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Chapter 21: Problem 41

Derive the scalar form of the rotational equation of motion about the \(x\) axis if \(\Omega \neq \omega\) and the moments and products of inertia of the body are constant with respect to time.

Short Answer

Expert verified
The scalar form of the rotational equation of motion about the \(x\) axis is \(M = I\omega\)

Step by step solution

01

Understanding the terms

Understand the given: The moments and products of inertia are constant over time. Angular velocity vector (\(\Omega\)) and angular velocity (\(\omega\)) are different, and \(\Omega 鈮� \omega\). The rotational equation of motion is sought for the x-axis.
02

Start with the general equation

Use the general rotational motion equation, given by the second derivative of the vector representing position with respect to time (often expressed as \(\frac{d^2\vec{r}}{dt^2}\)) equals to \(\vec{\Omega} \times (\vec{\Omega} \times \vec{r}) - 2\vec{\Omega} \times \frac{d\vec{r}}{dt}\).
03

Transform to scalar form and specify axis

In scalar form, the equation becomes: \(\frac{dI}{dt} = \Omega x M - I x \omega\) where I is moment of inertia, \(\Omega\) is the angular velocity vector, M is the net torque on the body, and \(\omega\) is the angular velocity. Now, focus only on the x-axis, assuming the moments of inertia along y and z axis have no contribution towards rotation around the x-axis.
04

Apply given conditions

Given that the moments and products of inertia are constant regarding time, this leads to a derivation of \(\frac{dI}{dt} = 0\). Therefore, for the x-axis, the equation becomes: \(\Omega_x(I\omega - M) = 0\).
05

Simplify and solve

As it is given \(\Omega 鈮� \omega\) or \(\Omega_x 鈮� \omega_x\), \(\Omega_x 鈮� 0\), therefore the simplified equation of motion about the \(x\) axis becomes \(I\omega - M = 0\) or \(M = I\omega\) which is scalar form of the rotational equation of motion about the \(x\) axis.

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Most popular questions from this chapter

The blades of a wind turbine spin about the shaft \(S\) with a constant angular speed of \(\omega_{s}\), while the frame precesses about the vertical axis with a constant angular speed of \(\omega_{p-}\) Determine the \(x, y\), and \(z\) components of moment that the shaft exerts on the blades as a function of \(\theta\). Consider each blade as a slender rod of mass \(m\) and length \(l\).

The circular disk has a weight of \(15 \mathrm{lb}\) and is mounted on the shaft \(A B\) at an angle of \(45^{\circ}\) with the horizontal. Determine the angular velocity of the shaft when \(t=3 \mathrm{~s}\) if a constant torque \(M=2 \mathrm{lb} \cdot \mathrm{ft}\) is applied to the shaft. The shaft is originally spinning at \(\omega_{1}=8 \mathrm{rad} / \mathrm{s}\) when the torque is applied.

The football has a mass of \(450 \mathrm{~g}\) and radii of gyration about its axis of symmetry ( \(z\) axis) and its transverse axes ( \(x\) or \(y\) axis) of \(k_{z}=30 \mathrm{~mm}\) and \(k_{x}=k_{y}=50 \mathrm{~mm}\), respectively. If the football has an angular momentum of \(H_{G}=0.02 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\), determine its precession \(\dot{\phi}\) and spin \(\dot{\psi}\). Also, find the angle \(\beta\) that the angular velocity vector makes with the \(z\) axis

If a body contains no planes of symmetry, the principal moments of inertia can be determined mathematically. To show how this is done, consider the rigid body which is spinning with an angular velocity \(\omega\), directed along one of its principal axes of inertia. If the principal moment of inertia about this axis is \(I\), the angular momentum can be expressed as \(\mathbf{H}=I \omega=I \omega_{x} \mathbf{i}+I \omega_{y} \mathbf{j}+I \omega_{z} \mathbf{k} .\) The components of \(\mathbf{H}\) may also be expressed by Eqs. \(21-10\), where the inertia tensor is assumed to be known. Equate the \(\mathbf{i}, \mathbf{j}\), and \(\mathbf{k}\) components of both expressions for \(\mathbf{H}\) and consider \(\omega_{x}, \omega_{y}\), and \(\omega_{z}\) to be unknown. The solution of these three equations is obtained provided the determinant of the coefficients is zero. Show that this determinant, when expanded, yields the cubic equation $$ \begin{aligned} &P^{3}-\left(I_{x x}+I_{y y}+I_{z}\right) I^{2} \\ &\quad+\left(I_{x x} I_{y y}+I_{y y} I_{z z}+I_{z z} I_{x x}-I_{x y}^{2}-I_{y z}^{2}-I_{z x}^{2}\right) I \\ &\quad-\left(I_{x x} I_{y y} I_{z z}-2 I_{x y} I_{y z} I_{z x}-I_{x x} I_{y z}^{2}-I_{y y} I_{z x}^{2}-I_{z z} I_{x y}^{2}\right)=0 \end{aligned} $$ The three positive roots of \(I\), obtained from the solution of this equation, represent the principal moments of inertia \(I_{x}\), \(I_{y}\), and \(I_{z}\).

The satellite has a mass of \(1.8 \mathrm{Mg}\), and about axes passing through the mass center \(G\) the axial and transverse radii of gyration are \(k_{z}=0.8 \mathrm{~m}\) and \(k_{t}=1.2 \mathrm{~m}\), respectively. If it is spinning at \(\omega_{s}=6 \mathrm{rad} / \mathrm{s}\) when it is launched, determine its angular momentum. Precession occurs about the \(Z\) axis.
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