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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.

Short Answer

Expert verified
(a) \( \omega(t) = 0.400 + 3 \times 0.0120 \times t^2 \). (b) 0.400 rad/s. (c) Instantaneous: 1.300 rad/s, Average: 0.7 rad/s. These differ due to non-linear acceleration.

Step by step solution

01

Expression for Angular Velocity

The angular velocity \( \omega(t) \) is the derivative of the angle \( \theta(t) \) with respect to time. Thus, we have: \[ \omega(t) = \frac{d\theta}{dt} = \frac{d}{dt} (\gamma t + \beta t^3) \]Differentiating gives:\[ \omega(t) = \gamma + 3 \beta t^2 \]
02

Initial Angular Velocity

To find the initial angular velocity, substitute \( t = 0 \) into the expression for \( \omega(t) \):\[ \omega(0) = \gamma + 3 \beta (0)^2 = \gamma = 0.400 \, \text{rad/s} \]
03

Instantaneous Angular Velocity at t = 5 s

Substitute \( t = 5.00 \,s \) into the expression for \( \omega(t) \) to find the instantaneous angular velocity:\[ \omega(5) = 0.400 + 3(0.0120)(5)^2 \]Calculate:\[ \omega(5) = 0.400 + 3 \times 0.0120 \times 25 = 0.400 + 0.900 = 1.300 \, \text{rad/s} \]
04

Average Angular Velocity Over Interval

The average angular velocity \( \omega_{av-z} \) over the interval from \( t = 0 \) to \( t = 5 \) is given by:\[ \omega_{av-z} = \frac{\theta(5) - \theta(0)}{5 - 0} \]Calculate \( \theta(5) \):\[ \theta(5) = 0.400 \times 5 + 0.0120 \times (5)^3 = 2.0 + 1.5 = 3.5 \, \text{radians} \] Since \( \theta(0) = 0 \),\[ \omega_{av-z} = \frac{3.5 - 0}{5} = 0.7 \, \text{rad/s} \]
05

Compare Averages

The average of the instantaneous angular velocities at \( t = 0 \) and \( t = 5 \) is:\[ \frac{\omega(0) + \omega(5)}{2} = \frac{0.400 + 1.300}{2} = 0.850 \, \text{rad/s} \]This differs from \( \omega_{av-z} = 0.7 \, \text{rad/s} \) because the average angular velocity is affected by the acceleration term \( \beta t^3 \), which changes the rate of rotation non-linearly over time.

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Key Concepts

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

Understanding Rotational Kinematics
Rotational kinematics is the branch of physics that deals with the motion of objects that rotate or spin. Objects like merry-go-rounds rotate around an axis, and rotational kinematics helps describe this motion.
  • Key terms include angular position, angular velocity, and angular acceleration.
  • The equations and concepts are similar to those used in linear kinematics but are adapted for rotation.
By knowing the angular position of an object over time, we can calculate its angular velocity and acceleration, giving us a complete picture of its rotational motion.
Exploring Angular Displacement
Angular displacement refers to the change in the angle through which an object has rotated, measured in radians. It helps us understand how much rotation has occurred.For a merry-go-round, if the initial angle is zero, the angular displacement can be calculated using the function \[ \theta(t) = \gamma t + \beta t^3 \]where \( \gamma \) and \( \beta \) are coefficients that represent the constant and timedependent components of the motion, respectively.
  • This equation integrates time to determine how the angle changes as the object turns.
  • Understanding angular displacement helps in predicting future positions and in calculating other dynamic factors like velocity and acceleration.
Instantaneous Velocity Clarified
Instantaneous velocity is the speed of an object at a specific moment in time. For rotational motion, this is called angular velocity and is denoted as \( \omega(t) \).The angular velocity indicates how quickly the object is rotating at any given point, and can be derived by differentiating the expression for angular displacement:\[ \omega(t) = \frac{d\theta}{dt} = \gamma + 3\beta t^2 \]
  • This formula reveals that angular velocity changes with time due to the term \( 3\beta t^2 \).
  • At \( t = 5\) seconds, for instance, this function gives us the precise rotational speed at that moment, helping gauge performance or synchronization in mechanical systems.
Calculating Average Velocity
Average angular velocity over a period gives a broad picture of how an object has been rotating, smoothing out its speed variations into a single value.This is calculated by the change in angular displacement over time:\[ \omega_{av-z} = \frac{\theta(5) - \theta(0)}{5} \]
  • This formula computes the average rate of rotation from start to finish of a given time span.
  • Crucially, this value can differ from the mean of instantaneous velocities at the start and end times due to the nature of non-uniform motion.
Understanding both instantaneous and average angular velocities allows us to analyze rotational motion in different contexts, from simple machinery to complex engineering systems.

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Most popular questions from this chapter

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 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 linearly with time, and is \(+\)4.00 rad/s at \(t =\) 7.00 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?

An airplane propeller is rotating at 1900 rpm (rev/min). (a) Compute the propeller's angular velocity in rad/s. (b) How many seconds does it take for the propeller to turn through 35 \(^\circ\)?

(a) What angle in radians is subtended by an arc 1.50 m long on the circumference of a circle of radius 2.50 m? What is this angle in degrees? (b) An arc 14.0 cm long on the circumference of a circle subtends an angle of 128 \(^\circ\). What is the radius of the circle? (c) The angle between two radii of a circle with radius 1.50 m is 0.700 rad. What length of arc is intercepted on the circumference of the circle by the two radii?

How I Scales. If we multiply all the design dimensions of an object by a scaling factor \(f\), its volume and mass will be multiplied by \(f ^3\). (a) By what factor will its moment of inertia be multiplied? (b) If a \(\frac{1}{48}\)-scale model has a rotational kinetic energy of 2.5 J, what will be the kinetic energy for the full-scale object of the same material rotating at the same angular velocity?

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