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Chapter 8: Problem 33

A large grinding wheel in the shape of a solid cylinder of radius \(0.330 \mathrm{~m}\) is free to rotate on a frictionless, vertical axle. A constant tangential force of \(250 \mathrm{~N}\) applied to its edge causes the wheel to have an angular acceleration of \(0.940 \mathrm{rad} / \mathrm{s}^{2}\). (a) What is the moment of inertia of the wheel? (b) What is the mass of the wheel? (c) If the wheel starts from rest, what is its angular velocity after \(5.00\) s have elapsed, assuming the force is acting during that time?

Short Answer

Expert verified
The moment of inertia of the wheel is approximately 87.77 kg m^2, its mass is approximately 160.8 kg, and its angular velocity after 5.00 s is approximately 4.70 rad/s.

Step by step solution

01

Calculate the moment of inertia

We can start by solving part (a) of the question, which asks for the moment of inertia of the wheel. We know the angular acceleration (\(\alpha = 0.940 \mathrm{rad} / \mathrm{s}^{2}\)) and the tangential force (\(F_{t} = 250 \mathrm{~N}\)). The torque (\(\tau\)) of the wheel can be calculated by multiplying the tangential force by the radius of the wheel, \(\tau = F_{t} \cdot r = 250 N \cdot 0.33 m = 82.5 \mathrm{~Nm}\). According to Newton's second law for rotation, the torque is also equal to the moment of inertia (\(I\)) times the angular acceleration, \(\tau = I \cdot \alpha\). Therefore, we can solve for the moment of inertia, \(I = \tau / \alpha = 82.5 Nm / 0.940 rad/s^2 = 87.77 \mathrm{~kg \cdot m^2}\).
02

Calculate the mass of the wheel

The moment of inertia of a solid cylinder rotating about its axis is given by the formula \(I = \frac{1}{2} m r^2\), where \(m\) is the mass and \(r\) the radius. Equating this formula with the previously obtained moment of inertia allows us to solve for the mass: \(87.77 kg m^2 = \frac{1}{2} m (0.33 m)^2\), which gives \(m = 160.8 \mathrm{kg}\).
03

Find the angular velocity after 5.00 s

Lastly, we find the angular velocity (\(\omega\)) of the wheel after 5.00 s. The angular velocity can be found using the formula \(\omega = \omega_0 + \alpha t\), where \(\omega_0\) is the initial angular velocity, \(\alpha\) the angular acceleration, and \(t\) the time. Since the wheel starts from rest, \(\omega_0 = 0 rad/s\). Substituting these values gives \(\omega = 0 rad/s + 0.940 rad/s^2 \cdot 5.00 s = 4.70 rad/s\).

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Key Concepts

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

Moment of Inertia
The moment of inertia is a fundamental concept in physics, particularly in angular motion. It measures how much resistance an object has to changes in its rotational motion. Think of it as the rotational equivalent of mass in linear motion.
To calculate the moment of inertia (\(I\)) of a solid cylinder, we typically use the formula \(I = \frac{1}{2} m r^2\) for objects rotating about their central axis. This formula stems from the integral calculus applied within the object's geometry, showing how mass is distributed relative to the axis.
  • In this exercise, the torque exerted by the tangential force is used in combination with angular acceleration to find the moment of inertia.
  • Recognizing how the formula \(\tau = I \cdot \alpha\) (Torque = Moment of Inertia x Angular Acceleration) connects these values is crucial for solving problems in rotation dynamics.
Torque
Torque is what sets objects rotating. It's the measure of the force that can cause an object to rotate about an axis. You can think of it as a force applied at a distance. In this case, the distance is the radius of the cylinder.
The formula to compute torque \(\tau\) is \(\tau = F_{t} \cdot r\), where \(F_{t}\) is the tangential force and \(r\) is the radius.
  • For our grinding wheel, a tangential force of 250 N is applied, resulting in a torque of 82.5 Nm after multiplying by the radius 0.33 m.
  • Understanding how to manipulate this formula helps in predicting and controlling rotational movements effectively.
Angular Acceleration
Angular acceleration (\(\alpha\)) is the rate of change of angular velocity. It tells you how quickly an object is speeding up or slowing down its rotation. In our scenario, the wheel is subjected to an angular acceleration of \(0.94 \, \mathrm{rad/s^2}\).
This concept is directly linked to Newton's second law for rotation, \(\tau = I \cdot \alpha\), connecting angular acceleration to torque and moment of inertia.
  • This relationship signifies that the greater the torque applied, the greater the angular acceleration, assuming the moment of inertia remains constant.
  • In practical applications, understanding angular acceleration assists in controlling the rate at which machines like grinding wheels operate.
Solid Cylinder
A solid cylinder is one of the most common shapes in physics problems involving rotation. Its geometry allows for straightforward calculations of moment of inertia and rotational dynamics. In our scenario, the cylinder represents the grinding wheel, which is 0.33 meters in radius.
The moment of inertia of a solid cylinder rotating about its central axis is calculated with \(I = \frac{1}{2} m r^2\). This formula accounts for how mass is evenly distributed along its body.
  • This mass distribution simplifies analyzing and predicting rotational behaviors under applied forces.
  • The problem gives insight into real-world applications, such as in engine designs and equipment, where solid cylinder forms are prevalent.

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Most popular questions from this chapter

A solid, uniform disk of radius \(0.250 \mathrm{~m}\) and mass \(55.0 \mathrm{~kg}\) rolls down a ramp of length \(4.50 \mathrm{~m}\) that makes an angle of \(15.0^{\circ}\) with the horizontal. The disk starts from rest from the top of the ramp. Find (a) the speed of the disk's center of mass when it reaches the bottom of the ramp and (b) the angular speed of the disk at the bottom of the ramp.

The Iron Cross When a gymnast weighing \(750 \mathrm{~N}\) executes the iron cross as in Figure \(\mathrm{P} 8.85 \mathrm{a}\), the primary muscles involved in supporting this position are the latissimus dorsi ("lats") and the pectoralis major ("pecs"). The rings exert an upward force on the arms and support the weight of the gymnast. The force exerted by the shoulder joint on the arm is labeled \(\overrightarrow{\mathbf{F}}_{x}\) while the two muscles exert a total force \(\overrightarrow{\mathbf{F}}_{m}\) on the arm. Estimate the magnitude of the force \(\overrightarrow{\mathbf{F}}_{m}\). Note that one ring supports half the weight of the gymnast, which is \(375 \mathrm{~N}\) as indicated in Figure \(\mathrm{P} 8.85 \mathrm{~b}\). Assume that the force \(\overrightarrow{\mathbf{F}}_{\mathrm{m}}\) acts at an angle of \(45^{\circ}\) below the horizontal at a distance of \(4.0 \mathrm{~cm}\) from the shoulder joint. In your estimate, take the distance from the shoulder joint to the hand to be \(70 \mathrm{~cm}\) and ignore the weight of the arm.

The top in Figure P8.49 has a moment of inertia of \(4.00\) \(\times 10^{-4} \mathrm{~kg} \cdot \mathrm{m}^{2}\) and is initially at rest. It is free to rotate about a stationary axis \(A A^{\prime}\). A string wrapped around a peg along the axis of the top is pulled in such a manner as to maintain a constant tension of \(5.57 \mathrm{~N}\) in the string. If the string does not slip while wound around the peg, what is the angular speed of the top after \(80.0 \mathrm{~cm}\) of string has been pulled off the peg? Hint: Consider the work that is done.

A solid, horizontal cylinder of mass \(10.0 \mathrm{~kg}\) and radius \(1.00 \mathrm{~m}\) rotates with an angular speed of \(7.00 \mathrm{rad} / \mathrm{s}\) about a fixed vertical axis through its center. A \(0.250-\mathrm{kg}\) piece of putty is dropped vertically onto the cylinder at a point \(0.900 \mathrm{~m}\) from the center of rotation and sticks to the cylinder. Determine the final angular speed of the system.

According to the manual of a certain car, a maximum torque of magnitude \(65.0 \mathrm{~N} \cdot \mathrm{m}\) should be applied when tightening the lug nuts on the vehicle. If you use a wrench of length \(0.330 \mathrm{~m}\) and you apply the force at the end of the wrench at an angle of \(75.0^{\circ}\) with respect to a line going from the lug nut through the end of the handle, what is the magnitude of the maximum force you can exert on the handle without exceeding the recommendation?
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