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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?

Short Answer

Expert verified
The maximum value of the original kinetic energy of disk A so as not to exceed the maximum allowed value of the thermal energy will be \(\frac{1}{2}I_A \omega_A^2 + 2400J\).

Step by step solution

01

Setup the problem

We know that angular momentum is conserved in this system. We also know that all the kinetic energy of disk A will be transformed into thermal energy. Start by defining the initial moments of inertia and angular speeds as \(I_A\) and \(I_B\) for disk A and disk B respectively, while \(\omega_A\) will denote the original angular speed of disk A. Given that \(I_A = I_B/3\), due to the differences in materials of the two disks.
02

Calculate the total initial angular momentum

For the total initial angular momentum (\(L_i\)) of the system before the clutch is connected, we only consider disk A since disk B is initially at rest. So, \(L_i = I_A \omega_A\).
03

Calculate the final angular speed

After the clutch is connected, the disks will rotate at a common angular speed (\(\omega_f\)). As angular momentum is conserved, the total final angular momentum (\(L_f\)) of the system is given by \(L_f = (I_A + I_B) \omega_f = (I_A+I_A*3)\omega_f = 4I_A \omega_f\). Set this equal to \(L_i\) and solve for the final angular speed: \(\omega_f = \frac{I_A \omega_A}{4I_A} = \frac{\omega_A}{4}\)
04

Calculate the total final kinetic energy

The total final kinetic energy (\(K_f\)) after the clutch is connected is \(K_f = \frac{1}{2}(I_A + I_B) \omega_f^2 = \frac{1}{2} * 4I_A * (\frac{\omega_A}{4})^2 = \frac{1}{2}I_A \omega_A^2\)
05

Calculate the total initial kinetic energy

The total initial kinetic energy (\(K_i\)) of the system is determined by disk A since disk B is initially at rest. So, \(K_i = \frac{1}{2}I_A \omega_A^2\)
06

Calculate the maximum value of initial kinetic energy

The difference between the initial and final kinetic energy will be transformed into thermal energy. Hence the maximum value of the initial kinetic energy (\(K_i\)) should not exceed the sum of the final kinetic energy and the maximum thermal energy allowed which is \(2400J\). Therefore, \(K_i = K_f + thermal energy = \frac{1}{2}I_A \omega_A^2 + 2400\)

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Key Concepts

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

Conservation of Angular Momentum
Understanding the conservation of angular momentum is crucial in mechanical systems, particularly when designing components like flywheels and clutches. Angular momentum, a conserved quantity in a closed system, refers to the product of a rotating object's moment of inertia and its angular velocity. In the absence of external torques, the total angular momentum remains constant, even when internal forces act, such as a clutch engaging two rotating discs.

In the context of the exercise, disk A and disk B form a system with their shaft, and once they are connected, their combined angular momentum equals the initial angular momentum of just disk A. The application of this principle allows us to calculate the new angular speed after the connection, which significantly aids in predicting the behavior of mechanical systems during operations where kinetic energy is transformed and thermal energy is generated.
Kinetic Energy Transformation
When discussing kinetic energy transformation, we are referring to the process where kinetic energy, the energy of motion, changes form but is conserved in terms of its overall value within a system. This is intimately linked with the conservation of angular momentum in rotational systems.

In mechanical systems, such as the flywheel and clutch system in our exercise, kinetic energy can be transformed into other forms of energy, like thermal energy due to friction when two surfaces come into contact. The trick in managing this transformation efficiently lies in understanding how much kinetic energy an object has initially and predicting how much of that energy will convert into thermal energy. For our task, ensuring that the initial kinetic energy of disk A does not exceed the energy capacity of the system to absorb it in the form of heat is crucial for safe and efficient operation.
Flywheel Design Principles
Flywheels are designed to store and release energy efficiently, and their design principles involve careful consideration of materials and moments of inertia. A flywheel's capacity to store kinetic energy depends on its moment of inertia and the square of its angular velocity.

In designing a flywheel, engineers aim to maximize the moment of inertia within the constraints of size, weight, and the material's strength. The step-by-step solution highlighted that disk A’s moment of inertia is a third of disk B’s, indicating the choice of lighter material for disk A. This strategic choice impacts the system's ability to manage the thermal energy produced upon clutch engagement. Minimizing thermal energy through design is essential as excessive heating can lead to material fatigue and failure. Therefore, the design must balance the kinetic energy storage capabilities with the maximum permissible thermal energy to ensure durability and optimal performance of the flywheel system.

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Most popular questions from this chapter

A \(50.0 \mathrm{~kg}\) grindstone is a solid disk \(0.520 \mathrm{~m}\) in diameter. You press an ax down on the rim with a normal force of \(160 \mathrm{~N}\) (Fig. \(\mathbf{P} 10.58\) ). The coefficient of kinetic friction between the blade and the stone is \(0.60,\) and there is a constant friction torque of \(6.50 \mathrm{~N} \cdot \mathrm{m}\) between the axle of the stone and its bearings. (a) How much force must be applied tangentially at the end of a crank handle \(0.500 \mathrm{~m}\) long to bring the stone from rest to 120 rev \(/ \min\) in \(9.00 \mathrm{~s} ?\) (b) After the grindstone attains an angular speed of 120 rev \(/ \mathrm{min}\), what tangential force at the end of the handle is needed to maintain a constant angular speed of 120 rev \(/ \min ?\) (c) How much time does it take the grindstone to come from \(120 \mathrm{rev} / \mathrm{min}\) to rest if it is acted on by the axle friction alone?

What fraction of the total kinetic energy is rotational for the following objects rolling without slipping on a horizontal surface? (a) A uniform solid cylinder; (b) a uniform sphere; (c) a thin-walled, hollow sphere; (d) a hollow cylinder with outer radius \(R\) and inner radius \(R / 2\).

A block with mass \(m\) is revolving with linear speed \(v_{1}\) in a circle of radius \(r_{1}\) on a frictionless horizontal surface (see Fig. E10.42). The string is slowly pulled from below until the radius of the circle in which the block is revolving is reduced to \(r_{2}\). (a) Calculate the tension \(T\) in the string as a function of \(r\), the distance of the block from the hole. Your answer will be in terms of the initial velocity \(v_{1}\) and the radius \(r_{1} .\) (b) Use \(W=\int_{r_{1}}^{r_{2}} \overrightarrow{\boldsymbol{T}}(r) \cdot d \overrightarrow{\boldsymbol{r}}\) to calculate the work done by \(\vec{T}\) when \(r\) changes from \(r_{1}\) to \(r_{2}\). (c) Compare the results of part (b) to the change in the kinetic energy of the block.

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?

If the body's center of mass were not placed on the rotational axis of the turntable, how would the person's measured moment of inertia compare to the moment of inertia for rotation about the center of mass? (a) The measured moment of inertia would be too large; (b) the measured moment of inertia would be too small; (c) the two moments of inertia would be the same; (d) it depends on where the body's center of mass is placed relative to the center of the turntable.

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