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A wheel of diameter 40.0 cm starts from rest and rotates with a constant angular acceleration of 3.00 rad/s\(^2\). Compute the radial acceleration of a point on the rim for the instant the wheel completes its second revolution from the relationship (a) \(a_{rad} = \omega^2r\) and (b) \(a_{rad} = v^2/r\)

Short Answer

Expert verified
The radial acceleration is approximately 15.08 m/s² for both (a) and (b).

Step by step solution

01

Find Angular Displacement Required

To find when the wheel completes its second revolution, calculate the total angular displacement. Since one revolution is \(2\pi\) rad, two revolutions will be \(4\pi\) rad.
02

Calculate Final Angular Velocity

Use the formula for angular motion: \(\omega^2 = \omega_0^2 + 2\alpha \theta\), where \(\omega_0 = 0\) (starts from rest), \(\alpha = 3.00\,\text{rad/s}^2\), and \(\theta = 4\pi\). Solving for \(\omega\), we get:\[\omega^2 = 0 + 2 \times 3.00 \times 4\pi\]\[\omega^2 = 24\pi\]\.\[\omega \approx \sqrt{24\pi} \approx 8.68 \, \text{rad/s}\].
03

Find Radius of the Wheel

The diameter of the wheel is given as 40.0 cm. Thus, the radius \(r\) is half of the diameter: \(r = 20.0\, \text{cm} = 0.20\, \text{m}\).
04

Calculate Radial Acceleration Using \(a_{rad} = \omega^2 r\)

Using the formula \(a_{rad} = \omega^2 r\) and \(\omega = 8.68 \, \text{rad/s}\), \(r = 0.20 \, \text{m}\):\[a_{rad} = (8.68)^2 \times 0.20 \approx 15.08 \, \text{m/s}^2\].
05

Calculate Linear Velocity

The linear velocity \(v\) is related to the angular velocity by \(v = \omega r\). Thus,:\[v = 8.68 \times 0.20 \approx 1.74 \, \text{m/s}\].
06

Calculate Radial Acceleration Using \(a_{rad} = v^2/r\)

Using the formula \(a_{rad} = v^2 / r\) with \(v = 1.74 \, \text{m/s}\) and \(r = 0.20 \, \text{m}\), calculate:\[a_{rad} = \frac{(1.74)^2}{0.20} \approx 15.08 \, \text{m/s}^2\].

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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 a measure of how quickly an object's rotational speed changes. It is represented by the symbol \(\alpha \), and is typically measured in radians per second squared (rad/s\(^2\)). In the context of the rotating wheel starting from rest with a constant angular acceleration of 3.00 rad/s\(^2\), the wheel's speed increases gradually until it reaches the desired rotational speed. Angular acceleration can be calculated using the formula:
  • \( \alpha = \frac{\Delta \omega}{\Delta t} \)
This formula denotes that angular acceleration is the change in angular velocity (\( \Delta \omega \)) over the change in time (\( \Delta t \)). In this exercise, since the wheel starts from rest, the initial angular velocity (\( \omega_0 \)) is zero. Thus, the formula effectively shows us how the wheel spins faster, resulting in an increasing angular velocity.
Angular Displacement
Angular displacement refers to the angle through which an object rotates about a fixed point. In our exercise, the wheel undergoes an angular displacement when it turns completely around. Measured in radians, angular displacement is calculated by multiplying the number of complete revolutions by \( 2\pi \) radians per revolution. As given in the step by step solution, to find the angular displacement of a wheel that completes its second revolution, we compute:
  • Total angular displacement \( \theta = 2 \times 2\pi \text{ rad} = 4\pi \text{ rad} \)
This measurement tells us how far around the circle the wheel has moved. Understanding angular displacement is crucial when linking how linear motion translates to circular motion, as each revolution brings the wheel closer to a full circle.
Angular Velocity
Angular velocity defines how fast an object rotates or revolves relative to another point, measured in radians per second (rad/s). Think of it as the rate at which the angle changes with time. In the case of the wheel in the problem, the angular velocity was initially zero, as it started from rest. It then increased due to the wheel's constant angular acceleration. To calculate the final angular velocity (\( \omega \)) when the wheel finishes its second rotation, we use the formula:
  • \( \omega^2 = \omega_0^2 + 2\alpha \theta \)
With the values provided, solving gives \( \omega \approx 8.68 \, \text{rad/s} \), indicating how fast the wheel is spinning relative to its center. This angular velocity also helps derive the radial acceleration, essential to analyze the motion of any point on the wheel's rim.
Linear Velocity
Linear velocity is the rate at which an object moves along a path. In circular motion, it's derived from the angular velocity and connects rotational and linear movement. To find the linear velocity (\( v \)) of a point on the rim of our wheel, we use the relationship with the angular velocity:
  • \( v = \omega r \)
Given that \( \omega = 8.68 \, \text{rad/s} \) and the radius \( r = 0.20 \, \text{m} \), we calculate \( v \approx 1.74 \, \text{m/s} \). This value reveals the speed at which a point on the edge of the wheel is moving in a straight line. Linear velocity is crucial in understanding the mechanics of objects in circular path movements. It exemplifies how angular metrics convert into practical speeds on physical paths, aiding in computing radial acceleration using \( a_{rad} = \frac{v^2}{r} \). This seamless conversion highlights the interconnected nature of rotational and linear motion.

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Most popular questions from this chapter

According to the shop manual, when drilling a 12.7-mm-diameter hole in wood, plastic, or aluminum, a drill should have a speed of 1250 rev/min. For a 12.7-mm-diameter drill bit turning at a constant 1250 rev/min, find (a) the maximum linear speed of any part of the bit and (b) the maximum radial acceleration of any part of the bit.

Find the moment of inertia of a hoop (a thin-walled, hollow ring) with mass \(M\) and radius \(R\) about an axis perpendicular to the hoop’s plane at an edge.

Energy is to be stored in a 70.0-kg flywheel in the shape of a uniform solid disk with radius \(R =\) 1.20 m. To prevent structural failure of the flywheel, the maximum allowed radial acceleration of a point on its rim is 3500 m/s\(^2\). What is the maximum kinetic energy that can be stored in the flywheel?

At \(t\) \(=\) 3.00 s a point on the rim of a 0.200-m-radius wheel has a tangential speed of 50.0 m/s as the wheel slows down with a tangential acceleration of constant magnitude 10.0 m/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?

(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?

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