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Chapter 6: Problem 37

The motor \(M\) is used to hoist the 12,000 -lb stadium panel (centroidal radius of gyration \(\bar{k}=6.5 \mathrm{ft}\) ) into position by pivoting the panel about its corner \(A\). If the motor is capable of producing 5000 lb-ft of torque, what pulley diameter \(d\) will give the panel an initial counterclockwise angular acceleration of \(1.5 \mathrm{deg} / \mathrm{sec}^{2} ?\) Neglect all friction.

Short Answer

Expert verified
The pulley diameter should be approximately 0.069 ft.

Step by step solution

01

Convert Units

Before solving the problem, convert the angular acceleration from degrees per second squared to radians per second squared. 1 degree = \(\frac{\pi}{180}\) radians, therefore \[1.5 \frac{\text{deg}}{\text{sec}^2} = 1.5 \times \frac{\pi}{180} \frac{\text{rad}}{\text{sec}^2} \approx 0.02618 \frac{\text{rad}}{\text{sec}^2}\]
02

Calculate Moment of Inertia

Use the formula for the moment of inertia \(I\) about point A, where \(I = I_G + md^2\). However, for a pivot about the corner, use \(I = m{\bar{k}}^2\) since it simplifies calculation for a body rotating about a corner. \[m = \frac{12000}{32.2} \approx 372.67 \, \text{slug} \] \[I = m{\bar{k}}^2 = 372.67 \times 6.5^2 \approx 15726.76 \, \text{slug} \cdot \text{ft}^2\]
03

Apply Torque Equation

The net torque \(\tau\) is the product of the moment of inertia \(I\) and angular acceleration \(\alpha\): \[\tau = I\alpha = 15726.76 \times 0.02618 \approx 411.87 \,\text{lb-ft}\]To maintain this torque with the pulley, we equate the motor torque (5000 lb-ft) to the effective torque,\[\tau = T - r F = 0\] or \[5000 = r \times mg \]Therefore, \[d = \frac{2 \times 411.87}{12000} = 0.0686 \, \text{ft}\]

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Key Concepts

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

Rotational Dynamics
Rotational dynamics explores the behavior of rotating bodies under various forces and torques. It extends the concepts of linear motion to rotational motion, incorporating angular counterparts to concepts like velocity, acceleration, and force.
The fundamental principles of rotational dynamics are analogous to Newton's laws of motion but revolve around rotations.
  • Angular displacement, velocity, and acceleration describe the movement of a rotating object.
  • Torque is the rotational analogue of force, responsible for changes in rotational motion.
  • The equations of motion in rotational dynamics often include terms for angular variables and consider the properties of the rotating body, such as its moment of inertia.
Understanding rotational dynamics is essential in engineering mechanics, particularly when analyzing systems where components move in a circular path.
Moment of Inertia
The moment of inertia, denoted by the symbol I, is a measure of an object's resistance to changes in its rotational motion. It plays a similar role to mass in linear dynamics.
The moment of inertia depends on the distribution of an object's mass relative to the axis of rotation.
  • In calculations, it's determined by using the formula: \[ I = ext{sum of}(mr^2) \]
  • Where m is the mass of individual particles of the object, and r is the distance of each particle from the axis of rotation.
  • For more complex shapes, specific formulas adapt to the shape's geometry and the axis of rotation.
In this exercise, the moment of inertia about point A was calculated using the centroidal radius of gyration, allowing for a direct assessment of rotational resistance.
Torque Calculation
Torque is the measure of how much a force acting on an object causes that object to rotate. Its magnitude is calculated using the cross product of the force and the radius at which it is applied.
This is expressed mathematically as:
  • \[ \tau = r \times F \]
  • Here, \(\tau\) is the torque, \(r\) is the radius, and \(F\) is the force applied perpendicular to the radius.
In the exercise, torque created by the motor is essential for determining how much force can be exerted to produce the required angular motion. Calculations focus on setting motor torque equal to the effective torque needed based on the mass and radius provided, ensuring the stadium panel achieves the desired angular acceleration.
Angular Acceleration
Angular acceleration, designated usually by \(\alpha\), is the rate of change of angular velocity over time. It's an important concept in rotational motion as it quantifies how quickly an object is speeding up or slowing down its rotation.
  • In terms of rotation, it can be calculated using:\[ \alpha = \frac{\Delta \omega}{\Delta t} \]
  • Where \(\Delta \omega\) is change in angular velocity, and \(\Delta t\) is the time interval during which this change occurs.
  • In our problem, angular acceleration was converted from degrees/second\(^2\) to radians/second\(^2\) to maintain standard units for computation.
This conversion is vital as it allows for consistency across all calculations, enabling accurate results in determining how quickly the motor must operate to achieve the desired speed of panel rotation.

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Most popular questions from this chapter

Each of the uniform bars \(O A\) and \(O B\) has a mass of \(2 \mathrm{kg}\) and is freely hinged at \(O\) to the vertical shaft, which is given an upward acceleration \(a=g / 2\). The links which connect the light collar \(C\) to the bars have negligible mass, and the collar slides freely on the shaft. The spring has a stiffness \(k=130 \mathrm{N} / \mathrm{m}\) and is uncompressed for the position equivalent to \(\theta=0 .\) Calculate the angle \(\theta\) assumed by the bars under conditions of steady acceleration.

The 75 -kg flywheel has a radius of gyration about its shaft axis of \(\bar{k}=0.50 \mathrm{m}\) and is subjected to the torque \(M=10\left(1-e^{-t}\right) \mathrm{N} \cdot \mathrm{m},\) where \(t\) is in seconds. If the flywheel is at rest at time \(t=0\) determine its angular velocity \(\omega\) at \(t=3\) s.

The wad of clay of mass \(m\) is initially moving with a horizontal velocity \(v_{1}\) when it strikes and sticks to the initially stationary uniform slender bar of mass \(M\) and length \(L\). Determine the final angular velocity of the combined body and the \(x\) -component of the linear impulse applied to the body by the pivot \(O\) during the impact.

The semicircular disk of mass \(m\) and radius \(r\) is released from rest at \(\theta=0\) and rotates freely in the vertical plane about its fixed bearing at \(O .\) Derive expressions for the \(n\) - and \(t\) -components of the force \(F\) on the bearing as functions of \(\theta\)

The sheave of 400 -mm radius has a mass of \(50 \mathrm{kg}\) and a radius of gyration of \(300 \mathrm{mm}\). The sheave and its 100 -kg load are suspended by the cable and the spring, which has a stiffness of \(1.5 \mathrm{kN} / \mathrm{m}\) If the system is released from rest with the spring initially stretched \(100 \mathrm{mm}\), determine the velocity of \(O\) after it has dropped \(50 \mathrm{mm}\)
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