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A bicycle wheel has an initial angular velocity of \(1.50 \mathrm{rad} / \mathrm{s}\) (a) If its angular acceleration is constant and equal to \(0.200 \mathrm{rad} / \mathrm{s}^{2},\) what is its angular velocity at \(t=2.50 \mathrm{~s} ?\) (b) Through what angle has the wheel turned between \(t=0\) and \(t=2.50 \mathrm{~s} ?\)

Short Answer

Expert verified
The final angular velocity at t = 2.50 s is 2.00 rad/s and the wheel has turned an angle of 4.375 rad between t = 0 and t = 2.50 s

Step by step solution

01

Calculation of Final Angular Velocity

In this step, the final angular velocity is calculated by plugging the values into the equation \(\omega_f = \omega_i + \alpha t\). Given that \(\omega_i\) = 1.50 rad/s, \(\alpha\) = 0.200 rad/s\(^2\), and \(t\) = 2.50 s, the angular velocity at \(t\) = 2.50 s is therefore \(\omega_f\) = 1.50 rad/s + 0.200 rad/s\(^2\) * 2.50 s = 2.00 rad/s
02

Calculation of Angular Displacement

In this step, the total angular displacement is calculated by plugging the values into the equation \(\theta = \omega_i t + 0.5 * \alpha * t^2\). Given that \(\omega_i\) = 1.50 rad/s, \(\alpha\) = 0.200 rad/s\(^2\), and \(t\) = 2.50 s, the angular displacement is therefore \(\theta\) = 1.50 rad/s * 2.50 s + 0.5 * 0.200 rad/s\(^2\) * (2.50 s)\(^2\) = 3.75 rad + 0.625 rad = 4.375 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 is the rate at which the angular velocity of an object changes with time. It is a vector quantity, meaning it has both magnitude and direction. In our bicycle wheel example, the angular acceleration is constant at a value of \(0.200 \text{ rad/s}^2\). This implies that for every second, the angular velocity of the wheel increases by \(0.200 \text{ rad/s}\).

To calculate the angular velocity after a certain time with constant angular acceleration, we use the equation:

\[ \omega_f = \omega_i + \alpha t \]

Here, \(\omega_f\) is the final angular velocity, \(\omega_i\) is the initial angular velocity, \(\alpha\) is the angular acceleration, and \(t\) is the time over which the acceleration occurs. In practice, understanding and applying this concept allows one to predict future motion given current motion conditions and how they change over time.
Angular Displacement
Angular displacement refers to the change in the angle as an object rotates about a point. It's the angle through which a point or line has been rotated in a specified sense about a specified axis. For our bicycle wheel, we are interested in how much the wheel has turned from the starting position, measured in radians.

To find the total angular displacement, we can use the formula:

\[ \theta = \omega_i t + \frac{1}{2} \alpha t^2 \]

Here, \(\theta\) represents the angular displacement, \(\omega_i\) is the initial angular velocity, \(\alpha\) is the angular acceleration, and \(t\) is the time duration. This formula is derived from the kinematic equations for linear motion, adapted for rotational motion by substituting linear displacement with angular displacement and linear acceleration with angular acceleration.

Understanding angular displacement is beneficial for interpreting how far the wheel has turned during its motion, which is crucial for calculating distances travelled and for navigational purposes in circular paths.
Equations of Rotational Motion
The equations of rotational motion are analogs of the equations used for linear motion. They describe the relationship between angular displacement, angular velocity, angular acceleration, and time. For objects in rotational motion with constant acceleration, these equations provide a powerful tool to analyze various rotational dynamics.

Key equations include:
  • The angular version of velocity-time: \(\omega_f = \omega_i + \alpha t\)
  • The angular displacement-time: \(\theta = \omega_i t + \frac{1}{2} \alpha t^2\)
  • The angular version of the equation that relates initial velocity, final velocity, displacement, and acceleration (without time): \(\omega_f^2 = \omega_i^2 + 2\alpha \theta\)

Using these equations, one can calculate any unknown variable if the others are known, much like with their linear counterparts. Having a firm grasp of these equations is essential for solving problems involving anything from simple wheels to complex rotating machinery.

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Most popular questions from this chapter

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.

A roller in a printing press turns through an angle \(\begin{array}{llll}\theta(t) & \text { given } & \text { by } & \theta(t)=\gamma t^{2}-\beta t^{3}, & \text { where } & \gamma=3.20 \mathrm{rad} / \mathrm{s}^{2} & \text { and }\end{array}\) \(\beta=0.500 \mathrm{rad} / \mathrm{s}^{3} .\) (a) Calculate the angular velocity of the roller as a function of time. (b) Calculate the angular acceleration of the roller as a function of time. (c) What is the maximum positive angular velocity, and at what value of \(t\) does it occur?

A circular saw blade with radius \(0.120 \mathrm{~m}\) starts from rest and turns in a vertical plane with a constant angular acceleration of \(2.00 \mathrm{rev} / \mathrm{s}^{2} .\) After the blade has turned through \(155 \mathrm{rev},\) a small piece of the blade breaks loose from the top of the blade. After the piece breaks loose, it travels with a velocity that is initially horizontal and equal to the tangential velocity of the rim of the blade. The piece travels a vertical distance of \(0.820 \mathrm{~m}\) to the floor. How far does the piece travel horizontally, from where it broke off the blade until it strikes the floor?

A child is pushing a merry-go-round. The angle through which the merry-go- round has turned varies with time according to \(\theta(t)=\gamma t+\beta t^{3}, \quad\) where \(\quad \gamma=0.400 \mathrm{rad} / \mathrm{s} \quad\) and \(\quad \beta=0.0120 \mathrm{rad} / \mathrm{s}^{3}\) (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 \(t=5.00 \mathrm{~s}\) and the average angular velocity \(\omega_{\mathrm{av}-z}\) for the time interval \(t=0\) to \(t=5.00 \mathrm{~s}\) Show that \(\omega_{\mathrm{av}-z}\) is \(n o t\) equal to the average of the instantaneous angular velocities at \(t=0\) and \(t=5.00 \mathrm{~s},\) and explain.

A high-speed flywheel in a motor is spinning at 500 rpm when a power failure suddenly occurs. The flywheel has mass \(40.0 \mathrm{~kg}\) and diameter \(75.0 \mathrm{~cm}\). The power is off for \(30.0 \mathrm{~s}\), and during this time the flywheel slows due to friction in its axle bearings. During the time the power is off, the flywheel makes 200 complete revolutions. (a) At what rate is the flywheel spinning when the power comes back on? (b) How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on, and how many revolutions would the wheel have made during this time?

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