Rigid Body Dynamics Questions and Answers

Answer: B

The principal moments are \(\frac{M(R_{1}^{2} + R_{2}^{2})}{2}\), \(\frac{M(R_{1}^{2} + R_{2}^{2})}{4}\), \(\frac{M(R_{1}^{2} + R_{2}^{2})}{4}\).

About the axis through the centre perpendicular to the disc, an annulus of mass M has

\[ \begin{aligned} I_{z} &= \frac{M}{\pi(R_{1}^{2} - R_{2}^{2})}\int_{R_{2}}^{R_{1}} 2\pi r^{3}dr \\ &= \frac{M(R_{1}^{2} + R_{2}^{2})}{2} \end{aligned} \]

For a flat body the perpendicular axis theorem gives \(I_{x} + I_{y} = I_{z}\), and by symmetry \(I_{x} = I_{y} = \frac{M(R_{1}^{2} + R_{2}^{2})}{4}\).

The options with \(R_{1}^{2} - R_{2}^{2}\) come from subtracting moments without keeping the total mass fixed, and the options with 8 break the perpendicular axis theorem.

Answer: D

A rigid body has 6 degrees of freedom.

A free rigid body in space is fixed by the position of one of its points, three coordinates (translation), and by its orientation about that point, three angles such as the Euler angles (rotation). Equivalently, three non-collinear points have 9 coordinates tied by 3 distance constraints, leaving \(9 - 3 = 6\).

1 is a body rotating about a fixed axis, 2 a particle on a surface, and 3 a free particle in space or a rigid body with one point fixed.

Answer: B

The \(z\)-component of the angular velocity is constant.

Euler's torque-free equation for the third principal axis is

\[ \begin{aligned} I_{zz}\dot{\omega}_{z} &= (I_{xx} - I_{yy}) \\ &\quad \times \omega_{x}\omega_{y} \end{aligned} \]

With \(I_{xx} = I_{yy}\) the right side vanishes, so \(\dot{\omega}_{z} = 0\) and \(\omega_{z}\) is constant.

It is the other two components that perform simple harmonic motion (they precess about the symmetry axis), so option A describes \(\omega_{x}\) and \(\omega_{y}\), not \(\omega_{z}\). \(\omega_{z}\) need not be zero, and it does not vary with time.

Answer: C

The radius of gyration is \((I/M)^{1/2}\).

The radius of gyration k is defined as the distance at which the whole mass could be placed to give the same moment of inertia: \(I = Mk^{2}\). Solving, \(k = \sqrt{I/M}\).

Only this option has the dimension of length. \((I/M)^{2}\) has dimension length to the fourth power, and the options with M/I are inverse lengths or worse.

Answer: B

Such a body is called a rotor.

When two principal moments are equal and the third is zero (\(I_{A} = 0\), \(I_{B} = I_{C}\)), all the mass lies on one axis, as in a diatomic or linear polyatomic molecule such as HCl or CO\(_{2}\). This is the rigid rotor of rotational spectroscopy, with energy levels \(E_{J} = \frac{\hbar^{2}}{2I}J(J + 1)\).

A symmetric top has two equal moments and a third that is non-zero (\(NH_{3}\)), a spherical top has all three equal (\(CH_{4}\)), and an asymmetric top has all three different (\(H_{2}O\)).

Answer: A

A rigid body with one point fixed has three degrees of freedom.

A free rigid body in space has six: three for the position of a reference point and three for its orientation. Fixing one point removes the three translational ones, leaving the three rotational coordinates, usually the Euler angles \(\phi\), \(\theta\), \(\psi\). This is the setting of the spinning top and of Euler's equations of rigid-body motion.

Six is the free body. With two coordinates the body could not reach every orientation about the fixed point, and one degree of freedom is a body rotating about a fixed axis, which needs two fixed points, not one.

Answer: A

The inverse is the transpose of A, option A.

A is orthogonal (its rows are orthonormal), so \(A^{-1} = A^{T}\). Transposing swaps the off-diagonal entries:

\[ A^{T} = \begin{pmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

As a check, \(AA^{T} = I\) because \(\cos^{2}\theta + \sin^{2}\theta = 1\). It is the rotation by \(-\theta\) about the z-axis, undoing the rotation by \(\theta\). Options C and D rotate about the x-axis instead, and option B permutes the rows.

Answer: A

The angular velocity is \(\sqrt{3g/2l}\).

Energy is conserved. The centre of mass, at \(l/2\) from O, falls by \(\frac{l}{2}(1 - \cos 60^{\circ}) = \frac{l}{4}\) as the stick turns through 60 degrees:

\[ \begin{aligned} mg\frac{l}{4} &= \frac{1}{2}\cdot\frac{ml^{2}}{3}\omega^{2} \\ \omega^{2} &= \frac{3g}{2l} \end{aligned} \]

\(\sqrt{2g/3l}\) inverts the factor \(\frac{3}{2}\), and \(\sqrt{2g/l}\) and \(\sqrt{g/2l}\) do not satisfy the energy equation for a 60 degree turn.

Answer: C

The moment of inertia about the x-axis is \(I_{xx} = \sum_{i} m_{i}(y_{i}^{2} + z_{i}^{2})\).

The moment of inertia about an axis uses the squared distance of each particle from that axis. The distance of \((x_{i}, y_{i}, z_{i})\) from the x-axis is \(\sqrt{y_{i}^{2} + z_{i}^{2}}\), which gives option C.

\(\sum m_{i}(x_{i}^{2} + y_{i}^{2})\) is \(I_{zz}\) and \(\sum m_{i}(x_{i}^{2} + z_{i}^{2})\) is \(I_{yy}\). A negative sum like \(-\sum m_{i} x_{i}^{2}\) resembles a product of inertia such as \(I_{xy} = -\sum m_{i} x_{i}y_{i}\), which carries the minus sign in this convention, but a moment of inertia is never negative.