Chapter 18.3, Problem 18.110P

To determine

(a)

The condition for steady precision.

Answer to Problem 18.110P

The condition for steady precision is $Wc=\left(I{\omega }_z-I^{\prime }\dot{\varphi }\cos \theta \right)\varphi$.

Explanation of Solution

Given information:

The below figure represent the schematic diagram of the system.

Figure-(1)

Write the expression angular velocity.

$\omega =-\dot{\varphi }\sin \theta i+\left(\dot{\psi }+\dot{\varphi }\cos \theta \right)k$ ........ (I)

Here, the precision rate is $\dot{\varphi }$, the spin rate is $\dot{\psi }$, the angle of the axis $BC$ of the sphare from vertical is $\theta$ and the unit vector in $x$ direction is $i$ and the unit vector in $z$ direction is $k$.

Write the expression of angular momentum about the centroid $G$.

$H_G=I_x{\omega }_{x}i+I_y{\omega }_{y}j+I_z{\omega }_{z}k$ ........ (II)

Here, the moment of inertia about the $x$ axis is $I_x$, the moment of inertia about the $y$ axis is $I_y$, the moment of inertia about the $z$ axis is $I_z$, the angular velocity about the $x$ axis is ${\omega }_x$, the angular velocity about the $y$ axis is ${\omega }_y$, the angular velocity about the $z$ axis is ${\omega }_z$ and the unit vector in $y$ direction is $j$.

Write the expression of angular momentum about the centroid $G$.

$H_G=-\left(I^{\prime }\dot{\varphi }\sin \theta \right)i+I{\omega }_{z}k$ ........ (III)

Here, the inertia about the reference frame is $I^{\prime }$.

Write the expression of angular velocity about the reference frame.

$\Omega =-\dot{\varphi }\sin \theta i+\dot{\varphi }\cos \theta k$ ........ (IV)

Write the expression of the total moment.

${\dot{H}}_G=\Omega \times H_G$ ......... (V)

Substitute $-\dot{\varphi }\sin \theta i+\dot{\varphi }\cos \theta k$ for $\Omega$ and $-\left(I^{\prime }\dot{\varphi }\sin \theta \right)i+I{\omega }_{z}k$ for $H_G$ in Equation (V).

$\begin{array}{c}{\dot{H}}_G=\left(-\dot{\varphi }\sin \theta i+\dot{\varphi }\cos \theta k\right)\times \left\{-\left(I^{\prime }\dot{\varphi }\sin \theta \right)i+I{\omega }_{z}k\right\}\\ =\left[\begin{array}{l}\left(-\dot{\varphi }\sin \theta i\right)\times \left(-I^{\prime }\dot{\varphi }\sin \theta \right)i+\left(-\dot{\varphi }\sin \theta i\right)\left(I{\omega }_{z}k\right)+\\ \left(\dot{\varphi }\cos \theta k\right)\times \left(-I^{\prime }\dot{\varphi }\sin \theta \right)i+\left(\dot{\varphi }\cos \theta k\right)\times \left(I{\omega }_{z}k\right)\end{array}\right]\\ =\left[-\left(\dot{\varphi }\sin \theta \right)\left(I{\omega }_z\right)j-\left(\dot{\varphi }\cos \theta \right)\left(-I^{\prime }\dot{\varphi }\sin \theta \right)j\right]\\ =\left(I^{\prime }\dot{\varphi }\sin \theta \cos \theta -I{\omega }_z\sin \theta \right)\varphi j\end{array}$ ........ (VI)

Write the expression of moment about $G$.

${\dot{H}}_G=-Wc\sin \theta j$ ......... (VII)

Here, the weight is $W$.

Substitute $-Wc\sin \theta j$ for ${\dot{H}}_G$ in Equation (VI).

$\begin{array}{l}-Wc\sin \theta j=\left(I^{\prime }\dot{\varphi }\sin \theta \cos \theta -I{\omega }_z\sin \theta \right)\varphi j\\ -Wc=\left(\frac{I^{\prime }\dot{\varphi }\sin \theta \cos \theta }{\sin \theta }-\frac{I{\omega }_z\sin \theta }{\sin \theta }\right)\varphi \\ Wc=\left(I{\omega }_z-I^{\prime }\dot{\varphi }\cos \theta \right)\varphi \end{array}$ ....... (VIII)

Conclusion:

The condition for steady precision is $Wc=\left(I{\omega }_z-I^{\prime }\dot{\varphi }\cos \theta \right)\varphi$.

To determine

(b)

The condition of steady precision if the rate of spin of the top is very large compared with its rate of precision.

Answer to Problem 18.110P

The condition of steady precision if the rate of spin of the top is very large compared with its rate of precision is $I\dot{\psi }\dot{\varphi }$.

Explanation of Solution

Write the expression of $z$ component of the angular velocity.

${\omega }_z=\varphi \cos \theta +\dot{\psi }$ ......... (IX)

Substitute $\varphi \cos \theta +\dot{\psi }$ for ${\omega }_z$ in Equation (VIII).

$\begin{array}{c}Wc=\left\{I\left(\varphi \cos \theta +\dot{\psi }\right)-I^{\prime }\dot{\varphi }\cos \theta \right\}\dot{\varphi }\\ =\left\{I\varphi \cos \theta +I\dot{\psi }-I^{\prime }\dot{\varphi }\cos \theta \right\}\dot{\varphi }\\ =\left\{I\dot{\psi }-\left(I^{\prime }-I\right)\varphi \cos \theta \right\}\dot{\varphi }\end{array}$ ........ (X)

Here, the term $\left(I^{\prime }-I\right)\varphi \cos \theta$ in Equation (X) is negligible, because the $\dot{\varphi }<<\dot{\psi }$.

$Wc=I\dot{\psi }\dot{\varphi }$ ....... (XI)

Conclusion:

The condition of steady precision if the rate of spin of the top is very large compared with its rate of precision is $I\dot{\psi }\dot{\varphi }$.

To determine

(c)

The percentage error.

Answer to Problem 18.110P

The percentage error is $-7.95\%$.

Explanation of Solution

Given information:

Mass of the top is $85\text{g}$, the radius of gyration of the top with respect to its symmetry is $21\text{mm}$, the radius of gyration of the top with respect to transverse axis is $45\text{mm}$, the distance between the point of precession and the centre of the gravity is $37.5\text{mm}$, the rate of the spin is $1800\text{rpm}$, and the angle between vertical and the axis of the top is $30°$.

Write the expression of weight.

$W=mg$ ........ (XII)

Here, the acceleration due to gravity is $g$.

Write the expression of pin rate.

$\dot{\psi }=\frac{2\pi N}{60}$ ........ (XIII)

Here, the number of revolution is $N$.

Write the expression of inertia about the transverse axis.

$I^{\prime }=m{k}_z{}^2$ ........ (XIV)

Here, the radius of gyration is $z$ direction is $k_z$.

Write the expression of inertia.

$I=m{k}_x{}^2$ ........ (XV)

Here, the radius of gyration is $x$ direction is $k_x$.

Write the expression of percentage error.

$\%\text{error=}\left(\frac{{\dot{\varphi }}_{\text{before}}-{\dot{\varphi }}_{\text{after}}}{{\dot{\varphi }}_{\text{after}}}\right)\times 100$ ........ (XVI)

Calculation:

Substitute $85\text{g}$ for $m$ and $9.81{\text{m}/\text{s}}^2$ for $g$ in Equation (XII).

$\begin{array}{c}W=\left(85\text{g}\right)\times \left(\frac{1\text{kg}}{1000\text{g}}\right)\left(9.81{\text{m}/\text{s}}^2\right)\\ =\left(0.085\text{kg}\right)\left(9.81{\text{m}/\text{s}}^2\right)\\ =\left(0.83385\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\right)\times \left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\\ =0.83385\text{N}\end{array}$

Substitute $1800\text{rpm}$ for $N$ in Equation (XIII).

$\begin{array}{c}\dot{\psi }=\frac{2\pi \left(1800\text{rpm}\right)}{60}\\ =\frac{11309.7335}{60}\text{rad}/\text{s}\\ =188.496\text{rad}/\text{s}\end{array}$

Substitute $85\text{g}$ for $m$ and $45\text{mm}$ for $k$ in Equation (XIV).

$\begin{array}{c}{I}^{\prime }=\left(85\text{g}\right){\left(45\text{mm}\right)}^2\\ =\left(85\text{g}\right)\times \left(\frac{1\text{kg}}{1000\text{g}}\right){\left\{\left(45\text{mm}\right)\times \left(\frac{1\text{m}}{1000\text{mm}}\right)\right\}}^2\\ =0.000172125\text{kg}\cdot {\text{m}}^{\text{2}}\end{array}$

Substitute $85\text{g}$ for $m$ and $21\text{mm}$ for $k$ in Equation (XIV).

$\begin{array}{c}I=\left(85\text{g}\right){\left(21\text{mm}\right)}^2\\ =\left(85\text{g}\right)\times \left(\frac{1\text{kg}}{1000\text{g}}\right){\left\{\left(21\text{mm}\right)\times \left(\frac{1\text{m}}{1000\text{mm}}\right)\right\}}^2\\ =0.000037485\text{kg}\cdot {\text{m}}^{\text{2}}\end{array}$

Substitute $0.83385\text{N}$ for $W$, $37.5\text{mm}$ for $c$, $0.000037485\text{kg}\cdot {\text{m}}^{\text{2}}$ for $I$, $0.000172125\text{kg}\cdot {\text{m}}^{\text{2}}$ for $I^{\prime }$, $188.496\text{rad}/\text{s}$ for $\dot{\psi }$ and $30°$ for $\theta$ in Equation (X).

$\begin{array}{l}\left(0.83385\text{N}\right)\left(37.5\text{mm}\right)=\left[\left\{\begin{array}{l}\left(0.000037485\text{kg}\cdot {\text{m}}^{\text{2}}\right)\left(188.496\text{rad}/\text{s}\right)\\ -\left(\begin{array}{l}0.000172125\text{kg}\cdot {\text{m}}^{\text{2}}-\\ 0.000037485\text{kg}\cdot {\text{m}}^{\text{2}}\end{array}\right)\varphi \cos 30°\end{array}\right\}\dot{\varphi }\right]\\ \left[\begin{array}{l}\left(0.83385\text{N}\right)\times \left(\frac{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}{1\text{N}}\right)\\ \times \left(37.5\text{mm}\right)\times \left(\frac{1\text{m}}{1000\text{mm}}\right)\end{array}\right]=\left[\left\{\begin{array}{l}\left(0.000037485\text{kg}\cdot {\text{m}}^{\text{2}}\right)\left(188.496\text{rad}/\text{s}\right)\\ -\left(\begin{array}{l}0.000172125\text{kg}\cdot {\text{m}}^{\text{2}}-\\ 0.000037485\text{kg}\cdot {\text{m}}^{\text{2}}\end{array}\right)\varphi \cos 30°\end{array}\right\}\dot{\varphi }\right]\\ \left(0.0001166\text{kg}\cdot {\text{m}}^{\text{2}}\right){\dot{\varphi }}^2-\left(0.0070658\text{kg}\cdot {\text{m}}^2/\text{s}\right)\dot{\varphi }+0.031269{\text{m}}^2/{\text{s}}^{\text{2}}=0\\ \dot{\varphi }=\left[\frac{\begin{array}{l}-\left(-0.0070658\text{kg}\cdot {\text{m}}^2/\text{s}\right)\pm \\ \sqrt{{\left(-0.0070658\text{kg}\cdot {\text{m}}^2/\text{s}\right)}^2-4\left(0.0001166\text{kg}\cdot {\text{m}}^{\text{2}}\right)\left(0.031269{\text{m}}^2/{\text{s}}^{\text{2}}\right)}\end{array}}{2\times \left(0.0001166\text{kg}\cdot {\text{m}}^{\text{2}}\right)}\right]\end{array}$

$\begin{array}{l}\dot{\varphi }=\frac{\left(0.0070658\text{kg}\cdot {\text{m}}^2/\text{s}\right)\pm 0.005945\text{kg}\cdot {\text{m}}^2/\text{s}}{2\times \left(0.0001166\text{kg}\cdot {\text{m}}^{\text{2}}\right)}\\ \dot{\varphi }=55.79\text{rad}/\text{s}\\ \dot{\varphi }=4.807\text{rad}/\text{s}\end{array}$

Substitute $0.83385\text{N}$ for $W$, $37.5\text{mm}$ for $c$, $0.000037485\text{kg}\cdot {\text{m}}^{\text{2}}$ for $I$, $0.000172125\text{kg}\cdot {\text{m}}^{\text{2}}$ for $I^{\prime }$, $188.496\text{rad}/\text{s}$ for $\dot{\psi }$ and $30°$ for $\theta$ in Equation (XI).

$\begin{array}{l}\left(0.83385\text{N}\right)\left(37.5\text{mm}\right)=\left(0.000037485\text{kg}\cdot {\text{m}}^{\text{2}}\right)\left(188.496\text{rad}/\text{s}\right)\dot{\varphi }\\ \dot{\varphi }=\frac{\left(0.83385\text{N}\right)\times \left(\frac{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}{1\text{N}}\right)\left(37.5\text{mm}\right)\times \left(\frac{1\text{m}}{1000\text{mm}}\right)}{\left(0.000037485\text{kg}\cdot {\text{m}}^{\text{2}}\right)\left(188.496\text{rad}/\text{s}\right)}\\ \dot{\varphi }=\frac{0.031269\text{kg}\cdot {\text{m}}^2/{\text{s}}^{\text{2}}}{0.00706577\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}}\\ \dot{\varphi }=4.425\text{rad}/\text{s}\end{array}$

Substitute $4.425\text{rad}/\text{s}$ for ${\dot{\varphi }}_{\text{before}}$ and $4.807\text{rad}/\text{s}$ for ${\dot{\varphi }}_{\text{after}}$ in Equation (XVI).

$\begin{array}{c}\%\text{error}=\left(\frac{4.425\text{rad}/\text{s}-4.807\text{rad}/\text{s}}{4.807\text{rad}/\text{s}}\right)\times 100\\ =-\left(\frac{0.382\text{rad}/\text{s}}{4.807\text{rad}/\text{s}}\right)\times 100\\ =-7.95\%\end{array}$

Conclusion:

The percentage error is $-7.95\%$.