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Chapter 10: Problem 20

\(\cdot\) The flywheel of a motor has a mass of 300.0 \(\mathrm{kg}\) and a moment of inertia of 580 \(\mathrm{kg} \cdot \mathrm{m}^{2} .\) The motor develops a constant torque of \(2000.0 \mathrm{N} \cdot \mathrm{m},\) and the flywheel starts from rest. (a) What is the angular acceleration of the flywheel? (b) What is its angular velocity after it makes 4.00 revolutions? (c) How much work is done by the motor during the first 4.00 revolutions?

Short Answer

Expert verified
(a) Angular acceleration: 3.4483 rad/s虏. (b) Angular velocity: 23.332 rad/s. (c) Work done: 50265.5 J.

Step by step solution

01

Understanding Torque and Angular Acceleration

To find the angular acceleration \( \alpha \), use the relation between torque \( \tau \) and angular acceleration: \( \tau = I \times \alpha \), where \( I \) is the moment of inertia. Given \( \tau = 2000.0 \, \mathrm{N} \cdot \mathrm{m} \) and \( I = 580 \, \mathrm{kg} \cdot \mathrm{m}^{2} \), rearrange to solve for \( \alpha \): \( \alpha = \frac{\tau}{I} \). Substitute the known values: \( \alpha = \frac{2000.0}{580} = 3.4483 \, \mathrm{rad/s^2} \).
02

Finding Angular Displacement in Radians

Convert revolutions to radians for part (b) and (c) calculations. 1 revolution equals \( 2\pi \) radians, thus 4.00 revolutions equal \( 4 \times 2\pi = 8\pi \) radians.
03

Calculating Angular Velocity

Use the formula for angular velocity \( \omega \) at the end of a displacement with constant angular acceleration: \( \omega^2 = \omega_0^2 + 2\alpha\theta \). Here, \( \omega_0 = 0 \) (starts from rest), \( \alpha = 3.4483 \, \mathrm{rad/s^2} \), and \( \theta = 8\pi \). \( \omega^2 = 0 + 2 \times 3.4483 \times 8\pi \), \( \omega^2 = 173.0256\pi \), \( \omega = \sqrt{173.0256\pi} \approx 23.332 \, \mathrm{rad/s} \).
04

Calculating Work Done by the Motor

Work done \( W \) by the motor is given by \( W = \tau \times \theta \), substituting \( \tau = 2000.0 \, \mathrm{N} \cdot \mathrm{m} \) and \( \theta = 8\pi \) radians, \( W = 2000.0 \times 8\pi \), \( W = 16000\pi \approx 50265.5 \, \mathrm{J} \).

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Key Concepts

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

Torque
Torque is the rotational equivalent of linear force. It measures how effectively a force can cause an object to rotate about an axis. Think of it like the twisting action you apply when turning a screwdriver or opening a jar lid. The formula to calculate torque \( \tau \) is given by: \[\tau = r \times F \times \sin(\theta) \]where:
  • \( r \) is the distance from the axis of rotation to where the force is applied (the lever arm).
  • \( F \) is the force applied.
  • \( \theta \) is the angle between the force and the lever arm.
In the context of our exercise, the motor applies a constant torque of \( 2000.0 \,\mathrm{N} \cdot \mathrm{m} \), which causes the flywheel to start spinning from rest, demonstrating how torque is vital in starting rotational motion.
Moment of Inertia
The moment of inertia is a measure of an object's resistance to changes in its rotational motion. Similar to how mass characterizes an object's resistance to linear acceleration, the moment of inertia plays this role in rotational dynamics. The formula for moment of inertia \( I \) depends on the object's shape and the axis of rotation. For a point mass, it's expressed as: \[ I = \sum m_i r_i^2 \]where:
  • \( m_i \) is the mass of each point in the object.
  • \( r_i \) is the distance of each mass point from the axis of rotation.
In our problem, the flywheel has a given moment of inertia of \( 580 \, \mathrm{kg} \cdot \mathrm{m}^2 \), indicating it's somewhat spread out from the axis, impacting how quickly it can spin up under applied torque.
Angular Acceleration
Angular acceleration \( \alpha \) describes how quickly the rotational velocity of an object changes. It鈥檚 analogous to linear acceleration, but in terms of rotational motion. The formula connecting torque and angular acceleration is:\[ \tau = I \times \alpha \]This relationship highlights how both the applied torque and the moment of inertia influence angular acceleration. Solving for \( \alpha \) in our exercise involves:\[ \alpha = \frac{\tau}{I} = \frac{2000.0}{580} \approx 3.4483 \, \mathrm{rad/s^2} \]This value tells us how effectively the flywheel is accelerating rotationally due to the motor's applied torque.
Angular Velocity
Angular velocity \( \omega \) is a measure of how fast something rotates, usually expressed in radians per second. It indicates the angle an object sweeps through per unit of time. To find final angular velocity after a given angular displacement, the formula is:\[ \omega^2 = \omega_0^2 + 2\alpha\theta \]where:
  • \( \omega_0 \) is the initial angular velocity (0 in this case, as it starts from rest).
  • \( \alpha \) is the angular acceleration.
  • \( \theta \) is the angular displacement in radians.
Applying the values from our problem, we find:\[ \omega^2 = 2 \times 3.4483 \times 8\pi \]\[ \omega = \sqrt{173.0256\pi} \approx 23.332 \, \mathrm{rad/s} \]Thus, after 4 revolutions, the flywheel rotates at an angular velocity of approximately \( 23.332 \, \mathrm{rad/s} \).
Work and Energy in Rotational Motion
In rotational motion, work is done when a torque causes an angular displacement. The work \( W \) can be calculated using:\[ W = \tau \times \theta \]where \( \theta \) is the angular displacement in radians. The concept is comparable to linear work, where force acts over distance. From the exercise, substituting \( \tau = 2000.0 \, \mathrm{N} \cdot \mathrm{m} \) and converting 4 revolutions to radians \( 8\pi \):\[ W = 2000.0 \times 8\pi \]\[ W \approx 50265.5 \, \mathrm{J} \]This calculation shows us how much work the motor has performed in spinning the flywheel through its initial 4 revolutions, highlighting the energy required in rotational motions.

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Most popular questions from this chapter

A large wooden turntable in the shape of a flat disk has a radius of 2.00 \(\mathrm{m}\) and a total mass of 120 \(\mathrm{kg}\) . The turntable is initially rotating at 3.00 \(\mathrm{rad} / \mathrm{s}\) about a vertical axis through its center. Suddenly, a 70.0 kg parachutist makes a soft landing on the turntable at a point on its outer edge. Find the angular speed of the turntable after the parachutist lands. (Assume that you can treat the parachutist as a particle.)

Two people carry a heavy electric motor by placing it on a light board 2.00 m long. One person lifts at one end with a force of \(400 \mathrm{N},\) and the other lifts the opposite end with a force of 600 \(\mathrm{N}\) . (a) What is the weight of the motor, and where along the board is its center of gravity located? (b) Suppose the board is not light but weighs \(200 \mathrm{N},\) with its center of gravity at its center, and the two people each exert the same forces as before. What is the weight of the motor in this case, and where is its center of gravity located?

What would be the radial (centripetal) acceleration of an astronaut standing on the inner surface of the space station, which is at a distance of 240 \(\mathrm{m}\) from the axis of rotation, once the space station reaches its final angular velocity of 0.20 \(\mathrm{rad} / \mathrm{s}\) ? A. 0 \(\mathrm{m} / \mathrm{s}^{2}\) B. 4.8 \(\mathrm{m} / \mathrm{s}^{2}\) \(\mathrm{C}, 9.6 \mathrm{m} / \mathrm{s}^{2}\) D. 48 \(\mathrm{m} / \mathrm{s}^{2}\)

Supporting an injured arm: I. A 650 N person must have her injured arm supported, with the upper arm horizontal and the fore- arm vertical. (See Figure \(10.76 .\) ) According to biomedical tables and direct measurements, her upper arm is 26 \(\mathrm{cm}\) long (measured from the shoulder joint), accounts for 3.50\(\%\) of her body weight, and has a center of mass 13.0 \(\mathrm{cm}\) from her shoulder joint. Her forearm (including the hand) is 34.0 \(\mathrm{cm}\) long, makes up 3.25\(\%\) of her body weight, and has a center of mass 43.0 \(\mathrm{cm}\) from her shoulder joint. (a) Where is the center of mass of the person's arm when it is supported as shown? (b) What weight \(W\) is needed to support her arm? (c) Find the horizontal and vertical components of the force that the shoulder joint exerts on her arm.

\bullet A small block on a frictionless horizontal surface has a mass of 0.0250 \(\mathrm{kg}\) . It is attached to a massless cord passing through a hole in the surface. (See Figure \(10.54 .\) The block is originally revolving at a distance of 0.300 \(\mathrm{m}\) from the hole with an angular speed of 1.75 rad/s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.150 \(\mathrm{m} .\) You may treat the block as a particle. (a) Is angular momentum conserved? Why or why not? (b) What is the new angular speed? (c) Find the change in kinetic energy of the block. (d) How much work was done in pulling the cord?
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