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Chapter 10: Problem 67

A wheel with radius \(0.0600 \mathrm{~m}\) rotates about a horizontal frictionless axle at its center. The moment of inertia of the wheel about the axle is \(2.50 \mathrm{~kg} \cdot \mathrm{m}^{2}\). The wheel is initially at rest. Then at \(t=0 \mathrm{a}\) force \(F=(5.00 \mathrm{~N} / \mathrm{s}) t\) is applied tangentially to the wheel and the wheel starts to rotate. What is the magnitude of the force at the instant when the wheel has turned through 8.00 revolutions?

Short Answer

Expert verified
The magnitude of the force when the wheel has completed 8 revolutions can be found by substituting the given values into the equations used in the steps. This would give us a numerical answer for the force.

Step by step solution

01

- Calculate the Total Angle Turned

First, we need to calculate how much the wheel has rotated in radians. We know that one full revolution equates to \(2\pi\) radians. So, 8 revolutions will be \(8 * 2\pi\) radians.
02

- Determine the Angular Acceleration

Applying the equations of circular motion, we know that under constant angular acceleration \(\alpha\), the change in the angle \(\theta\) is equal to \(0.5 * \alpha * t^{2}\), where \(t\) is the time taken. Since we need to find \(t\) when \(\theta\) is \(8 * 2\pi\), we can rearrange the equation to \(\alpha = \frac{2\theta}{t^{2}}\)
03

- Use the Torque Equation

We know that the torque \(T\) of the object is \(T = I*\alpha\), where \(I\) is the moment of inertia and \(\alpha\) is the angular acceleration. As the force applied is tangential and its lever arm is the radius of the wheel, we can also express the torque as \(T=F*r\). Equating the two, we get \(F= \frac{I*\alpha}{r}\)
04

- Determine the Applied Force

From the equation obtained in Step 3, we can substitute the values of \(I\), \(\alpha\) and \(r\) into the equation to get the value of \(F\). We then substitute the force at time \(t\) which is \(F=(5.00 \mathrm{~N} / \mathrm{s}) t\), find \(t\) and multiply by \(5.00 \mathrm{~N} / \mathrm{s}\), to get the required force.

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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, symbolized by the letter 'I', is a measure of how difficult it is to change the rotational motion of an object. It鈥檚 akin to mass in linear dynamics, but for rotational motion. The moment of inertia depends on the mass of the object and how that mass is distributed relative to the axis of rotation.

For a point mass, the moment of inertia is given by the product of the mass (m) and the square of the perpendicular distance (r) from the axis of rotation: \[ I = mr^2 \.\] In our exercise, the moment of inertia for the wheel is provided as 2.50 kg路m虏. This value plays a pivotal role when we apply a force to rotate the wheel, as it dictates how much torque is needed for a certain angular acceleration.
Angular Acceleration
Angular acceleration, denoted by the Greek letter alpha \(\alpha\), is a measure of how quickly the rotational velocity of an object is changing. It describes the rate at which an object spins faster or slower. The unit of angular acceleration is radians per second squared (rad/s虏).

For practical use, angular acceleration is found using the relation \[ \alpha = \frac{\Delta\omega}{\Delta t} \.\] Where \(\Delta\omega\) is the change in angular velocity and \(\Delta t\) is the time interval over which this change occurs. However, in our wheel scenario, we use the kinematic equation for constant angular acceleration that relates angle, angular acceleration, and time: \[ \theta = 0.5 \cdot \alpha \cdot t^2 \.\] Solving this relation enables us to find the angular acceleration when the wheel has turned through a known angle, such as the 8.00 revolutions in the given problem.
Torque
Torque, often represented as 'T', is the rotational equivalent of force. It reflects how much force is causing an object to rotate about an axis. If you've ever used a wrench to tighten a bolt, you've applied torque. The larger the force or the longer the wrench, the greater the torque.

The formula for torque when a force is applied tangentially (as it is in the wheel problem) is \[ T = F \cdot r \.\] where 鈥楩鈥� is the force and 'r' is the radius of the wheel. In rotational dynamics, there's a direct link between torque and angular acceleration, given by: \[ T = I \cdot \alpha \.\] Combining these equations, we can find the force applied to the wheel based on its moment of inertia and angular acceleration. By understanding torque, we can ascertain that as the tangential force increases over time, per the equation \(F=(5.00 \mathrm{~N} / \mathrm{s}) t\), we'd expect the torque 鈥� and therefore the wheel's angular acceleration 鈥� to increase proportionally. This correlation is critical for solving the wheel's rotational dynamics in our exercise.

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Most popular questions from this chapter

The moment of inertia of the empty turntable is \(1.5 \mathrm{~kg} \cdot \mathrm{m}^{2}\). With a constant torque of \(2.5 \mathrm{~N} \cdot \mathrm{m},\) the turntable-person system takes \(3.0 \mathrm{~s}\) to spin from rest to an angular speed of \(1.0 \mathrm{rad} / \mathrm{s} .\) What is the person's moment of inertia about an axis through her center of mass? Ignore friction in the turntable axle. (a) \(2.5 \mathrm{~kg} \cdot \mathrm{m}^{2}\) (b) \(6.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\) (c) \(7.5 \mathrm{~kg} \cdot \mathrm{m}^{2} ;\) (d) \(9.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\).

A large turntable with radius \(6.00 \mathrm{~m}\) rotates about a fixed vertical axis, making one revolution in \(8.00 \mathrm{~s}\). The moment of inertia of the turntable about this axis is \(1200 \mathrm{~kg} \cdot \mathrm{m}^{2}\). You stand, barefooted, at the rim of the turntable and very slowly walk toward the center, along a radial line painted on the surface of the turntable. Your mass is \(70.0 \mathrm{~kg}\). since the radius of the turntable is large, it is a good approximation to treat yourself as a point mass. Assume that you can maintain your balance by adjusting the positions of your feet. You find that you can reach a point \(3.00 \mathrm{~m}\) from the center of the turntable before your feet begin to slip. What is the coefficient of static friction between the bottoms of your feet and the surface of the turntable?

A uniform solid disk made of wood is horizontal and rotates freely about a vertical axle at its center. The disk has radius \(0.600 \mathrm{~m}\) and mass \(1.60 \mathrm{~kg}\) and is initially at rest. A bullet with mass \(0.0200 \mathrm{~kg}\) is fired horizontally at the disk, strikes the rim of the disk at a point perpendicular to the radius of the disk, and becomes embedded in its rim, a distance of \(0.600 \mathrm{~m}\) from the axle. After being struck by the bullet, the disk rotates at \(4.00 \mathrm{rad} / \mathrm{s}\). What is the horizontal velocity of the bullet just before it strikes the disk?

Under some circumstances, a star can collapse into an extremely dense object made mostly of neutrons and called a neutron star. The density of a neutron star is roughly \(10^{14}\) times as great as that of ordinary solid matter. Suppose we represent the star as a uniform, solid, rigid sphere, both before and after the collapse. The star's initial radius was \(7.0 \times 10^{5} \mathrm{~km}\) (comparable to our sun); its final radius is \(16 \mathrm{~km}\). If the original star rotated once in 30 days, find the angular speed of the neutron star.

A uniform, \(4.5 \mathrm{~kg},\) square, solid wooden gate \(1.5 \mathrm{~m}\) on each side hangs vertically from a frictionless pivot at the center of its upper edge. A \(1.1 \mathrm{~kg}\) raven flying horizontally at \(5.0 \mathrm{~m} / \mathrm{s}\) flies into this door at its center and bounces back at \(2.0 \mathrm{~m} / \mathrm{s}\) in the opposite direction. (a) What is the angular speed of the gate just after it is struck by the unfortunate raven? (b) During the collision, why is the angular momentum conserved but not the linear momentum?
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