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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

An electric turntable \(0.750 \mathrm{~m}\) in diameter is rotating about a fixed axis with an initial angular velocity of \(0.250 \mathrm{rev} / \mathrm{s}\) and a constant angular acceleration of \(0.900 \mathrm{rev} / \mathrm{s}^{2}\). (a) Compute the angular velocity of the turntable after \(0.200 \mathrm{~s}\). (b) Through how many revolutions has the turntable spun in this time interval? (c) What is the tangential speed of a point on the rim of the turntable at \(t=0.200 \mathrm{~s} ?\) (d) What is the magnitude of the resultant acceleration of a point on the rim at \(t=0.200 \mathrm{~s} ?\)

A sphere with radius \(R=0.200 \mathrm{~m}\) has density \(\rho\) that decreases with distance \(r\) from the center of the sphere according to \(\rho=3.00 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}-\left(9.00 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{4}\right) r .\) (a) Calculate the total mass of the sphere. (b) Calculate the moment of inertia of the sphere for an axis along a diameter.

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 \(v=1.25 \mathrm{~m} / \mathrm{s} .\) Because the radius of the track varies as it spirals outward, the angular speed of the disc must change as the \(\mathrm{CD}\) 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 \(\mathrm{CD}, r_{0}\) 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 \(d s=r d \theta .\) Using the above expression for \(r(\theta),\) integrate \(d s\) 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 vi. 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 \(25.0 \mathrm{~mm}\), the track radius increases by \(1.55 \mu \mathrm{m}\) per revolution, and the playing time is \(74.0 \mathrm{~min} .\) 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 \(t=74.0 \mathrm{~min}\)

The rotating blade of a blender turns with constant angular acceleration \(1.50 \mathrm{rad} / \mathrm{s}^{2}\). (a) How much time does it take to reach an angular velocity of \(36.0 \mathrm{rad} / \mathrm{s},\) starting from rest? (b) Through how many revolutions does the blade turn in this time interval?

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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