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Chapter 9: Problem 28

At \(t=3.00\) s a point on the rim of a 0.200 -m-radius wheel has a tangential speed of 50.0 \(\mathrm{m} / \mathrm{s}\) as the wheel slows down with a tangential acceleration of constant magnitude 10.0 \(\mathrm{m} / \mathrm{s}^{2}\) . (a) Calculate the wheel's constant angular acceleration. (b) Calculate the angular velocities at \(t=3.00\) s and \(t=0 .\) (c) Through what angle did the wheel turn between \(t=0\) and \(t=3.00\) s? (d) At what time will the radial acceleration equal \(g ?\)

Short Answer

Expert verified
(a) 50.0 rad/s虏, (b) 蠅(3s)=250 rad/s, 蠅(0s)=100 rad/s, (c) 胃=525 rad, (d) never reaches g.

Step by step solution

01

Understand the Given Data

We are given that at \( t = 3.00 \) s, a point on the rim of a wheel (radius \( R = 0.200 \) m) has a tangential speed \( v = 50.0 \ \mathrm{m/s} \) and a constant tangential acceleration \( a_t = 10.0 \ \mathrm{m/s}^2 \). We need to calculate the angular acceleration, angular velocities at \( t = 3.00 \) s and \( t = 0 \) s, the angle turned between \( t = 0 \) and \( t = 3.00 \) s, and the time when radial acceleration equals gravity \( g = 9.81 \ \mathrm{m/s}^2 \).
02

Calculate Angular Acceleration (a)

The angular acceleration \( \alpha \) is related to tangential acceleration \( a_t \) by the formula: \( a_t = R \alpha \). Solving for \( \alpha \), we have:\[ \alpha = \frac{a_t}{R} = \frac{10.0 \ \mathrm{m/s}^2}{0.200 \ \mathrm{m}} = 50.0 \ \mathrm{rad/s}^2 \]
03

Calculate Angular Velocities (b)

The angular velocity \( \omega \) is related to tangential speed \( v \) by the equation: \( v = R \omega \). At \( t = 3.00 \) s, \(\omega_1 = \frac{v}{R} = \frac{50.0 \ \mathrm{m/s}}{0.200 \ \mathrm{m}} = 250 \ \mathrm{rad/s} \).We can find \( \omega_0 \) at \( t=0 \) s using the kinematic equation \( \omega_1 = \omega_0 + \alpha t \). Solving for \( \omega_0 \):\[ \omega_0 = \omega_1 - \alpha t = 250 \ \mathrm{rad/s} - (50.0 \ \mathrm{rad/s}^2)(3.00 \ \mathrm{s}) = 100 \ \mathrm{rad/s} \]
04

Calculate Angle Turned (c)

The angle \( \theta \) turned can be calculated using the equation: \( \theta = \omega_0 t + \frac{1}{2}\alpha t^2 \).Substituting the known values, we have:\[\theta = (100 \ \mathrm{rad/s})(3.00 \ \mathrm{s}) + \frac{1}{2}(50.0 \ \mathrm{rad/s}^2)(3.00 \ \mathrm{s})^2 \= 300 \ \mathrm{rad} + \frac{1}{2}(50.0)(9) \= 300 + 225 \= 525 \ \mathrm{rad}\]
05

Calculate Time When Radial Acceleration Equals g (d)

Radial acceleration \( a_r \) is given by \( a_r = R\omega^2 \). Set \( a_r = g \), which gives:\[ g = R(\omega_0 + \alpha t)^2 \]\[ 9.81 = 0.200(100 + 50t)^2 \]Solving the quadratic equation, \( (100 + 50t)^2 = \frac{9.81}{0.200} \), we find:\[ (100 + 50t)^2 = 49.05 \100 + 50t = \sqrt{49.05} \50t = \sqrt{49.05} - 100 \t = \frac{\sqrt{49.05} - 100}{50} \]Finally, solving with \( \sqrt{49.05} \approx 7.004 \), we get:\[ t \approx \frac{7.004 - 100}{50} \]This gives \( t \approx \) negative value, which doesn't make sense, so it means the radial acceleration never equals \( g \) in this case. The tangential component dominates.

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Key Concepts

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

Tangential Acceleration
Tangential acceleration is a measure of how quickly a wheel's tangential velocity changes over time. It acts along the edge of the wheel, in the direction of its motion.
The unit for tangential acceleration is meters per second squared (\(\mathrm{m/s^2}\)).
  • The formula to calculate tangential acceleration (\(a_t\)) is straightforward: \(a_t = R \alpha\), where \(R\) is the radius of the wheel and \(\alpha\) is the angular acceleration.
  • In our example, we are given a constant tangential acceleration of 10.0 \(\mathrm{m/s^2}\). This helps to determine how quickly the wheel's tangential speed is decreasing, as the wheel is slowing down.
  • This constant tangential acceleration is crucial to find the angular acceleration of the wheel.
Angular Velocity
Angular velocity describes how fast an object rotates or revolves relative to another point, usually the center of a circle. It is measured in radians per second (\(\mathrm{rad/s}\)).
  • To find angular velocity, we use the relationship: \(v = R \omega\), where \(v\) is tangential speed and \(\omega\) is angular velocity.
  • At \(t = 3.00\) s, the angular velocity \(\omega_1\) for a wheel with a tangential speed of 50.0 \(\mathrm{m/s}\) and a radius of 0.200 m can be calculated as \(\omega_1 = \frac{v}{R} = \frac{50.0}{0.200} = 250\) \(\mathrm{rad/s}\).
  • To find the initial angular velocity \(\omega_0\), we use \(\omega_1 = \omega_0 + \alpha t\), solving for \(\omega_0\) gives 100 \(\mathrm{rad/s}\) for this exercise.
Radial Acceleration
Radial acceleration, also known as centripetal acceleration, points towards the center of the circle in which the object is moving. It ensures that the object follows a circular path instead of moving off tangentially.
The unit for radial acceleration is meters per second squared (\(\mathrm{m/s^2}\)).
  • The formula to calculate radial acceleration \(a_r\) is \(a_r = R \omega^2\), where \(R\) is the radius and \(\omega\) is the angular velocity.
  • Setting the radial acceleration equal to gravitational acceleration \(g = 9.81 \mathrm{m/s^2}\) allows us to determine when the forces acting on the object equal those of gravity.
  • In this exercise, it turns out the radial acceleration never reaches \(g\) because the tangential deceleration causes the wheel to slow substantially before that happens.
Angular Displacement
Angular displacement measures how much an object has rotated or moved along the circle. It is measured in radians and gives an idea of the total angle through which the object has turned.
  • The equation often used is \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2\), which takes into account both the initial angular velocity \(\omega_0\) and the angular acceleration \(\alpha\).
  • In the given exercise, for \(t = 3.00\) s, \(\theta\) is calculated as \(525 \ \mathrm{rad}\), indicating in radians the amount the wheel turned during that time interval.
  • Understanding angular displacement helps visualize the rotation path and progress of the object over time.

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Most popular questions from this chapter

A uniform disk of radius \(R\) is cut in half so that the remaining half has mass \(M\) (Fig. E9.34a).(a) What is the moment of inertia of this half about an axis perpendicular to its plane through point \(A\) ? (b) Why did your answer in part (a) come out the same as if this were a complete disk of mass \(M ?\) (c) What would be the moment of inertia of a quarter disk of mass \(M\) and radius \(R\) about an axis perpendicular to its plane passing through point \(B(\) Fig. \(E 9.34 b) ?\)

A fan blade rotates with angular velocity given by \(\omega_{z}(t)=\gamma-\beta t^{2},\) where \(\gamma=5.00 \mathrm{rad} / \mathrm{s}\) and \(\beta=0.800 \mathrm{rad} / \mathrm{s}^{3} .\) \begin{equation}\begin{array}{l}{\text { (a) Calculate the angular acceleration as a function of time. }} \\ {\text { (b) Calculate the instantaneous angular acceleration } \alpha_{z} \text { at } t=3.00 \text { s }}\end{array}\end{equation} \begin{equation} \begin{array}{l}{\text { and the average angular acceleration } \alpha_{\mathrm{av}-z} \text { for the time interval }} \\ {t=0 \text { to } t=3.00 \text { s. How do these two quantities compare? If they }} \\ {\text { are different, why are they different? }}\end{array} \end{equation}

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 0.300 \(\mathrm{rad} / \mathrm{s}^{2}\) , what is its angular velocity at \(t=2.50 \mathrm{s} ?\) (b) Through what angle has the wheel turned between \(t=0\) and \(t=2.50 \mathrm{s} ?\)

The flywheel of a gasoline engine is required to give up 500 \(\mathrm{J}\) of kinetic energy while its angular velocity decreases from 650 \(\mathrm{rev} / \mathrm{min}\) to 520 \(\mathrm{rev} / \mathrm{min}\) . What moment of inertia is required?

An electric fan is turned off, and its angular velocity decreases uniformly from 500 rev/min to 200 rev/min in 4.00 s. (a) Find the angular acceleration in \(\mathrm{rev} / \mathrm{s}^{2}\) and the number of revolutions made by the motor in the \(4.00-\) interval. (b) How many more seconds are required for the fan to come to rest if the angular acceleration remains constant at the value calculated in part (a)?
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