Rigid Body Dynamics Questions and Answers

Q: A disc of radius \(R_{1}\) having uniform surface density has a concentric hole of radius \(R_{2} \lt R_{1}\). If its mass is M, the principal moments of inertia are

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.

Q: The degrees of freedom of a rigid body are :

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.

Q: Euler's equation of motion for a rigid body about a fixed point in the absence of any net torque and \(I_{xx} = I_{yy}\) imply that the \(z\)-component of the angular velocity :

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.

Q: The radius of gyration of a rigid body of mass M and moment of inertia I is :

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.

Q: A rigid body, whose two of the principal moments of inertia are equal and the third is zero, is called as :

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 (\(H_{2}O\)).

A disc of radius \(R_{1}\) having uniform surface density has a concentric hole of radius \(R_{2} \lt R_{1}\). If its mass is M, the principal moments of inertia are

A.\(\frac{M(R_{1}^{2} - R_{2}^{2})}{2}\), \(\frac{M(R_{1}^{2} - R_{2}^{2})}{4}\), \(\frac{M(R_{1}^{2} - R_{2}^{2})}{4}\)
B.\(\frac{M(R_{1}^{2} + R_{2}^{2})}{2}\), \(\frac{M(R_{1}^{2} + R_{2}^{2})}{4}\), \(\frac{M(R_{1}^{2} + R_{2}^{2})}{4}\)
C.\(\frac{M(R_{1}^{2} + R_{2}^{2})}{2}\), \(\frac{M(R_{1}^{2} + R_{2}^{2})}{4}\), \(\frac{M(R_{1}^{2} + R_{2}^{2})}{8}\)
D.\(\frac{M(R_{1}^{2} - R_{2}^{2})}{2}\), \(\frac{M(R_{1}^{2} - R_{2}^{2})}{4}\), \(\frac{M(R_{1}^{2} - R_{2}^{2})}{8}\)

Exam Year(s)

IIT JAM - 2018,

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.

Discuss in Forum

The degrees of freedom of a rigid body are :

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

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.

Discuss in Forum

Euler's equation of motion for a rigid body about a fixed point in the absence of any net torque and \(I_{xx} = I_{yy}\) imply that the \(z\)-component of the angular velocity :

A.satisfies simple harmonic motion
B.is constant
C.is always zero
D.is a function of time

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

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.

Discuss in Forum

The radius of gyration of a rigid body of mass M and moment of inertia I is :

A.\((I/M)^{2}\)
B.\((M/I)^{2}\)
C.\((I/M)^{1/2}\)
D.\((M/I)^{1/2}\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

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.

Discuss in Forum

A rigid body, whose two of the principal moments of inertia are equal and the third is zero, is called as :

A.Asymmetrical top
B.Rotor
C.Spherical top
D.Symmetrical top

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

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 (\(H_{2}O\)).

Rigid Body Dynamics Questions and Answers

Topics

Other Subjects

Tags: Rigid Body Dynamics Questions and Answers

About Us

Rigid Body Dynamics Questions and Answers

Topics

Other Subjects

Tags: Rigid Body Dynamics Questions and Answers

About Us

Stay informed about all important Dates & Notifications