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Disk \(A\) has a mass of \(6 \mathrm{kg}\) and an initial angular velocity of \(360 \mathrm{rpm}\) clockwise; disk \(B\) hass of \(3 \mathrm{kg}\) and is initially at rest. The disks are brought together by applying a horizontal force of magnitude \(20 \mathrm{N}\) to the axle of disk \(A .\) Knowing that \(\mu_{k}=0.15\) between the disks and neglecting bearing friction, determine \((a)\) the angular acceleration of each disk, \((b)\) the final angular velocity of each disk.

Short Answer

Expert verified
(a) \( \alpha_A = \frac{1}{r}, \alpha_B = \frac{2}{r} \). (b) \( \omega_f = 25.13 \, \text{rad/s} \) for both disks.

Step by step solution

01

Calculate Initial Angular Velocity

Convert the initial angular velocity of disk \(A\) from rotations per minute (rpm) to radians per second (rad/s). We use the formula:\[ \omega_i = \frac{360 \times 2\pi}{60} \]Calculating this gives:\[ \omega_i = 37.7 \text{ rad/s} \].
02

Determine Frictional Force

Calculate the frictional force using the coefficient of kinetic friction \(\mu_k = 0.15\) and the normal force \(N\) which is due to the applied force on the axle of disk A. The frictional force \(f_k\) is given by:\[ f_k = \mu_k \times 20 = 3 \text{ N}. \]
03

Calculate Torque Due to Friction

Calculate the torque \(\tau\) due to friction. The torque is the product of the frictional force and radius of the disk. Assuming disk A and B have the same radius \(r\), torque on disk A is:\[ \tau_A = f_k \times r \text{ and same for disk B, } \tau_B = f_k \times r \].
04

Calculate Angular Acceleration

Apply Newton's second law for rotation, \(\tau = I\alpha\), where \(I\) is the moment of inertia and \(\alpha\) is the angular acceleration. Moment of inertia for a disk is \(I = \frac{1}{2}m r^2\):\[ I_A = \frac{1}{2}(6)r^2, \quad I_B = \frac{1}{2}(3)r^2. \]Thus, angular acceleration for disks are:\[ \alpha_A = \frac{f_k \cdot r}{I_A} = \frac{3r}{3r^2} = \frac{1}{r}, \quad \alpha_B = \frac{f_k \cdot r}{I_B} = \frac{3r}{1.5r^2} = \frac{2}{r}. \]
05

Determine Final Angular Velocity

Since the disks reach a common velocity due to friction after time:\( t \) when they stop experiencing relative slippage, the final angular velocity \(\omega_f\) can be determined by the angular deceleration and conservation of angular momentum:\[ I_A \cdot \omega_i + I_B \cdot 0 = (I_A + I_B) \cdot \omega_f. \]This gives:\[ 3r^2 \cdot 37.7 = 4.5r^2 \omega_f \Rightarrow \omega_f = \frac{3 \times 37.7}{4.5} \approx 25.13 \, \text{rad/s} \].

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Key Concepts

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

Angular Acceleration
Angular acceleration is a measure of how quickly the angular velocity of an object changes with time. In the case of rotating disks, this concept plays a critical role in understanding how the disks speed up or slow down when forces are applied.
To find the angular acceleration, we use the formula \( \alpha = \frac{\tau}{I} \), where \( \tau \) is the torque applied, and \( I \) is the moment of inertia. Torque is the rotational equivalent of force, and moment of inertia determines how much the object resists changes in its rotational motion.
  • For disk A, the angular acceleration \( \alpha_A \) is calculated using its moment of inertia \( I_A = \frac{1}{2}m r^2 \) and torque due to friction \( \tau_A \). Given the frictional force \( f_k \) and radius \( r \), the expression becomes \( \alpha_A = \frac{f_k \cdot r}{I_A} \).
  • Similarly, for disk B, the angular acceleration \( \alpha_B \) is effectively \( \alpha_B = \frac{f_k \cdot r}{I_B} \).
Understanding angular acceleration helps predict how the motion of objects like spinning disks changes over time when external forces and torques are involved.
Angular Velocity
Angular velocity represents how fast an object spins around its axis. It is often expressed in radians per second (rad/s).
In the exercise, we start with disk A having an initial angular velocity and disk B initially at rest. The process of friction causes the disks to eventually reach a common angular velocity.
Here's how it's analyzed:
  • The initial angular velocity of disk A was calculated by converting rotations per minute (rpm) to radians per second using \( \omega_i = \frac{360 \times 2\pi}{60} \).
  • After the frictional interactions, both disks eventually share a final angular velocity. This is due to the conservation of angular momentum, where the total initial angular momentum equals the total final angular momentum.
  • The formula \( I_A \cdot \omega_i + I_B \cdot 0 = (I_A + I_B) \cdot \omega_f \) is used to find the final shared angular velocity \( \omega_f \).
This shared velocity is crucial because it represents the stable spinning rate the system reaches. It is a clear indicator of how momentum and external forces influence rotational motion.
Moment of Inertia
Moment of inertia is a fundamental property of rotating objects that quantifies how difficult it is to change their rotational state. Think of it as the rotational equivalent of mass, which affects how objects move or resist force in linear motion.
For a disk, the moment of inertia \( I \) is determined by the formula \( I = \frac{1}{2}mr^2 \), where \( m \) is the mass of the disk, and \( r \) is its radius. The moment of inertia depends both on the amount of mass and how far that mass is distributed from the axis of rotation.
  • In our problem, each disk's moment of inertia is calculated separately. Disk A's moment is \( I_A = \frac{1}{2}(6)r^2 \) and disk B's is \( I_B = \frac{1}{2}(3)r^2 \).
  • This calculation is significant when finding the angular accelerations, as it directly influences the angular acceleration formula \( \alpha = \frac{\tau}{I} \).
Understanding moment of inertia helps engineers and scientists design and predict the behavior of rotating systems, thereby ensuring the stability and control of various mechanical and engineering applications.

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Most popular questions from this chapter

The \(100-\) mm-radius brake drum is attached to a flywheel that is not shown. The drum and flywheel together have a mass of \(300 \mathrm{kg}\) and a radius of gyration of \(600 \mathrm{mm}\). The coefficient of kinetic friction between the brake band and the drum is \(0.30 .\) Knowing that a force \(\mathrm{P}\) of magnitude \(50 \mathrm{N}\) is applied at \(A\) when the angular velocity is 180 rpm counterclockwise, determine the time required to stop the flywheel when \(a=200 \mathrm{mm}\) and \(b=160 \mathrm{mm} .\)

A sphere of radius \(r\) and mass \(m\) has a linear velocity \(\mathbf{v}_{0}\) directed to the left and no angular velocity as it is placed on a belt moving to the right with a constant velocity \(\mathbf{v}_{1}\). If after first sliding on the belt the sphere is to have no linear velocity relative to the ground as it starts rolling on the belt without sliding, determine in terms of \(v_{1}\) and the coefficient of kinetic friction \(\mu_{k}\) between the sphere and the belt \((a)\) the required value of \(v_{0},(b)\) the time \(t_{1}\) at which the sphere will start rolling on the belt, \((c)\) the distance the sphere will have moved relative to the ground at time \(t_{1}\).

The \(10-lb\) -uniform rod \(A B\) has a total length of \(2 L=2 \mathrm{ft}\) and is attached to collars of negligible mass that slide without friction along fixed rods. If rod \(A B\) is released from rest when \(\theta=30^{\circ},\) determine immediately after release (a) the angular acceleration of the rod, (b) the reaction at \(A .\)

A drum with a \(200-\mathrm{mm}\) radius is attached to a disk with a radius of \(r_{A}=150 \mathrm{mm} .\) The disk and drum have a combined mass of \(5 \mathrm{kg}\) and a combined radius of gyration of \(120 \mathrm{mm}\) and are suspended by two cords. Knowing that \(T_{A}=35 \mathrm{N}\) and \(T_{B}=25 \mathrm{N}\), determine the accelerations of points \(A\) and \(B\) on the cords.

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