is:\[\theta = \omega_0 t + \frac{1}{2} \alpha t^2\]We'll use this formula to solve for \( \omega_0 \).
03

Plug in Known Values and Solve for \(\omega_0\)

Substitute the known values into the equation: \[30.0 = \omega_0 (4.00) + \frac{1}{2} (2.25) (4.00)^2\]Calculate \( \frac{1}{2} \times 2.25 \times 16.0 \):\[= \frac{1}{2} \times 36.0\]\[= 18.0 \text{ rad}\]Thus, the equation becomes:\[30.0 = 4.00\omega_0 + 18.0\]
04

Isolate \(\omega_0\) and Solve

Subtract 18.0 from both sides:\[30.0 - 18.0 = 4.00\omega_0\]\[12.0 = 4.00\omega_0\]Divide both sides by 4.00:\[\omega_0 = \frac{12.0}{4.00} = 3.0 \, \text{rad/s}\]
05

Conclusion

The initial angular velocity of the wheel at the beginning of the 4-second interval is \( 3.0 \, \text{rad/s} \). Ensure the result makes sense in the context of the given data.

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Key Concepts

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

Angular Acceleration
Angular acceleration is a concept that describes the rate of change of angular velocity over time. Imagine it as how quickly something is spinning up or slowing down as it rotates. In this exercise, the turntable has a constant angular acceleration of \(2.25 \ \text{rad/s}^2\). That means with each passing second, the velocity increases by \(2.25 \ \text{rad/s}\). Angular acceleration can be positive, causing the object to spin faster, or negative, causing it to slow down.

Key points about angular acceleration include:
  • It is typically denoted by the symbol \(\alpha\).
  • Its units are radians per second squared \((\text{rad/s}^2)\).
  • It is a vector, meaning it has both magnitude and direction.
In rotational motion, just as in linear motion with linear acceleration, angular acceleration allows us to predict how the velocity and position of a rotating object will change over time.
Angular Displacement
Angular displacement refers to the change in angle as an object moves along a circular path. In the context of this problem, the turntable rotates through an angle of \(30.0 \ \text{rad}\) over a period of 4 seconds, which is its angular displacement during the time interval. This concept is similar to linear displacement in straight-line motion but for rotational movement.

Important aspects of angular displacement are:
  • It is represented by the symbol \(\theta\).
  • It is measured in radians, a unit derived from the radius and circumference of a circle.
  • Unlike angular velocity or acceleration, which are rates of change, angular displacement is a measure of distance in angular terms.
Understanding angular displacement helps predict how far and in what direction an object has rotated within a certain timeframe.
Initial Angular Velocity
Initial angular velocity is the angular speed at which a rotating object starts its motion. In the exercise, it represents the speed of the turntable at the very beginning of the 4-second period. It's what we're trying to find out from the given values using the angular kinematics equation.

Some essentials about initial angular velocity are:
  • It is denoted by \(\omega_0\).
  • Measured in radians per second \((\text{rad/s})\).
  • It can be determined using the equation \(\theta = \omega_0 t + \frac{1}{2} \alpha t^2\), given angular displacement, time, and angular acceleration.
In solving the exercise, by substituting into the kinematics equation, the initial angular velocity of the wheel is found to be \(3.0 \ \text{rad/s}\), confirming the start rotation speed before any acceleration occurred. Understanding initial angular velocity is crucial as it sets the stage for how an object begins its rotational journey.

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Most popular questions from this chapter

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.

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?

The angular velocity of a flywheel obeys the equation \(\omega_z\)(\(t\)) \(= A + Bt^2\), where \(t\) is in seconds and \(A\) and \(B\) are constants having numerical values 2.75 (for \(A\)) and 1.50 (for \(B\)). (a) What are the units of \(A\) and \(B\) if \(\omega_z\) is in rad/s? (b) What is the angular acceleration of the wheel at (i) \(t = 0\) and (ii) \(t =\) 5.00 s? (c) Through what angle does the flywheel turn during the first 2.00 s? (\(Hint\): See Section 2.6.)

On a compact disc (CD), music is coded in a pattern of tiny pits arranged in a track that spirals outward toward the rim of the disc. As the disc spins inside a CD player, the track is scanned at a constant \(linear\) speed of \(v =\) 1.25m/s. Because the radius of the track varies as it spirals outward, the angular speed of the disc must change as the CD is played. (See Exercise 9.20.) Let's see what angular acceleration is required to keep \(v\) constant. The equation of a spiral is \(r(\theta) = r_0 + \beta\theta\), where \(r_0\) is the radius of the spiral at \(\theta =\) 0 and \(\beta\) is a constant. On a CD, \(r_0\) is the inner radius of the spiral track. If we take the rotation direction of the CD to be positive, \(\beta\) must be positive so that \(r\) increases as the disc turns and \(\theta\) increases. (a) When the disc rotates through a small angle \(d\theta\), the distance scanned along the track is \(ds = rd\theta\). Using the above expression for \(r(\theta)\), integrate \(ds\) to find the total distance \(s\) scanned along the track as a function of the total angle \(\theta\) through which the disc has rotated. (b) Since the track is scanned at a constant linear speed \(v\), the distance s found in part (a) is equal to \(vt\). Use this to find \(\theta\) as a function of time. There will be two solutions for \(\theta\) ; choose the positive one, and explain why this is the solution to choose. (c) Use your expression for \(\theta(t)\) to find the angular velocity \(\omega_z\) and the angular acceleration \(\alpha_z\) as functions of time. Is \(\alpha_z\) constant? (d) On a CD, the inner radius of the track is 25.0 mm, the track radius increases by 1.55 mm per revolution, and the playing time is 74.0 min. Find \(r_0, \beta,\) and the total number of revolutions made during the playing time. (e) Using your results from parts (c) and (d), make graphs of \(\omega_z\) (in rad/s) versus \(t\) and \(\alpha_z\) (in rad/s\(^2\)) versus \(t\) between \(t =\) 0 and \(t =\) 74.0 min. \(\textbf{The Spinning eel.}\) American eels (\(Anguilla\) \(rostrata\)) are freshwater fish with long, slender bodies that we can treat as uniform cylinders 1.0 m long and 10 cm in diameter. An eel compensates for its small jaw and teeth by holding onto prey with its mouth and then rapidly spinning its body around its long axis to tear off a piece of flesh. Eels have been recorded to spin at up to 14 revolutions per second when feeding in this way. Although this feeding method is costly in terms of energy, it allows the eel to feed on larger prey than it otherwise could.

A thin, rectangular sheet of metal has mass \(M\) and sides of length \(a\) and \(b\). Use the parallel-axis theorem to calculate the moment of inertia of the sheet for an axis that is perpendicular to the plane of the sheet and that passes through one corner of the sheet.

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