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Chapter 9: Problem 26

At \(t=3.00 \mathrm{~s}\) a point on the rim of a \(0.200-\mathrm{m}\) -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 \(t=3.00 \mathrm{~s}\) and \(t=0 .\) (c) Through what angle did the wheel turn between \(t=0\) and \(t=3.00 \mathrm{~s} ?\) (d) At what time will the radial acceleration equal \(g ?\)

Short Answer

Expert verified
The wheel's constant angular acceleration is \(50~\mathrm{rad/s}^2\). The angular velocities at \(t=3.00\mathrm{s}\) and \(t=0\) are \(250~\mathrm{rad/s}\) and \(400~\mathrm{rad/s}\) respectively. The wheel turned through an angle of \(1425~\mathrm{rad}\) between \(t=0\) and \(t=3.00\mathrm{s}\). The radial acceleration equals \(g\) at a time of approximately \(7.86\mathrm{s}\).

Step by step solution

01

Calculate the angular acceleration

Angular acceleration (\(\alpha\)) can be obtained by dividing tangential acceleration (\(a_t\)) by radius (r). Therefore, using \(\alpha = \frac{a_t}{r}\), where \(a_t\) = 10.0 m/s\(^2\) and r = 0.200 m: \(\alpha = \frac{10.0~ \mathrm{m/s}^2}{0.200 ~\mathrm{m}} = 50.0 ~\mathrm{rad/s}^2\)
02

Calculate the initial and final angular velocities

Angular velocity (\(\omega\)) can be calculated by dividing tangential velocity (\(v_t\)) by radius (r). The angular velocity at \(t=3.00~s\) (\(\omega_f\)) is obtained as \(\omega_f = \frac{v_t}{r} = \frac{50.0~\mathrm{m/s}}{0.200~\mathrm{m}}= 250.0~\mathrm{rad/s}\)The angular velocity at \(t=0\) (\(\omega_i\)) is found by using the formula \(\omega_i = \omega_f + \alpha t\)Therefore, substituting for \(\omega_f = 250.0~\mathrm{rad/s}\), \(\alpha = 50.0 ~\mathrm{rad/s}^2\), and \(t = 3.00~s\):\(\omega_i = 250.0~\mathrm{rad/s} + 50.0 ~\mathrm{rad/s}^2 * 3.00 ~s = 400~\mathrm{rad/s}\)
03

Calculate the angle turned

The angle turned (\(\theta\)) within the time interval can be calculated using the equation \(\Theta = \omega_i t + 0.5*\alpha*t^2\)Substituting for \(\omega_i = 400 ~\mathrm{rad/s}\), \(\alpha = 50.0 ~\mathrm{rad/s}^2\), and \(t = 3.00 ~s\):\(\Theta = 400 ~\mathrm{rad/s} * 3.00 ~s + 0.5*50.0 ~\mathrm{rad/s}^2*(3.00 ~s)^2 = 1200 \mathrm{rad} + 225\mathrm{rad} = 1425 \mathrm{rad}\)
04

Calculate the time when radial acceleration equals g

Radial acceleration (\(a_r\)) is given by the formula \(a_r = \omega^2 * r\)Setting \(a_r = g\), where \(g = 9.81 ~m/s^2\), we can solve for \(\omega\) and use the equation \(\omega = \omega_i - \alpha t\) to find the time (t):From \(\omega^2 = \frac{g}{r}\), we find \(\omega = \sqrt{\frac{9.81 ~m/s^2}{0.200 m}} ~= 7 ~\mathrm{rad/s}\)Substituting \(\omega = 7 ~\mathrm{rad/s}\), \(\omega_i = 400 ~\mathrm{rad/s}\), and \(\alpha = 50 ~\mathrm{rad/s}^2\) in the equation \(\omega = \omega_i - \alpha t\)We find \(t = \frac{\omega_i - \omega}{\alpha} = \frac{400 - 7}{50} ~= 7.86~s\)

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Key Concepts

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

Tangential Speed
Tangential speed, or the linear speed of a point moving along a circular path, is calculated by multiplying the radius of the circle by the angular velocity. It's important to understand that as a point moves along the circumference of a circle, it covers a linear distance which can be represented by its tangential speed. This concept is crucial in understanding the motion of any point on a rotating object.

For example, to find the tangential speed of a point on a wheel, you can use the formula: \( v_t = \rho \times \frac{\theta}{t} \), where \(v_t\) is the tangential speed, \(\rho\) is the radius of the wheel, \(\theta\) is the angle the wheel has turned, and \(t\) is the time taken. For an object moving with a constant angular velocity, tangential speed remains consistent. However, when there is angular acceleration involved, the tangential speed changes over time.
Tangential Acceleration
Tangential acceleration is the measure of how quickly the tangential speed of a point on a rotating object changes. It is directly proportional to the angular acceleration and the radius of the circle. The formula used to calculate tangential acceleration is \( a_t = \rho \times \text{alpha} \), where \(a_t\) is tangential acceleration, \(\rho\) is the radius, and \(\text{alpha}\) is the angular acceleration.

Understanding tangential acceleration is vital when analyzing situations where rotational speeds are changing, such as in the case of the wheel slowing down in our exercise. Here, knowing that the tangential acceleration is constant allows us to infer information about the consistent rate at which the wheel's rotation is slowing down. When working through problems involving tangential acceleration, it helps to visualize the linear counterpart to the rotational motion to better grasp how changes in angular speed affect tangential speed.
Angular Velocity
Angular velocity refers to the rate of change of the angular position of an object, essentially describing how fast an object is rotating. It is usually measured in radians per second (rad/s). To determine the angular velocity of any given point on a rotating object, you can divide the tangential speed by the radius of the circle using the formula: \(\text{omega} = \frac{v_t}{\rho}\), where \(\text{omega}\) is the angular velocity, \(v_t\) is the tangential speed, and \(\rho\) is the radius.

Our exercise provides an excellent practice on how to calculate angular velocity at different instances when an angular acceleration is present. As the angular velocity changes over time due to angular acceleration, understanding both these concepts is fundamental when studying rotational dynamics. Angular velocity plays a key role in various physical phenomena and engineering applications, making it a cornerstone concept in physics.

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Most popular questions from this chapter

A bucket of mass \(m\) is tied to a massless cable that is wrapped around the outer rim of a uniform pulley of radius \(R,\) on a frictionless axle, similar to the system shown in Fig. E9.47. In terms of the stated variables, what must be the moment of inertia of the pulley so that it always has half as much kinetic energy as the bucket?

A physics student of mass \(43.0 \mathrm{~kg}\) is standing at the edge of the flat roof of a building, \(12.0 \mathrm{~m}\) above the sidewalk. An unfriendly \(\operatorname{dog}\) is running across the roof toward her. Next to her is a large wheel mounted on a horizontal axle at its center. The wheel, used to lift objects from the ground to the roof, has a light crank attached to it and a light rope wrapped around it; the free end of the rope hangs over the edge of the roof. The student grabs the end of the rope and steps off the roof. If the wheel has radius \(0.300 \mathrm{~m}\) and a moment of inertia of \(9.60 \mathrm{~kg} \cdot \mathrm{m}^{2}\) for rotation about the axle, how long does it take her to reach the sidewalk, and how fast will she be moving just before she lands? Ignore friction in the axle.

A uniform disk has radius \(R_{0}\) and mass \(M_{0}\). Its moment of inertia for an axis perpendicular to the plane of the disk at the disk's center is \(\frac{1}{2} M_{0} R_{0}^{2}\). You have been asked to halve the disk's moment of inertia by cutting out a circular piece at the center of the disk. In terms of \(R_{0}\), what should be the radius of the circular piece that you remove?

You are to design a rotating cylindrical axle to lift \(800 \mathrm{~N}\) buckets of cement from the ground to a rooftop \(78.0 \mathrm{~m}\) above the ground. The buckets will be attached to a hook on the free end of a cable that wraps around the rim of the axle; as the axle turns, the buckets will rise. (a) What should the diameter of the axle be in order to raise the buckets at a steady \(2.00 \mathrm{~cm} / \mathrm{s}\) when it is turning at \(7.5 \mathrm{rpm} ?\) (b) If instead the axle must give the buckets an upward acceleration of \(0.400 \mathrm{~m} / \mathrm{s}^{2},\) what should the angular acceleration of the axle be?

A thin uniform rod of mass \(M\) and length \(L\) is bent at its center so that the two segments are now perpendicular to each other. Find its moment of inertia about an axis perpendicular to its plane and passing through (a) the point where the two segments meet and (b) the midpoint of the line connecting its two ends.
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