Chapter 10, Problem 10.33P

(a)

To determine

Prove that the kinetic energy can be written as, $T=\frac{1}{2}{\displaystyle \sum m_{\alpha }}\left[{\left(\omega r_{\alpha }\right)}^2-{\left(\omega \cdot r_{\alpha }\right)}^2\right]$.

Answer to Problem 10.33P

it is proved that the kinetic energy can be written as, $T=\frac{1}{2}{\displaystyle \sum m_{\alpha }}\left[{\left(\omega r_{\alpha }\right)}^2-{\left(\omega \cdot r_{\alpha }\right)}^2\right]$.

Explanation of Solution

Let the mass of each particle of the rigid body is $m_a$ and each particle is rotating about an  axis at a distance from them.

Write the expression for the angular velocity of rotation about an axis of each particle.

    $V_{\alpha }=\omega \times r_{\alpha }$        (I)

Here, $V_{\alpha }$ is the angular velocity of rotation about an axis of each particle, $r_{\alpha }$ is the distance traveled.

Write the expression for the kinetic energy.

    $T_{\alpha }=\frac{1}{2}{m}_{\alpha }{v}_{\alpha }^2$        (II)

Here, $T_{\alpha }$ is the kinetic energy, $m_{\alpha }$ is the mass of the particle.

Use equation (I) in (II),

    $T=\frac{1}{2}{m}_{\alpha }{\left(\omega \times r_{\alpha }\right)}^2$        (III)

The expansion of the vector ${\left(a\times b\right)}^2$,

Let $\theta$ is the angle between vector $a$ and $b$,

    ${\left(a\times b\right)}^2={\left(ab\sin \theta \widehat{n}\right)}^2$

Here, $n$ is the unit vector in the direction of $a\times b$.

    $\begin{array}{c}{\left(a\times b\right)}^2={\left(ab\sin \theta \right)}^2\widehat{n}\cdot \widehat{n}\\ =a^2{b}^2{\sin }^2\theta \\ =a^2{b}^2\left(1-{\cos }^2\theta \right)\\ =a^2{b}^2-a^2{b}^2{\cos }^2\theta \end{array}$

    ${\left(a\times b\right)}^2=a^2{b}^2-{\left(a\cdot b\right)}^2$        (IV)

Apply equation (IV) in (III),

    $T_{\alpha }=\frac{1}{2}{m}_{\alpha }\left[{\left(\omega r_{\alpha }\right)}^2-{\left(\omega \cdot r_{\alpha }\right)}^2\right]$

The total kinetic energy of the rigid body is,

    $\begin{array}{c}T={\displaystyle \sum T_{\alpha }}\\ =\frac{1}{2}{\displaystyle \sum m_{\alpha }}\left[{\left(\omega r_{\alpha }\right)}^2-{\left(\omega \cdot r_{\alpha }\right)}^2\right]\end{array}$

Conclusion:

Therefore, it is proved that the kinetic energy can be written as, $T=\frac{1}{2}{\displaystyle \sum m_{\alpha }}\left[{\left(\omega r_{\alpha }\right)}^2-{\left(\omega \cdot r_{\alpha }\right)}^2\right]$.

(b)

To determine

Prove that the angular momentum $L$ of the body can be written as $L={\displaystyle \sum m_{\alpha }}\left[\omega r_{\alpha }^2-r_{\alpha }\left(\omega \cdot r_{\alpha }\right)\right]$

Answer to Problem 10.33P

It is proved that the angular momentum $L$ of the body can be written as $L={\displaystyle \sum m_{\alpha }}\left[\omega r_{\alpha }^2-r_{\alpha }\left(\omega \cdot r_{\alpha }\right)\right]$

Explanation of Solution

Write the expression for the angular momentum of each particle.

    $l_{\alpha }=r_{\alpha }\times p_{\alpha }$        (V)

Here, $p_{\alpha }$ is the momentum of the particle.

The momentum of the particle is given by,

    $p_{\alpha }=m_{\alpha }{v}_{\alpha }$        (VI)

Use equation (I) in (VI),

    $p_a=m_a\left(\omega \times r_a\right)$        (VII)

Use equation (VII) in (V),

    $\begin{array}{c}{l}_{\alpha }=r_{\alpha }\times \left[m_{\alpha }\left(\omega \times r_{\alpha }\right)\right]\\ =m_{\alpha }\left[r_{\alpha }\times \left(\omega \times r_{\alpha }\right)\right]\end{array}$

According to the vector identity,

    $A\times \left(B\times C\right)=B\left(A\cdot C\right)-C\left(A\cdot B\right)$        (VIII)

Using equation (VIII)

    $\begin{array}{c}{l}_{\alpha }=m_{\alpha }\left[\omega \left(r_{\alpha }\cdot r_{\alpha }\right)-r_{\alpha }\left(r_{\alpha }\cdot \omega \right)\right]\\ =m_{\alpha }\left[\omega r_{\alpha }^2-r_{\alpha }\left(\omega \cdot r_{\alpha }\right)\right]\end{array}$

Conclusion:

The total angular momentum of the rigid body is

    $\begin{array}{c}L={\displaystyle \sum l_{\alpha }}\\ =m_{\alpha }\left[\omega r_{\alpha }^2-r_{\alpha }\left(\omega \cdot r_{\alpha }\right)\right]\end{array}$

Therefore, It is proved that the angular momentum $L$ of the body can be written as $L={\displaystyle \sum m_{\alpha }}\left[\omega r_{\alpha }^2-r_{\alpha }\left(\omega \cdot r_{\alpha }\right)\right]$

(c)

To determine

Using results from part (a) and (b), prove that $T=\frac{1}{2}\omega \cdot L=\frac{1}{2}\overline{\omega }I\omega$.

Answer to Problem 10.33P

Using results from part (a) and (b), it is proved that $T=\frac{1}{2}\omega \cdot L=\frac{1}{2}\overline{\omega }I\omega$.

Explanation of Solution

In order to prove $T=\frac{1}{2}\omega \cdot L$, from the RHS $\frac{1}{2}\omega \cdot L$ is given by the result from part (b),

    $\begin{array}{c}\frac{1}{2}\omega \cdot L=\frac{1}{2}\omega \cdot \left[{\displaystyle \sum m_{\alpha }}\left(\omega r_{\alpha }^2-r_{\alpha }\left(\omega \cdot r_{\alpha }\right)\right)\right]\\ =\frac{1}{2}\left[{\displaystyle \sum m_{\alpha }}\left(\omega \cdot \omega r_{\alpha }^2-\left(\omega \cdot r_{\alpha }\right)\left(\omega \cdot r_{\alpha }\right)\right)\right]\\ =\frac{1}{2}{\displaystyle \sum m_{\alpha }}\left[{\left(\omega r_{\alpha }\right)}^2-\left(\omega \cdot r_{\alpha }^2\right)\right]\\ =T\end{array}$

Thus it is proved that $T=\frac{1}{2}\omega \cdot L$.

The expression for the angular momentum is,

    $L=I\omega$

Using this, the above equation becomes,

    $T=\frac{1}{2}\omega \cdot \left(I\omega \right)$

Here, $I$ is the inertia tensor.

But the matrix form of $\omega \cdot I\omega$ is $\tilde{\omega }I\omega$

Here, $\tilde{\omega }$ is the is the row matrix of the column matrix $\omega$ and $I$. So that,

    $T=\frac{1}{2}\tilde{\omega }I\omega$

Conclusion:

Using results from part (a) and (b), it is proved that $T=\frac{1}{2}\omega \cdot L=\frac{1}{2}\overline{\omega }I\omega$.=

(d)

To determine

Show that with respect to the principal axes, $T=\frac{1}{2}\left({\lambda }_1{\omega }_1^2+{\lambda }_2{\omega }_2^2+{\lambda }_3{\omega }_3^2\right)$.

Answer to Problem 10.33P

It is Shown that with respect to the principal axes, $T=\frac{1}{2}\left({\lambda }_1{\omega }_1^2+{\lambda }_2{\omega }_2^2+{\lambda }_3{\omega }_3^2\right)$.

Explanation of Solution

Write the expression for the moment of inertia tensor $I$.

    $I=\mathrm{diag}\left({\lambda }_1,{\lambda }_2,{\lambda }_3\right)$

Generally, moment of inertia tensor can be written as,

    $I_{ij}={\lambda }_i{\delta }_{ij}$

Then the $I\omega$ is given by,

    $\begin{array}{c}I\omega =\left(\begin{array}{ccc}{\lambda }_1& 0& 0\\ 0& {\lambda }_2& 0\\ 0& 0& {\lambda }_3\end{array}\right)\left(\begin{array}{c}{\omega }_1\\ {\omega }_2\\ {\omega }_3\end{array}\right)\\ =\left(\begin{array}{l}{\lambda }_1{\omega }_1\hfill \\ {\lambda }_2{\omega }_2\hfill \\ {\lambda }_3{\omega }_3\hfill \end{array}\right)\end{array}$        (IX)

Then $\tilde{\omega }=\left({\omega }_1,{\omega }_2,{\omega }_3\right)$, using equation (IX),

    $\begin{array}{c}\tilde{\omega }I\omega =\left({\omega }_1{\omega }_2{\omega }_3\right)\left(\begin{array}{c}{\lambda }_1{\omega }_1\\ {\lambda }_2{\omega }_2\\ {\lambda }_3{\omega }_3\end{array}\right)\\ ={\lambda }_1{\omega }_1^2+{\lambda }_2{\omega }_2^2+{\lambda }_3{\omega }_3^2\end{array}$        (X)

Conclusion:

It is Shown that with respect to the principal axes, $T=\frac{1}{2}\left({\lambda }_1{\omega }_1^2+{\lambda }_2{\omega }_2^2+{\lambda }_3{\omega }_3^2\right)$.