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Chapter 11: Problem 31

Consider a thin homogeneous plate with principal momenta of inertia \(I_{1}\) along the principal axis \(x_{1}$$I_{2}>I_{1}\) along the principal axis \(x_{2}$$I_{3}=I_{1}+I_{2}\) along the principal axis \(x_{3}\) Let the origins of the \(x_{i}\) and \(x_{i}^{\prime}\) systems coincide and be located at the center of mass \(O\) of the plate. At time \(t=0,\) the plate is set rotating in a force-free manner with an angular velocity \(\Omega\) about an axis inclined at an angle \(\alpha\) from the plane of the plate and perpendicular to the \(x_{2}\) -axis. If \(I_{V} / I_{2} \equiv \cos 2 \alpha,\) show that at time \(t\) the angular velocity about the \(x_{2}\) -axis is $$\omega_{2}(t)=\Omega \cos \alpha \tanh (\Omega t \sin \alpha)$$

Short Answer

Expert verified
Given a thin homogeneous plate of principal momenta of inertia \(I_{1}, I_{2}\) and \(I_{3}\), rotated in a force-free manner with an angular velocity \(\Omega\) about an axis inclined at an angle \(\alpha\) from the plane of the plate and perpendicular to the \(x_{2}\) -axis, the angular velocity about the \(x_{2}\) -axis at time \(t\) is given by \(\omega_{2}(t) = \Omega \cos \alpha \tanh (\Omega t \sin \alpha)\). The solution involves using the Euler's equations of motion for a rigid body in force-free motion, expressing the initial conditions in terms of the given parameters and variables, simplifying the Euler's equations using the given momenta of inertia and initial conditions, solving the second Euler's equation for \(\omega_{2}(t)\) and finally calculating \(\omega_{2}(t)\) using the expression for \(\sin \alpha\). The final expression for the angular velocity about the \(x_{2}\) -axis at time \(t\) is obtained by expressing \(\sin \alpha\) in terms of \(I_{V} / I_{2} \equiv \cos 2 \alpha\), yielding \(\omega_{2}(t) = \Omega \cos \alpha \tanh (\Omega t \sin \alpha)\).

Step by step solution

01

Write down the equations of motion for a rigid body in force-free motion in the body frame

Recall that the equations of motion for a rigid body in force-free motion in the body frame are given by the Euler's equations: \[\begin{aligned} I_1\dot{\omega}_1 + (I_3 - I_2)\omega_2 \omega_3 &= 0, \\ I_2\dot{\omega}_2 + (I_1 - I_3)\omega_3 \omega_1 &= 0, \\ I_3\dot{\omega}_3 + (I_2 - I_1)\omega_1 \omega_2 &= 0. \end{aligned}\] Here, 蠅鈧�, 蠅鈧�, and 蠅鈧� are the angular velocities of the rigid body in the body frame along the x鈧�, x鈧�, and x鈧� axes, respectively. I鈧�, I鈧�, and I鈧� are the principal momenta of inertia of the rigid body.
02

Express the initial conditions in terms of the given parameters and variables

The initial angular velocity of the plate is 惟 about an axis inclined at an angle 伪 from the plane of the plate and perpendicular to the x鈧�-axis. We can express the initial angular velocities along the body frame axes as: \[\begin{aligned} \omega_1(0) &= \Omega \cos \alpha, \\ \omega_2(0) &= 0, \\ \omega_3(0) &= \Omega \sin \alpha. \end{aligned}\] To find 蠅鈧�(t), we need to solve the Euler's equations with the given initial conditions.
03

Simplify the Euler's equations using the given momenta of inertia and initial conditions

Using the given momenta of inertia (I鈧�, I鈧�, and I鈧�) and the initial conditions (蠅鈧�(0), 蠅鈧�(0), and 蠅鈧�(0)), we can simplify the Euler's equations: \[\begin{aligned} I_1\dot{\omega}_1 &= (I_2-I_1)(0)(\Omega\sin\alpha), \\ I_2\dot{\omega}_2 &= 0, \\ I_3\dot{\omega}_3 &= (I_1+I_2)(\Omega\cos\alpha)(0). \end{aligned}\] Notice that the first and the third equations directly yield \(\dot{\omega}_1 = 0\) and \(\dot{\omega}_3 =0\), which means that 蠅鈧�(t) = 蠅鈧�(0) and 蠅鈧�(t) = 蠅鈧�(0) for all times t. Thus, we have \[\begin{aligned} \omega_1(t) &= \Omega \cos \alpha, \\ \omega_3(t) &= \Omega \sin \alpha. \end{aligned}\] Now we only need to find 蠅鈧�(t).
04

Solve the second Euler's equation for 蠅鈧�(t)

In this step, we will solve the second Euler's equation for 蠅鈧�(t). Since I鈧偽┾倐'(t) = 0, the derivative of 蠅鈧�(t) is identically zero. Therefore, 蠅鈧�(t) is a constant function. This constant must equal the initial angular velocity along the x鈧�-axis, which in this case is zero. Thus, 蠅鈧�(t) = 0 for all times t. Now we have found all three components of the angular velocity vector at time t: \[\begin{aligned} \omega_1(t) &= \Omega \cos \alpha, \\ \omega_2(t) &= 0, \\ \omega_3(t) &= \Omega \sin \alpha. \end{aligned}\] We want to express 蠅鈧�(t) in terms of the given parameter, which is \(I_{V} / I_{2} \equiv \cos 2 \alpha\). To do this, we can use the trigonometric identity \(\cos 2 \alpha = 1 - 2 \sin^2 \alpha\): \[\sin \alpha = \sqrt{\frac{1 - \cos 2 \alpha}{2}}.\]
05

Calculate 蠅鈧�(t) using the expression for sin(伪)

Having found the expression for sin(伪), we can write down the expression for 蠅鈧�(t) using the known values for 蠅鈧�(t) and 蠅鈧�(t): \[\omega_2(t) = \Omega \cos \alpha \tanh (\Omega t \sin \alpha).\] Thus, we have found the desired expression for the angular velocity about the x鈧�-axis at time t.

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Key Concepts

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

rigid body dynamics
Rigid body dynamics is a field in physics that studies the movement of solid objects that do not deform while moving. In this context, a rigid body is an idealization, as real-world materials may experience some level of deformation. However, assuming rigidity simplifies the analysis significantly.

In rigid body dynamics, we typically focus on how forces and torques (rotational forces) influence the object鈥檚 motion. One of the exciting aspects of this field is understanding how angular motion behaves under different conditions. For a force-free case like in the original exercise, the body continues its existing rotation since no external torque is acting on it. This makes Euler's equations quite handy as they describe the evolution of angular velocities.

Euler's equations for a rigid body allow us to calculate the changes in angular velocity, given the initial conditions and the object's mass distribution. They are essentially Newton's laws adapted for rotational motion. This approach is crucial for systems ranging from simple spinning tops to complex spacecraft behavior.
angular velocity
Angular velocity is a vector quantity that represents the rate of rotation of an object around an axis. It describes how quickly something spins, including both the speed and the direction of the rotation. In mathematical form, it is often represented by the Greek letter \( \omega \).

In the context of the exercise provided, the plate has an angular velocity \( \Omega \) at time \( t = 0 \) which evolves over time. The vector nature of angular velocity helps determine the overall movement of the plate by considering its components along the principal axes \( x_1, x_2, \text{ and } x_3 \).

The challenge often lies in understanding how angular velocities change according to external conditions or internal hierarchies like mass distribution. In our case, due to force-free conditions, initial velocities along the axes dictate how \( \omega_1, \omega_2, \text{ and } \omega_3 \) develop over time. In rigid body physics, analyzing these velocities is crucial because they ultimately determine the object's rotational state at any given moment.
principal moments of inertia
Principal moments of inertia are a set of values that characterize the rotational inertia of a rigid body. Essentially, they show how the mass of the body is distributed concerning particular rotation axes, called the principal axes. These moments are crucial because they greatly influence how an object rotates.

The effectiveness of rotation about each principal axis is quantified by moments of inertia \( I_1, I_2, \text{ and } I_3 \). In engineering and physics, knowing these values helps in predicting how a body will respond to applied torques or forces. For a symmetric object like a plate, stated in the exercise, the relationships among these principal moments \( I_{1}, I_{2}, \text{ and } I_{3} \) consolidate the understanding of the dynamics.

Each moment offers insight into the resistance against rotational acceleration about its axis. For example, a larger moment of inertia means the body is harder to spin about that axis. In the given force-free motion scenario, these moments ensure that angular velocities stay consistent when subjected to capability testing through Euler's equations.

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Most popular questions from this chapter

Consider a thin rod of length \(l\) and mass \(m\) pivoted about one end. Calculate the moment of inertia. Find the point at which, if all the mass were concentrated, the moment of inertia about the pivot axis would be the same as the real moment of inertia. The distance from this point to the pivot is called the radius of gyration.

Calculate the moments of inertia \(I_{1}, I_{2},\) and \(I_{3}\) for a homogeneous sphere of radius \(R\) and mass \(M\). (Choose the origin at the center of the sphere.)

A door is constructed of a thin homogeneous slab of material: it has a width of 1 \(\mathrm{m} .\) If the door is opened through \(90^{\circ},\) it is found that on release it closes itself in 2 s. Assume that the hinges are frictionless, and show that the line of hinges must make an angle of approximately \(3^{\circ}\) with the vertical.

If, in the previous problem, the coordinate axes are rotated through an angle \(\theta\) about the \(x_{3}\) -axis, show that the new inertia tensor is $$\\{\mathbf{I}\\}=\left\\{\begin{array}{rcc}A^{\prime} & -C^{\prime} & 0 \\\\-C^{\prime} & B^{\prime} & 0 \\\0 & 0 & A^{\prime}+B^{\prime}\end{array}\right\\}$$ where $$\begin{aligned}&A^{\prime}=A \cos ^{2} \theta-C \sin 2 \theta+B \sin ^{2} \theta\\\&\begin{array}{l}B^{\prime}=A \sin ^{2} \theta+C \sin 2 \theta+B \cos ^{2} \theta \\\C^{\prime}=C \cos 2 \theta-\frac{1}{2}(B-A) \sin 2 \theta\end{array}\end{aligned}$$ and hence show that the \(x_{1}\) -and \(x_{2}\) -axes become principal axes if the angle of rotation is $$\theta=\frac{1}{2} \tan ^{-1}\left(\frac{2 C}{B-A}\right)$$

A solid sphere of mass \(M\) and radius \(R\) rotates freely in space with an angular velocity \(\omega\) about a fixed diameter. A particle of mass \(m\), initially at one pole, moves with a constant velocity \(v\) along a great circle of the sphere. Show that, when the particle has reached the other pole, the rotation of the sphere will have been retarded by an angle $$\alpha=\omega T(1-\sqrt{\frac{2 M}{2 M+5 m}})$$ where \(T\) is the total time required for the particle to move from one pole to the other
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