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A uniform disk with mass \(40.0 \mathrm{~kg}\) and radius \(0.200 \mathrm{~m}\) is pivoted at its center about a horizontal, frictionless axle that is stationary. The disk is initially at rest, and then a constant force \(F=30.0 \mathrm{~N}\) is applied tangent to the rim of the disk. (a) What is the magnitude \(v\) of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.200 revolution? (b) What is the magnitude \(a\) of the resultant acceleration of a point on the rim of the disk after the disk has turned through 0.200 revolution?

Short Answer

Expert verified
The tangential velocity \(v\) of a point on the rim of the disk after 0.200 revolution is approximately 3.27 m/s and the resultant acceleration \(a\) of the point is approximately 53.51 m/s^2.

Step by step solution

01

Calculating Angular Acceleration

To start, we calculate the angular acceleration. Torque is given by the equation \(Torque = Force * radius = 30.0 N * 0.200 m = 6 N.m\). Moment of inertia (I) for a uniform disc about an axis through its center and perpendicular to the plane of the disc is \(I = (1/2) * mass * radius^2 = (1/2) * 40 kg * (0.200)^2 m = 0.800 kg.m^2\). Using Torque = I * angular acceleration, we solve for angular acceleration (\(alpha\)): \(alpha = Torque/I = 6 N.m/0.800 kg.m^2 = 7.5 rad/s^2\).
02

Determining Tangential Velocity

We then compute the tangential velocity (v) after the disk has made 0.200 revolutions. We first convert the revolutions to radians as the angular units in all the formulas are in radians. 0.200 revolution is the same as \(0.200 * 2 * \pi = 0.40 \pi\) radians. Using the equation \(vel_{final}^2 = vel_{initial}^2 + 2 * alpha * Angular displacement\), where initial velocity is 0 as the disk was initially at rest, we have \(v = \sqrt{2 * alpha * Angular displacement} = \sqrt{2 * 7.5 rad/s * 0.40 pi rad} = 3.27 m/s\).
03

Calculating Resultant Acceleration

Finally, we calculate the resultant acceleration of a point on the rim of the disk. A point moving in a circle of radius has two accelerations - radial acceleration (\(a_r\) which is directed towards the center of the circle) and tangential acceleration (\(a_t\) which is directed along the tangent at the point). \(a_r = v^2/radius = (3.27 m/s)^2/0.200 m = 53.48 m/s^2\) and \(a_t = alpha * radius = 7.5 rad/s^2 * 0.200 m = 1.5 m/s^2\). The resultant acceleration is the vector sum of the tangential and radial accelerations, that is, \(a = \sqrt{(a_t^2 + a_r^2)} = \sqrt{(1.5 m/s^2)^2 + (53.48 m/s^2)^2} = 53.51 m/s^2\).

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Key Concepts

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

Angular Acceleration
To begin our exploration of rotational motion, let's delve into the concept of angular acceleration. Angular acceleration, often represented by the Greek letter alpha (\( \text{alpha} \)), is the rate at which the angular velocity of an object changes with respect to time. Imagine spinning a bike wheel faster and faster; the wheel's angular acceleration is the rate at which you're increasing its spin. In the context of the exercise, once we have the torque applied and the moment of inertia, we calculate angular acceleration using \( \text{alpha} = \frac{\text{Torque}}{\text{moment of inertia}} \). This value is pivotal because it describes how quickly the rotational speed of the disk increases. It's essential to note that angular acceleration can also be negative, indicating a decrease in rotational speed, much like when you apply the brakes on that spinning bike wheel.

Understanding angular acceleration is not just about the final value. It’s about how the initial state of rest (zero angular velocity) changes as a force is applied, and the implications this has for other aspects of motion, such as tangential velocity and the resultant acceleration at a point on the spinning object. This comprehension lays the groundwork for grasping other rotational dynamics.
Tangential Velocity
Imagine swinging a stone tied to a string in a circular path; tangential velocity is the speed of the stone moving along this circular path. This value is always tangent to the circle at any given point, much like a racer on a track. When dealing with a rotating object like our disk, the velocity of a point on its rim is called the tangential velocity (\(v\)). With the previous calculation of angular acceleration, we're ready to find this velocity after the disk has rotated 0.200 revolutions.

Relating Angular and Tangential Velocity

In rotational motion, the relationship between the angular acceleration and the tangential velocity is given by \( v = \sqrt{ 2 * \text{alpha} * \text{Angular displacement} } \). We use this formula, remembering to convert revolutions to radians, to find the tangential velocity. By doing so, we understand how the force applied not only accelerates the disk in terms of spin but also translates into a linear speed along its edge.

Diving deeper, tangential velocity is not constant across all points of the rotating object—rather, it's directly related to the radius at which a point lies from the center of rotation. Closer to the center, the speed is lower; further out, it's higher, just like the stone on the string travels faster as you extend your arm.
Moment of Inertia
In the world of rotational motion, moment of inertia is akin to mass in linear motion; it's a measure of how resistant an object is to changes in its rotation, often symbolized as \(I\). Material, mass distribution, and the axis about which an object rotates all play roles in determining the moment of inertia. For the uniform disk in our exercise, the moment of inertia is a calculated constant that depends on the disk's mass and radius, expressed as \( I = \frac{1}{2} * m * r^2 \).

Importance in Calculating Angular Acceleration

The moment of inertia is integral in determining angular acceleration because it represents the rotational 'mass.' Hence, when torque is applied, the greater the moment of inertia, the less angular acceleration produced for the same amount of force. This concept mirrors how a heavier object requires more force to achieve the same acceleration in linear motion. It's a fundamental property that dictates how an object responds to applied forces and contributes to the overall dynamics of rotational motion.

Applying this to our exercise, we see that the moment of inertia is the denominator in our angular acceleration calculation, emphasizing its inverse relationship with the rotational acceleration of the disk.
Resultant Acceleration
Now, let's untangle the concept of resultant acceleration at a point on the rim of a rotating disk. This quantity is a combination of two components: radial (or centripetal) and tangential acceleration. Radial acceleration points towards the center of the circle and keeps the object moving in a circular path, while tangential acceleration is aligned with the tangential velocity of the point, indicating change in the speed along the circle.

Calculating Components

For a point on our disk, radial acceleration is calculated using the formula \( a_r = \frac{v^2}{r} \), where r is the radius of the disk. Tangential acceleration is directly related to angular acceleration as \( a_t = \text{alpha} * r \). These calculations show us the individual effects of the force applied: one keeping the object in circular motion, the other changing its rotational speed.

To find the actual acceleration a point on the rim experiences, we take the square root of the sum of the squares of these two components (using the Pythagorean theorem), giving us the resultant acceleration. Understanding this helps students picture the combined effect of the radial 'pull' and the tangential 'push' on a point as the disk spins.
Uniform Circular Motion
Lastly, uniform circular motion is a type of motion in which an object travels in a circular path at a constant speed. The key element here is that while the linear speed is constant, velocity is not, because velocity is a vector quantity and has both magnitude and direction. The direction is constantly changing in circular motion.

Relation to Rotational Motion

While our disk experiences acceleration and therefore is not in uniform circular motion, the concept still helps us understand the nature of the motion it's undergoing. If the tangential velocity were constant (and no tangential acceleration), it'd be a perfect example of uniform circular motion. However, in this scenario, the force causes a constant angular acceleration, translating to a continuously increasing tangential velocity.

Recognizing uniform circular motion allows students to differentiate between scenarios where only radial acceleration occurs due to velocity's changing direction and non-uniform scenarios where additional tangential acceleration is present due to changing speed. The uniform circular motion is an idealized scenario that serves as a fundamental reference point for understanding the more complex motions exhibited by objects such as our rotating disk.

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Most popular questions from this chapter

An engine delivers 175 hp to an aircraft propeller at 2400 rev \(/\) min. (a) How much torque does the aircraft engine provide? (b) How much work does the engine do in one revolution of the propeller?

A solid cylinder with radius \(0.140 \mathrm{~m}\) is mounted on a frictionless, stationary axle that lies along the cylinder axis. The cylinder is initially at rest. Then starting at \(t=0\) a constant horizontal force of \(3.00 \mathrm{~N}\) is applied tangential to the surface of the cylinder. You measure the angular displacement \(\theta-\theta_{0}\) of the cylinder as a function of the time \(t\) since the force was first applied. When you plot \(\theta-\theta_{0}\) (in radians) as a function of \(t^{2}\left(\right.\) in \(\left.\mathrm{s}^{2}\right),\) your data lie close to a straight line. If the slope of this line is \(16.0 \mathrm{rad} / \mathrm{s}^{2},\) what is the moment of inertia of the cylinder for rotation about the axle?

The mechanism shown in Fig. \(\mathbf{P} \mathbf{1 0 . 6 4}\) is used to raise a crate of supplies from a ship's hold. The crate has total mass \(50 \mathrm{~kg} .\) A rope is wrapped around a wooden cylinder that turns on a metal axle. The cylinder has radius \(0.25 \mathrm{~m}\) and moment of inertia \(I=2.9 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about the axle. The crate is suspended from the free end of the rope. One end of the axle pivots on frictionless bearings; a crank handle is attached to the other end. When the crank is turned, the end of the handle rotates about the axle in a vertical circle of radius \(0.12 \mathrm{~m},\) the cylinder turns, and the crate is raised. What magnitude of the force \(\vec{F}\) applied tangentially to the rotating crank is required to raise the crate with an acceleration of \(1.40 \mathrm{~m} / \mathrm{s}^{2} ?\) (You can ignore the mass of the rope as well as the moments of inertia of the axle and the crank.)

In your job as a mechanical engineer you are designing a flywheel and clutch- plate system like the one in Example \(10.11 .\) Disk \(A\) is made of a lighter material than disk \(B\), and the moment of inertia of disk \(A\) about the shaft is one-third that of disk \(B .\) The moment of inertia of the shaft is negligible. With the clutch disconnected, \(A\) is brought up to an angular speed \(\omega_{0} ; B\) is initially at rest. The accelerating torque is then removed from \(A,\) and \(A\) is coupled to \(B\). (Ignore bearing friction.) The design specifications allow for a maximum of \(2400 \mathrm{~J}\) of thermal energy to be developed when the connection is made. What can be the maximum value of the original kinetic energy of disk \(A\) so as not to exceed the maximum allowed value of the thermal energy?

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