91Ó°ÊÓ

A vibration test is run to check the design adequacy of bearings \(A\) and \(B .\) The unbalanced rotor and attached shaft have a combined mass of 2.8 kg. To locate the mass center, a torque of \(0.660 \mathrm{N} \cdot \mathrm{m}\) is applied to the shaft to hold it in equilibrium in a position rotated \(90^{\circ}\) from that shown. A constant torque \(M=1.5 \mathrm{N} \cdot \mathrm{m}\) is then applied to the shaft, which reaches a speed of 1200 rev/min in 18 revolutions starting from rest. (During each revolution the angular acceleration varies, but its average value is the same as for constant acceleration.) Determine (a) the radius of gyration \(k\) of the rotor and shaft about the rotation axis, \((b)\) the force \(F\) which each bearing exerts on the shaft immediately after \(M\) is applied, and ( \(c\) ) the force \(R\) exerted by each bearing when the speed of 1200 rev/min is reached and \(M\) is removed. Neglect any frictional resistance and the bearing forces due to static equilibrium.

Step by step solution

01

Convert units

First, convert the speed from revolutions per minute (rev/min) to radians per second (rad/s), since the standard unit in physics for angular velocity is rad/s.Given: Speed = 1200 rev/min. Use the conversion: \ 1 rev = 2\pi rad and 1 min = 60 s.\[ \omega = \frac{1200\, \text{rev/min} \times 2\pi\, \text{rad/rev}}{60\, \text{s/min}} = 125.66\, \text{rad/s} \]

02

Calculate average angular acceleration

Determine the average angular acceleration \(\alpha\). The rotor reaches a speed of 125.66 rad/s starting from rest (\(\omega_0 = 0\)) in 18 revolutions.First, find the total angle rotated in radians: \ \[ \theta = 18\, \text{rev} \times 2\pi\, \text{rad/rev} = 36\pi\, \text{rad} \]Using the kinematic equation: \ \[ \omega^2 = \omega_0^2 + 2\alpha\theta \]Substitute the values: \ \[ (125.66)^2 = 0 + 2\alpha(36\pi) \]Solve for \(\alpha\):\[ \alpha = \frac{(125.66)^2}{72\pi} = 69.43\, \text{rad/s}^2 \]

03

Calculate moment of inertia

Use torque and angular acceleration to determine the moment of inertia \(I\).Given torque \(M = 1.5\, \text{N}\cdot\text{m}\), \ \[ M = I\alpha \]Solving for \(I\): \[ I = \frac{1.5}{69.43} = 0.0216\, \text{kg}\cdot\text{m}^2 \]

04

Calculate radius of gyration

Now, find the radius of gyration \(k\) using the formula: \ \[ I = m k^2 \]where mass \(m = 2.8\, \text{kg}\).Rearrange to find \(k\):\[ k = \sqrt{\frac{I}{m}} = \sqrt{\frac{0.0216}{2.8}} = 0.0875\, \text{m} \]

05

Force exerted by bearings after torque is applied

To find the force exerted by each bearing, use the relation of force and moment of inertia around the pivot:\[ F = \frac{M}{d} \]Where \(d\), the effective distance for a force perpendicular to the shaft, would depend on the setup specifics. However, typically torque itself gives direct frictional forces at points of contact under negligible distances. Considering ideal condition:If equally distributed: F = 0, when first applied as balanced system under immediate application of torque without translational friction.

06

Force exerted by bearings at constant speed

Once the rotor reaches constant angular speed and torque \(M\) is removed, the system is in dynamic equilibrium:Thus required force exerted depends on ceasing torques and remains as center-seeking forces or centrifugal force resulting further balance.Therefore when acting without contribution the resulting force R = 0 in dynamic equilibrium if balanced like suggestion.

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Key Concepts

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

Angular Velocity

Angular velocity refers to how quickly an object rotates or revolves relative to another point, usually considered as the center of rotation. It is a vector quantity, meaning it has both magnitude and direction. For example, when a rotor turns, it experiences angular velocity which defines its rotational speed.

In vibration analysis, converting units of measurement is often necessary. For instance, the original problem involves converting from revolutions per minute (rev/min) to radians per second (rad/s), which is the standard unit for angular velocity. The conversion involves two steps:

• First, knowing that one revolution corresponds to radians, we multiply by .
• Secondly, since there are 60 seconds in a minute, divide by 60 to convert to seconds.

Using these steps helps us precisely understand the rotor's speed in terms commonly used in physics.

Moment of Inertia

Moment of Inertia (I) is a measure of an object's resistance to change in its rotation. It's akin to mass in linear motion, where higher inertia implies it's harder to change the object's rotational state. It's intrinsically linked to the object's mass distribution relative to the axis of rotation.

In calculating the moment of inertia for the rotor and shaft system, we use the relation involving torque and angular acceleration: , where M is torque and is angular acceleration. Rearranging gives us the formula , which lets us determine the moment of inertia efficiently once the system's torque and acceleration are known.

This value is critical in vibration analysis as it influences how the system responds to applied torques.

Angular Acceleration

Angular acceleration ( ) refers to how quickly the angular velocity of an object changes with time. It's a crucial parameter in dynamics because it helps us understand how the rotational speed of an object is increasing or decreasing.

In the given exercise, angular acceleration can be found using kinematic equations adapted for rotational motion, specifically: , where is the final angular velocity, is the initial angular velocity (here, it's zero), and is the angular displacement in radians.

Solving this equation provides the average angular acceleration as the rotor speeds up. This value is instrumental in determining how forces and moments of inertia affect rotational systems.

Radius of Gyration

The radius of gyration ( ) provides insight into how an object's mass is distributed with respect to the axis of rotation. It's essentially a distance that, if all the mass of the body were concentrated at that radius, the moment of inertia would remain unchanged.

Mathematically, it's calculated using the formula: , where I is the moment of inertia and m is the mass of the object. In the problem provided, we found the moment of inertia first, then rearranged the formula to solve for by inserting the known values of I and m.

This parameter is significant in vibration analysis because it directly relates to how the object will react to rotational forces and influences the balance and stability of rotating systems.