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Chapter 7: Problem 78

The \(6-\mathrm{kg}\) circular disk and attached shaft rotate at a constant speed \(\omega=10000\) rev/min. If the center of mass of the disk is \(0.05 \mathrm{mm}\) off center, determine the magnitudes of the horizontal forces \(A\) and \(B\) supported by the bearings because of the rotational imbalance.

Short Answer

Expert verified
The forces at bearings A and B are each 8.235 N.

Step by step solution

01

Convert rotational speed to radians per second

First, convert the rotational speed from revolutions per minute to radians per second using the formula: \[ \omega_{rad/s} = \omega_{rev/min} \times \frac{2\pi}{60} \]. Substitute the given \( \omega = 10000 \) rev/min: \[ \omega_{rad/s} = 10000 \times \frac{2\pi}{60} \approx 1047.2 \text{ rad/s}. \]
02

Determine the unbalanced mass

Calculate the unbalanced mass using the concept that for a body in rotational motion slightly off-center, the effective unbalanced mass is the total mass multiplied by the offset. Given that the center of mass is off by 0.05 mm, the unbalanced mass can be seen as a point mass at this distance: \[ m_{eff} = m \times d = 6 \times 0.00005 = 0.0003 \text{ kg m}. \]
03

Calculate centrifugal force due to imbalance

Centrifugal force due to the offset center is calculated using \( F_c = m_{eff} \times \omega^2 \times r \). Here, \( r = 0.00005 \text{ m} \) is the off-center distance: \[ F_c = 0.0003 \times (1047.2)^2 \times 0.00005 = 16.47 \text{ N}. \]
04

Determine forces at bearings A and B

Since the disk and shaft are symmetric and the unbalance acts in the horizontal plane, the horizontal forces at the bearings (\( A \) and \( B \)) due to the imbalance are equal. Therefore: \[ F_A = F_B = \frac{F_c}{2} = \frac{16.47}{2} = 8.235 \text{ N}. \]

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Key Concepts

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

Centrifugal Force
Centrifugal force is a fundamental concept in rotational dynamics that describes the apparent force felt by an object moving in a circular path. This force seems to "push" an object away from the center of its circular path.

In our example, which involves a rotating disk, the centrifugal force arises because the center of mass is slightly off-center. As the disk rotates, the unbalanced mass experiences a force directed outward.

This outward force can be calculated using the formula:
  • \( F_c = m_{eff} \times \omega^2 \times r \).
In this formula, \( m_{eff} \) is the effective unbalanced mass, \( \omega \) is the angular velocity in radians per second, and \( r \) is the radius or the offset distance.

Understanding centrifugal force is crucial for solving problems related to systems with rotational motion, especially when imbalances are present, causing additional stress on components such as bearings.
Rotational Imbalance
Rotational imbalance occurs when an object's mass is not evenly distributed around its axis of rotation. This imbalance can cause vibrations and additional forces during rotation.

In our scenario, the imbalance arises because the disk's center of mass is just 0.05 mm off its axis of rotation. This slight deviation can significantly affect the overall performance of the system, especially at high speeds.

To address rotational imbalances, engineers often assess:
  • The center of mass offset
  • The mass distribution
  • High-speed rotation effects
By calculating the effective unbalanced mass and using it to determine the centrifugal force, engineers can predict and mitigate the effects of this imbalance, ensuring smoother operation and less wear on components.
Bearing Forces
Bearing forces are essential to understand in any rotating system, as these components support and stabilize the rotating masses.

In the example provided, the horizontal forces \( A \) and \( B \) represent the support forces on the bearings caused by the rotational imbalance.

The forces at the bearings due to the imbalance were determined by calculating the centrifugal force and dividing it equally among the bearings. Therefore:
  • \( F_A = F_B = \frac{F_c}{2} \)
These forces are crucial to analyze to ensure that the bearings can withstand the loads imposed by the rotating parts. Properly assessing and compensating for these forces can prevent damage and extend the lifespan of the machinery.

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Most popular questions from this chapter

Each of the slender rods of length \(l\) and mass \(m\) is welded to the circular disk which rotates about the vertical \(z\) -axis with an angular velocity \(\omega .\) Each rod makes an angle \(\beta\) with the vertical and lies in a plane parallel to the \(y-z\) plane. Determine an expression for the angular momentum \(\mathbf{H}_{O}\) of the two rods about the origin \(O\) of the axes.

The thin circular disk of mass \(m\) and radius \(R\) is hinged about its horizontal tangent axis to the end of a shaft rotating about its vertical axis with an angular velocity \(\omega .\) Determine the steady-state angle \(\beta\) assumed by the plane of the disk with the vertical axis. Observe any limitation on \(\omega\) to ensure that \(\beta>0\)

An experimental car is equipped with a gyro stabilizer to counteract completely the tendency of the car to tip when rounding a curve (no change in normal force between tires and road). The rotor of the gyro has a mass \(m_{0}\) and a radius of gyration \(k\) and is mounted in fixed bearings on a shaft which is parallel to the rear axle of the car. The center of mass of the car is a distance \(h\) above the road, and the car is rounding an unbanked level turn at a speed \(v .\) At what speed \(p\) should the rotor turn and in what direction to counteract completely the tendency of the car to overturn for either a right or a left turn? The combined mass of car and rotor is \(m\)

The collar and clevis \(A\) are given a constant upward velocity of 8 in./sec for an interval of motion and cause the ball end of the bar to slide in the radial slot in the rotating disk. Determine the angular acceleration of the bar when the bar passes the position for which \(z=3\) in. The disk turns at the constant rate of 2 rad/sec.

The circular disk of radius \(r\) is mounted on its shaft which is pivoted at \(O\) so that it may rotate about the vertical \(z_{0}\) -axis. If the disk rolls at constant speed without slipping and makes one complete turn around the circle of radius \(R\) in time \(\tau\) determine the expression for the absolute angular velocity \(\omega\) of the disk. Use axes \(x-y-z\) which rotate around the \(z_{0}\) -axis. (Hint: The absolute angular velocity of the disk equals the angular velocity of the axes plus (vectorially) the angular velocity relative to the axes as seen by holding \(x-y-z\) fixed and rotating the circular disk of radius \(R\) at the rate of \(2 \pi / \tau\).)
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