2.3  My Final exam EME 121UC Davis, spring quarter 2011

  2.3.1  Problem 1
  2.3.2  Problem 2
  2.3.3  Problem 3
  2.3.4  Problem 4


2.3.1  Problem 1


The first step is to obtain an expression for the spring extension when the mass \(m\footnote{For ease of typing, will use $m$ to mean $m_{a}$}\) is moving and at an angle \(\theta \left ( t\right ) \). Let \(\delta \left ( t\right ) \) represent the current extension in spring at time \(t\), using kinematics as shown below the following relation is found\[ \delta =2L\sin \frac{\theta }{2}\] The diagram below gives a snap shot of the system at time \(t\)


The system has 2 DOF and they are \(x\) and \(\theta \).

Part (A) Velocity, acceleration and force diagrams




Generating equations of motion using \(F=ma\) Using the force and acceleration diagrams, equations of motions can now be generated. There are \(3\) unknowns: \(x,\theta ,T\), therefore 3 equations are required to solve for them. For mass \(m\), resolving along the radial direction results in\begin{equation} T+k\delta \cos \alpha =m\left ( L\dot{\theta }^{2}-\ddot{x}\sin \theta \right ) \tag{1} \end{equation} And resolving perpendicular to the radial direction gives\begin{equation} -k\delta \sin \alpha =m\left ( L\ddot{\theta }+\ddot{x}\cos \theta \right ) \tag{2} \end{equation} For mass \(M\), resolving in the vertical direction1 gives\begin{equation} F+T\sin \theta +k\delta \sin \alpha =M\ddot{x} \tag{3} \end{equation} Now the \(\sin \alpha \) and \(\cos \alpha \) terms are expressed as functions of \(\theta \). For \(\sin \alpha \), the law of sines can be used as follows\begin{align} \frac{\delta }{\sin \theta } & =\frac{L}{\sin \alpha }\nonumber \\ \sin \alpha & =\frac{L}{\delta }\sin \theta \tag{4} \end{align}

An expression for \(\cos \alpha \) as function of \(\theta \) is now found with the help of the following diagram


Therefore \begin{equation} \cos \alpha =\frac{L}{\delta }\left ( 1-\cos \theta \right ) \tag{5} \end{equation} Substituting Eqs. (4,5) into Eqs. (1,2,3) results in new set of three equations but without the angle \(\alpha \) explicitly appearing in the equations. Here are the 3 equations again after the above substitution. Eq. (1) becomes\begin{align} T+k\delta \left ( \frac{L}{\delta }\left ( 1-\cos \theta \right ) \right ) & =m\left ( L\dot{\theta }^{2}-\ddot{x}\sin \theta \right ) \nonumber \\ T+kL\left ( 1-\cos \theta \right ) & =m\left ( L\dot{\theta }^{2}-\ddot{x}\sin \theta \right ) \tag{1A} \end{align}

And Eq. (2) becomes\begin{align} -k\delta \left ( \frac{L}{\delta }\sin \theta \right ) & =m\left ( L\ddot{\theta }+\ddot{x}\cos \theta \right ) \nonumber \\ -kL\sin \theta & =m\left ( L\ddot{\theta }+\ddot{x}\cos \theta \right ) \tag{2A} \end{align}

And Eq. (3) becomes\begin{align} F+T\sin \theta +k\delta \left ( \frac{L}{\delta }\sin \theta \right ) & =M\ddot{x}\nonumber \\ F+T\sin \theta +kL\sin \theta & =M\ddot{x} \tag{3A} \end{align}

Equations (1A,2A,3A) contain 3 unknowns \(T,x\) and \(\theta \). In order to obtain equation of motion in only \(x\) and \(\theta \), the unknown \(T\) is eliminated. By solving for \(T\) from Eq. (1A) and substitute the result into Eq. (3A).

Eq. (1A) gives\[ T=m\left ( L\dot{\theta }^{2}-\ddot{x}\sin \theta \right ) -kL\left ( 1-\cos \theta \right ) \] Substituting the above into Eq. (3A) results in\begin{align*} F+\overset{T}{\overbrace{\left ( m\left ( L\dot{\theta }^{2}-\ddot{x}\sin \theta \right ) -kL\left ( 1-\cos \theta \right ) \right ) }}\sin \theta +kL\sin \theta & =M\ddot{x}\\ F+mL\dot{\theta }^{2}\sin \theta -m\ddot{x}\sin ^{2}\theta -kL\left ( 1-\cos \theta \right ) \sin \theta +kL\sin \theta & =M\ddot{x}\\ F+mL\dot{\theta }^{2}\sin \theta -kL\sin \theta +kL\cos \theta \sin \theta +kL\sin \theta & =M\ddot{x}+m\ddot{x}\sin ^{2}\theta \\ F+mL\dot{\theta }^{2}\sin \theta +kL\cos \theta \sin \theta & =\ddot{x}\left ( M+m\sin ^{2}\theta \right ) \end{align*}

Hence\begin{equation} \ddot{x}=\frac{F+mL\dot{\theta }^{2}\sin \theta +kL\cos \theta \sin \theta }{M+m\sin ^{2}\theta } \tag{6} \end{equation} Eq. (2A) gives\begin{align*} -kL\sin \theta & =m\left ( L\ddot{\theta }+\ddot{x}\cos \theta \right ) \\ mL\ddot{\theta } & =-kL\sin \theta -m\ddot{x}\cos \theta \\ \ddot{\theta } & =\frac{-kL\sin \theta -m\ddot{x}\cos \theta }{mL} \end{align*}

Hence\[ \ddot{\theta }=-\frac{k}{m}\sin \theta -\frac{\ddot{x}}{L}\cos \theta \] This completes part(A). The equations of motion are found as given in Eq. (6) for \(x\left ( t\right ) \) and Eq. (7) for \(\theta \left ( t\right ) \).

The equations are seen to be coupled. In part(B), they will be decoupled and first order form generated.


The 2 equations of motions found in part(A) are\begin{align} \ddot{x} & =\frac{F+mL\dot{\theta }^{2}\sin \theta +kL\cos \theta \sin \theta }{M+m\sin ^{2}\theta }\tag{1}\\ \ddot{\theta } & =-\frac{k}{m}\sin \theta -\frac{\ddot{x}}{L}\cos \theta \tag{2} \end{align}

Substituting (1) into (2) gives\[ \ddot{\theta }=-\frac{k}{m}\sin \theta -\frac{\cos \theta }{L}\left ( \frac{F+mL\dot{\theta }^{2}\sin \theta +kL\cos \theta \sin \theta }{M+m\sin ^{2}\theta }\right ) \] The equations are now decoupled. To convert to state space, 4 states variables are introduced as follows\[ \left . \begin{array} [c]{c}x_{1}=x\\ x_{2}=\theta \\ x_{3}=\dot{x}\\ x_{4}=\dot{\theta }\end{array} \right \} \overset{d/dt}{\rightarrow }\left . \begin{array} [c]{c}\dot{x}_{1}=x_{3}\\ \dot{x}_{2}=x_{4}\\ \dot{x}_{3}=\ddot{x}\\ \dot{x}_{4}=\ddot{\theta }\end{array} \right \} \rightarrow \begin{array} [c]{c}\dot{x}_{1}=x_{3}\\ \dot{x}_{2}=x_{4}\\ \dot{x}_{3}=\frac{F+mL\dot{\theta }^{2}\sin \theta +kL\cos \theta \sin \theta }{M+m\sin ^{2}\theta }\\ \dot{x}_{4}=-\frac{k}{m}\sin \theta -\frac{\cos \theta }{L}\left ( \frac{F+mL\dot{\theta }^{2}\sin \theta +kL\cos \theta \sin \theta }{M+m\sin ^{2}\theta }\right ) \end{array} \] Replacing all terms in the right side above by the state variables gives\begin{align*} \dot{x}_{1} & =x_{3}\\ \dot{x}_{2} & =x_{4}\\ \dot{x}_{3} & =\frac{F+mLx_{4}^{2}\sin x_{2}+kL\cos x_{2}\sin x_{2}}{M+m\sin ^{2}x_{2}}\\ \dot{x}_{4} & =-\frac{k}{m}\sin x_{2}-\frac{\cos x_{2}}{L}\left ( \frac{F+mLx_{4}^{2}\sin x_{2}+kL\cos x_{2}\sin x_{2}}{M+m\sin ^{2}x_{2}}\right ) \end{align*}

This completes problem 1.

2.3.2  Problem 2


Using the following velocity diagram generated in problem


The kinetic energy of the system is then given by\begin{align*} T & =\frac{1}{2}m\left ( \left [ L\dot{\theta }+\dot{x}\cos \theta \right ] ^{2}+\left [ \dot{x}\sin \theta \right ] ^{2}\right ) +\frac{1}{2}M\dot{x}^{2}\\ & =\frac{1}{2}m\left ( L^{2}\dot{\theta }^{2}+\dot{x}^{2}+2L\dot{\theta }\dot{x}\cos \theta \right ) +\frac{1}{2}M\dot{x}^{2} \end{align*}

The potential energy is due to the spring extension only given by\[ V=\frac{1}{2}k\delta ^{2}\] But \(\delta =2L\sin \frac{\theta }{2}\) as found in problem 1, therefore \begin{align*} V & =\frac{1}{2}k\left ( 2L\sin \frac{\theta }{2}\right ) ^{2}\\ & =2kL^{2}\sin ^{2}\frac{\theta }{2} \end{align*}

Hence the Lagrangian can be written as\begin{align*} L & =T-V\\ & =\frac{1}{2}m\left ( L^{2}\dot{\theta }^{2}+\dot{x}^{2}+2L\dot{\theta }\dot{x}\cos \theta \right ) +\frac{1}{2}M\dot{x}^{2}-2kL^{2}\sin ^{2}\frac{\theta }{2} \end{align*}

The system has two generalized coordinates \(x,\theta \).

Starting with \(\theta \) results in\[ \frac{\partial L}{\partial \dot{\theta }}=m\left ( L^{2}\dot{\theta }+L\dot{x}\cos \theta \right ) \] And\begin{align*} \frac{\partial L}{\partial \theta } & =\frac{1}{2}m\left ( -2L\dot{\theta }\dot{x}\sin \theta \right ) -4kL^{2}\left ( \sin \frac{\theta }{2}\right ) \left ( \cos \frac{\theta }{2}\right ) \frac{1}{2}\\ & =-mL\dot{\theta }\dot{x}\sin \theta -2kL^{2}\sin \frac{\theta }{2}\cos \frac{\theta }{2} \end{align*}

And\[ \frac{d}{dt}\frac{\partial L}{\partial \dot{\theta }}=m\left ( L^{2}\ddot{\theta }+L\ddot{x}\cos \theta -L\dot{x}\left ( \sin \theta \right ) \dot{\theta }\right ) \] Therefore, the equation of motion for \(\theta \) is\[ \frac{d}{dt}\frac{\partial L}{\partial \dot{\theta }}-\frac{\partial L}{\partial \theta }=Q_{\theta }\] But \(Q_{\theta }=0\) since no external force is acting to change \(\theta \) and the spring force has been accounted for in the potential \(V\), hence the above becomes\begin{align*} \frac{d}{dt}\frac{\partial L}{\partial \dot{\theta }}-\frac{\partial L}{\partial \theta } & =0\\ m\left ( L^{2}\ddot{\theta }+L\ddot{x}\cos \theta -L\dot{x}\left ( \sin \theta \right ) \dot{\theta }\right ) -\left ( -mL\dot{\theta }\dot{x}\sin \theta -2kL^{2}\sin \frac{\theta }{2}\cos \frac{\theta }{2}\right ) & =0\\ mL^{2}\ddot{\theta }+mL\ddot{x}\cos \theta -mL\dot{x}\dot{\theta }\sin \theta +mL\dot{\theta }\dot{x}\sin \theta +2kL^{2}\sin \frac{\theta }{2}\cos \frac{\theta }{2} & =0\\ mL^{2}\ddot{\theta }+mL\ddot{x}\cos \theta +2kL^{2}\sin \frac{\theta }{2}\cos \frac{\theta }{2} & =0 \end{align*}

Therefore\begin{align*} \ddot{\theta } & =\frac{-mL\ddot{x}\cos \theta }{mL^{2}}-\frac{2kL^{2}\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{mL^{2}}\\ & =-\left ( \frac{\ddot{x}}{L}\cos \theta +\frac{k}{m}2\sin \frac{\theta }{2}\cos \frac{\theta }{2}\right ) \end{align*}

But \(2\sin \frac{\theta }{2}\cos \frac{\theta }{2}=\sin \theta \footnote{Using $\sin 2A=2\sin A\cos A$ formula}\), then then above reduces to\begin{equation} \ddot{\theta }=-\frac{\ddot{x}}{L}\cos \theta -\frac{k}{m}\sin \theta \tag{1} \end{equation} Which is the same as was found in Eq. (1) in part(B) in the first problem when using the \(F=ma\) method.

Now the equation of motion for \(x\) will be found.\[ \frac{\partial L}{\partial \dot{x}}=m\left ( \dot{x}+L\dot{\theta }\cos \theta \right ) +M\dot{x}\] And\[ \frac{\partial L}{\partial x}=0 \] But \(Q_{x}\) is not zero this time. \(Q_{x}=F\) and it is positive since it adds energy to the system. This can found as follows\begin{align*} \Delta W & =\left ( \Delta x\right ) F\\ Q_{x} & =\frac{\Delta W}{\Delta x}=F \end{align*}

Continuing the derivation gives\[ \frac{d}{dt}\frac{\partial L}{\partial \dot{x}}=m\left ( \ddot{x}+L\ddot{\theta }\cos \theta -L\dot{\theta }^{2}\sin \theta \right ) +M\ddot{x}\] Hence the equation of motion for \(x\) is\begin{align*} \frac{d}{dt}\frac{\partial L}{\partial \dot{x}}-\frac{\partial L}{\partial x} & =Q_{\theta }\\ m\left ( \ddot{x}+L\ddot{\theta }\cos \theta -L\dot{\theta }^{2}\sin \theta \right ) +M\ddot{x} & =F \end{align*}

Or\begin{align} \ddot{x}\left ( m+M\right ) & =F+mL\dot{\theta }^{2}\sin \theta -mL\ddot{\theta }\cos \theta \nonumber \\ \ddot{x} & =\frac{F+mL\dot{\theta }^{2}\sin \theta -mL\ddot{\theta }\cos \theta }{m+M} \tag{2} \end{align}

These are the equations of motion as given above in Eqs. (1,2). To decouple these equations, substituting Eq. (1) into the RHS of Eq. (2) gives\[ \ddot{x}=\frac{F+mL\dot{\theta }^{2}\sin \theta -mL\overset{\ddot{\theta }}{\overbrace{\left ( -\frac{\ddot{x}}{L}\cos \theta -\frac{k}{m}\sin \theta \right ) }}\cos \theta }{m+M}\] Simplifying results in \begin{align*} \ddot{x} & =\frac{F+mL\dot{\theta }^{2}\sin \theta +m\ddot{x}\cos ^{2}\theta +Lk\sin \theta \cos \theta }{m+M}\\ \ddot{x}\left ( m+M\right ) -m\ddot{x}\cos ^{2}\theta & =F+mL\dot{\theta }^{2}\sin \theta +Lk\sin \theta \cos \theta \\ \ddot{x}\left ( m+M-m\cos ^{2}\theta \right ) & =F+mL\dot{\theta }^{2}\sin \theta +Lk\sin \theta \cos \theta \\ \ddot{x} & =\frac{F+mL\dot{\theta }^{2}\sin \theta +Lk\sin \theta \cos \theta }{m+M-m\cos ^{2}\theta } \end{align*}

But \(m+M-m\cos ^{2}\theta =m\left ( 1-\cos ^{2}\theta \right ) +M=m\left ( \sin ^{2}\theta \right ) +M\), hence the above becomes the decoupled equation of motion for \(x\)\begin{equation} \ddot{x}=\frac{F+mL\dot{\theta }^{2}\sin \theta +Lk\sin \theta \cos \theta }{M+m\sin ^{2}\theta } \tag{3} \end{equation} Comparing Eq. (3) above with Eq. (1) in part(B) of the first problem shows it is the same.

Therefore Eqs. (1,3) above are the equations of motion derived using Lagrangian method. They are verified to be the same equations of motion found using \(F=ma.\)

This completes problem 2.

2.3.3  Problem 3


The following 3D diagram shows the shopping cart with body fixed coordinates system oriented such as the \(U\) is positive in the backward direction. The right hand rule was used to draw the axes with the yaw, shown as \(\omega _{z}\) in the diagram, pointing in the positive direction as was given in the problem when viewed from top of the cart.


Starting from first principles, and using the rule\[ \frac{d}{dt}\mathbf{A}=\left ( \frac{d}{dt}\mathbf{A}\right ) _{\text{resolved}}+\mathbf{\omega \times A}\] The equation \(F=ma\) for the body is written, where \(\omega _{z}\) was replaced by \(r\) in the following equation since \(r\) is the more common term.\begin{align*} \mathbf{F} & =\frac{d}{dt}\mathbf{p}\\ & =\frac{d}{dt}\left ( m\mathbf{v}\right ) \\ & \mathbf{=}m\frac{d}{dt}\mathbf{v}\\ & =m\left [ \begin{pmatrix} \dot{U}\\ \dot{V}\\ \dot{Z}\end{pmatrix} +\begin{pmatrix} \omega _{x}\\ \omega _{y}\\ r \end{pmatrix} \otimes \begin{pmatrix} U\\ V\\ Z \end{pmatrix} \right ] \\ & =m\left [ \begin{pmatrix} \dot{U}\\ \dot{V}\\ \dot{Z}\end{pmatrix} +\det \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ \omega _{x} & \omega _{y} & r\\ U & V & Z \end{vmatrix} \right ] \\ & =m\left [ \begin{pmatrix} \dot{U}\\ \dot{V}\\ \dot{Z}\end{pmatrix} +\begin{pmatrix} \omega _{y}Z-rV\\ -\left ( \omega _{x}V-rU\right ) \\ \omega _{x}V-\omega _{y}U \end{pmatrix} \right ] \end{align*}

The only angular velocity present is \(r\) and \(\dot{Z}=0\) since there is no motion in the \(z\) direction, therefore the above simplifies to\[\begin{pmatrix} F_{x}\\ F_{y}\end{pmatrix} =m\begin{pmatrix} \dot{U}-rV\\ \dot{V}+rU \end{pmatrix} \] And since \(U\) is constant \(U_{0}\) and since there is no force in the \(x\) direction, the above simplifies to just the following equation\[ F_{y}=m\left ( \dot{V}+rU_{0}\right ) \] Where \(F_{y}=Y_{r}\) hence\begin{equation} Y_{r}=m\left ( \dot{V}+rU_{0}\right ) \tag{1} \end{equation} Now the torque equation \(\tau =I\omega \) on the rigid body is applied. Using the principle axes, the moment of inertia matrix is\[ I=\begin{pmatrix} I_{1} & 0 & 0\\ 0 & I_{2} & 0\\ 0 & 0 & I_{3}\end{pmatrix} \equiv \begin{pmatrix} I_{xx} & 0 & 0\\ 0 & I_{yy} & 0\\ 0 & 0 & I_{zz}\end{pmatrix} \] Hence\begin{align*} \mathbf{\tau } & =\frac{d}{dt}\left [ \begin{pmatrix} I_{1} & 0 & 0\\ 0 & I_{2} & 0\\ 0 & 0 & I_{3}\end{pmatrix}\begin{pmatrix} \omega _{x}\\ \omega _{y}\\ \omega _{z}\end{pmatrix} \right ] +\begin{pmatrix} \omega _{x}\\ \omega _{y}\\ r \end{pmatrix} \times \begin{pmatrix} I_{1} & 0 & 0\\ 0 & I_{2} & 0\\ 0 & 0 & I_{3}\end{pmatrix}\begin{pmatrix} \omega _{x}\\ \omega _{y}\\ r \end{pmatrix} \\\begin{pmatrix} \tau _{x}\\ \tau _{y}\\ \tau _{z}\end{pmatrix} & =\begin{pmatrix} I_{1} & 0 & 0\\ 0 & I_{2} & 0\\ 0 & 0 & I_{3}\end{pmatrix}\begin{pmatrix} \dot{\omega }_{x}\\ \dot{\omega }_{y}\\ \dot{r}\end{pmatrix} +\begin{pmatrix} \omega _{x}\\ \omega _{y}\\ r \end{pmatrix} \times \begin{pmatrix} I_{1}\omega _{x}\\ I_{2}\omega _{y}\\ I_{3}r \end{pmatrix} \\ & =\begin{pmatrix} I_{1}\dot{\omega }_{x}\\ I_{2}\dot{\omega }_{y}\\ I_{3}\dot{r}\end{pmatrix} +\det \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ \omega _{x} & \omega _{y} & r\\ I_{1}\omega _{x} & I_{2}\omega _{y} & I_{3}r \end{vmatrix} \\ & =\begin{pmatrix} I_{1}\dot{\omega }_{x}\\ I_{2}\dot{\omega }_{y}\\ I_{3}\dot{r}\end{pmatrix} +\begin{pmatrix} \omega _{y}\left ( I_{3}r\right ) -r\left ( I_{2}\omega _{y}\right ) \\ -\omega _{x}\left ( I_{3}r\right ) +r\left ( I_{1}\omega _{x}\right ) \\ \omega _{x}\left ( I_{2}\omega _{y}\right ) -\omega _{y}\left ( I_{1}\omega _{x}\right ) \end{pmatrix} \\ & =\begin{pmatrix} I_{1}\dot{\omega }_{x}\\ I_{2}\dot{\omega }_{y}\\ I_{3}\dot{r}\end{pmatrix} +\begin{pmatrix} \omega _{y}r\left ( I_{3}-I_{2}\right ) \\ \omega _{x}r\left ( I_{1}-I_{3}\right ) \\ \omega _{x}\omega _{y}\left ( I_{2}-I_{1}\right ) \end{pmatrix} \end{align*}

Since \(\omega _{y}=\omega _{x}=0\) the above simplifies to one torque equation\[ \tau _{z}=I_{3}\dot{r}\] Where \(\tau _{z}=bY_{r}\), hence the above becomes\begin{equation} bY_{r}=I_{3}\dot{r} \tag{2} \end{equation} Substituting Eq.(1) into (2) results in\begin{equation} bm\left ( \dot{V}+rU_{0}\right ) =I_{3}\dot{r} \tag{3} \end{equation} Now the no-slip condition is applied to the rear tire. Transferring the center of mass velocity to the back axial is illustrated in the following diagram


From the above it can be seen that the no-slip condition leads to the following result\[ V+br=0 \] Or\begin{equation} V=-br \tag{4} \end{equation} Substituting Eq. (4) back into Eq. (3) results in\begin{align*} bm\left ( -b\dot{r}+rU_{0}\right ) & =I_{3}\dot{r}\\ -mb^{2}\dot{r}+rbmU_{0} & =I_{3}\dot{r} \end{align*}

Hence\begin{equation} \dot{r}\left ( I_{3}+mb^{2}\right ) -\left ( bmU_{0}\right ) r=0 \tag{5} \end{equation} This is the equation of motion for \(r\). Assuming the solution is \(r=\bar{r}e^{st}\) then the above becomes\[ \bar{r}se^{st}\left ( I_{3}+mb^{2}\right ) -\left ( bmU_{0}\right ) \bar{r}e^{st}=0 \] For non-trivial solution, \(\bar{r}\neq 0\) and \(e^{st}\neq 0\) hence dividing by \(\bar{r}e^{st}\) gives\[ s\left ( I_{3}+mb^{2}\right ) -\left ( bmU_{0}\right ) =0 \] Therefore\[ s=\frac{bmU_{0}}{I_{3}+mb^{2}}\] The quantity in the denominator is positive always. Hence, \(U_{0}\) is assumed to be positive going backward, then this implies that \(s\) above is positive as all other quantities shown are positive.

Therefore it is concluded that the cart is unstable as the eigenvalue is positive.

This result agrees with the result given in class. Changing the axes orientation did not change the final conclusion, which is the cart will be unstable when moving in the rear direction.

This completes problem 3.

2.3.4  Problem 4




The above diagram shows the principal axes for the disk through its c.g. These axes are fixed on the disk and rotate with it. The principal moments of inertia matrix of the disk around its c.g. can be found by integration. They are shown here from tables\[ \left [ I_{cg}\right ] =\begin{bmatrix} I_{1} & 0 & 0\\ 0 & I_{2} & 0\\ 0 & 0 & I_{3}\end{bmatrix} =\begin{bmatrix} \frac{1}{12}M\left ( 3R^{2}+h^{2}\right ) & 0 & 0\\ 0 & \frac{MR^{2}}{2} & 0\\ 0 & 0 & \frac{1}{12}M\left ( 3R^{2}+h^{2}\right ) \end{bmatrix} \] To find the principal moments of inertia matrix around axes through \(0\), the parallel axes theorem is used. Hence\begin{align*} \left [ I_{0}\right ] & =\left [ I_{cg}\right ] +M\begin{bmatrix} L^{2} & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & L^{2}\end{bmatrix} \\ & =\begin{bmatrix} \frac{1}{12}M\left ( 3R^{2}+h^{2}\right ) & 0 & 0\\ 0 & \frac{MR^{2}}{2} & 0\\ 0 & 0 & \frac{1}{12}M\left ( 3R^{2}+h^{2}\right ) \end{bmatrix} +\begin{bmatrix} ML^{2} & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & ML^{2}\end{bmatrix} \end{align*}

Therefore\begin{align*} \left [ I_{0}\right ] & =\begin{bmatrix} I_{x} & 0 & 0\\ 0 & I_{y} & 0\\ 0 & 0 & I_{z}\end{bmatrix} \\ & =\begin{bmatrix} M\left ( \frac{R^{2}}{4}+\frac{h^{2}}{12}+L^{2}\right ) & 0 & 0\\ 0 & \frac{MR^{2}}{2} & 0\\ 0 & 0 & M\left ( \frac{R^{2}}{4}+\frac{h^{2}}{12}+L^{2}\right ) \end{bmatrix} \end{align*}

Part (b)

Viewing the disk from the front, looking at it when standing in front of the \(0y\) axis facing the disk, the following velocity components can be seen


The velocity \(L\omega _{0}\) is the instantaneous linear velocity of the c.g. of the disk due to the disk rotation around the post. The term \(R\omega _{disk}\) is the instantaneous linear velocity of the point where the disk touching the ground.

For no-slip these two velocities must be equal which means\[ R\omega _{disk}=L\omega _{0}\] Hence\[ \omega _{disk}=\frac{L}{R}\omega _{0}\]

Part (c)

The disk has kinetic energy due to the spins around its principal axes and due to orbital translation around the base post located at \(0\).


Therefore\begin{align*} T & =\frac{1}{2}M\left ( L\omega _{0}\right ) ^{2}+\frac{1}{2}I_{2}\omega _{disk}^{2}+\frac{1}{2}I_{3}\omega _{0}^{2}\\ & =\frac{1}{2}ML^{2}\omega _{0}^{2}+\frac{1}{2}\frac{MR^{2}}{2}\omega _{disk}^{2}+\frac{1}{2}M\left ( \frac{R^{2}}{4}+\frac{h^{2}}{12}+L^{2}\right ) \omega _{0}^{2} \end{align*}

Using the result found from (b) that \(\omega _{disk}=\frac{L}{R}\omega _{0}\) the above reduces to\begin{align*} T & =\frac{1}{2}ML^{2}\omega _{0}^{2}+\frac{MR^{2}}{4}\frac{L^{2}}{R^{2}}\omega _{0}^{2}+\frac{1}{2}M\left ( \frac{R^{2}}{4}+\frac{h^{2}}{12}+L^{2}\right ) \omega _{0}^{2}\\ & =\frac{h^{2}M\omega _{0}^{2}}{24}+ML^{2}\omega _{0}^{2}+\frac{1}{8}MR^{2}\omega _{0}^{2}+\frac{L^{2}MR^{2}\omega _{0}^{2}}{4R^{2}} \end{align*}

Therefore\[ T=\frac{M\omega _{0}^{2}}{24}\left ( h^{2}+30L^{2}+3R^{2}\right ) \]

Part (d)

The angular momentum of the disk around \(0\) is \[ \mathbf{H}_{0}=I_{0}\mathbf{\omega }\] where now the moment of inertia tensor used in the one relative to \(0\) and not relative to the \(c.g.\) of disk. Angular velocities do not change. Only the moment of inertia tensor is changed. This works since the disk has one point that is fixed in space at \(0\) as it rotates. This is due to the axial that extends from its center to point \(0\) to the disk, which is considered part of the disk, but contribute no inertia since it is assumed to be massless.

Therefore, the angular momentum around \(0\) can be written as\begin{align*} \mathbf{H}_{0} & =\begin{bmatrix} M\left ( \frac{R^{2}}{4}+\frac{h^{2}}{12}+L^{2}\right ) & 0 & 0\\ 0 & \frac{MR^{2}}{2} & 0\\ 0 & 0 & M\left ( \frac{R^{2}}{4}+\frac{h^{2}}{12}+L^{2}\right ) \end{bmatrix}\begin{bmatrix} 0\\ -\omega _{disk}\\ \omega _{0}\end{bmatrix} \\ & =\begin{bmatrix} 0\\ -\frac{MR^{2}}{2}\omega _{disk}\\ M\left ( \frac{R^{2}}{4}+\frac{h^{2}}{12}+L^{2}\right ) \omega _{0}\end{bmatrix} \end{align*}

Hence\[ \mathbf{H}_{0}=0\ \mathbf{i}-\left ( \frac{MR^{2}}{2}\omega _{disk}\right ) \ \mathbf{j}+M\left ( \frac{R^{2}}{4}+\frac{h^{2}}{12}+L^{2}\right ) \omega _{0}\mathbf{k}\] Therefore it magnitude is \[ \left \vert \mathbf{H}_{0}\right \vert =\sqrt{\left ( \frac{MR^{2}}{2}\omega _{disk}\right ) ^{2}+\left ( M\left ( \frac{R^{2}}{4}+\frac{h^{2}}{12}+L^{2}\right ) \omega _{0}\right ) ^{2}}\] Using the no-slip condition \(\omega _{disk}=\frac{L}{R}\omega _{0}\) the above simplifies to\begin{align*} \left \vert \mathbf{H}_{0}\right \vert & =\sqrt{\left ( \frac{MR^{2}}{2}\frac{L}{R}\omega _{0}\right ) ^{2}+\left ( M\left ( \frac{R^{2}}{4}+\frac{h^{2}}{12}+L^{2}\right ) \omega _{0}\right ) ^{2}}\\ \left \vert \mathbf{H}_{0}\right \vert & =\sqrt{\frac{M^{2}R^{2}L^{2}\omega _{0}^{2}}{4}+M^{2}\left ( \frac{R^{2}}{4}+\frac{h^{2}}{12}+L^{2}\right ) ^{2}\omega _{0}^{2}}\\ & =M\omega _{0}\sqrt{\frac{R^{2}L^{2}}{4}+\left ( \frac{R^{2}}{4}+\frac{h^{2}}{12}+L^{2}\right ) ^{2}} \end{align*}

The direction of \(\mathbf{H}_{0}\) is given in the diagram below



A constant \(\omega _{0}\) implies that the component \(H_{z}\) of the angular momentum is constant (in both magnitude and direction). Therefore the \(H_{y}\) component (due to the disk spin around \(y\) axis) is the only component that will be changing. The change comes due to change in direction  and not in magnitude as the disk rotates around the base.

The tip of the \(\mathbf{H}_{0}\) vector therefore travels along a circle above \(0\) at a distance given by magnitude of the constant \(H_{z}\) component. The vector \(\mathbf{H}_{0}\) will be rotating at the constant rate of \(\omega _{0}\) as illustrated in the following diagram


Looking at \(H_{y}\) component from above helps to determine the rate of change of angular momentum


The above diagram shows the instantaneous rate of change of the vector \(\mathbf{H}_{0}\). The vector \(\Delta \mathbf{H}\) is tangent to the circle at the tip of the \(\mathbf{H}_{0}\) vector. Hence \(\Delta \mathbf{H}\) points towards the \(x\) axis.

Since \(\mathbf{H}_{0}\) rotates at rate \(\omega _{0}\) this results in\[ \frac{\Delta \mathbf{H}_{0}}{\Delta t}=H_{y}\omega _{0}\mathbf{i}\] Where \(H_{y}=\frac{MR^{2}}{2}\omega _{disk}\) found earlier. (The minus sign has been accounted for by drawing the component in the negative direction above).

As can be seen from the diagram, the rate of change of the angular momentum (which is also a vector), always points in the positive \(x\) direction.  Therefore, in vector form, \(\mathbf{\dot{H}}_{0}\) can be written as\[ \mathbf{\dot{H}}_{0}=\begin{bmatrix} H_{y}\omega _{0}\\ 0\\ 0 \end{bmatrix} =\begin{bmatrix} \left ( \frac{MR^{2}}{2}\omega _{disk}\right ) \omega _{0}\\ 0\\ 0 \end{bmatrix} =\left ( \frac{MR^{2}}{2}\omega _{disk}\right ) \omega _{0}\ \mathbf{i}+0\ \mathbf{j}+0\mathbf{k}\] The following diagram shows the direction of \(\mathbf{\dot{H}}_{0}\) in the original diagram to help illustrate it better


Using the no-slip condition \(\omega _{disk}=\frac{L}{R}\omega _{0}\) in the above simplifies \(\mathbf{\dot{H}}_{0}\) resulting in\begin{align*} \mathbf{\dot{H}}_{0} & =\begin{bmatrix} \left ( \frac{MR^{2}}{2}\omega _{disk}\right ) \omega _{0}\\ 0\\ 0 \end{bmatrix} =\begin{bmatrix} \left ( \frac{MR^{2}}{2}\frac{L}{R}\omega _{0}\right ) \omega _{0}\\ 0\\ 0 \end{bmatrix} \\ & =\begin{bmatrix} \frac{MR}{2}L\omega _{0}^{2}\\ 0\\ 0 \end{bmatrix} \end{align*}

Therefore, since rate of change of angular momentum is not zero and has magnitude of \(\frac{MR}{2}L\omega _{0}^{2}\) and points in the positive \(x\) direction, and since \[ \frac{d}{dt}\mathbf{H}_{0}=\mathbf{\tau }_{0}\] Then there must be a torque \(\mathbf{\tau }_{0}\) acting around the \(x\) axis. In the next part this torque vector is verified and it cause outlined.


There exist a centripetal force due to rotation of the disk that points towards the \(0.\) This force vector has a moment arm which is the length \(R\) which is the height of the c.g. of the desk above the ground. Therefore, this produces a torque around the \(x\) axis as shown below.

Using the right hand rule, the torque vector can be seen to point in the \(x\) direction, which is the direction that the rate of change of angular momentum points to as would be expected.

The amount of the centripetal force is \[ F=Ma_{c}\] Where \(a_{c}\) is the disk acceleration towards the center \(0\) which is \(L\omega _{0}^{2}.\) Therefore\[ F=ML\omega _{0}^{2}\] Hence the torque magnitude is given by \[ \tau =MLR\omega _{0}^{2}\]


To verify that the above torque is equal to \(\frac{d}{dt}\mathbf{H}_{0}\), then applying the definition of \(\frac{d}{dt}\mathbf{H}_{0}\) and accounting for the rotation around \(0\) results in\begin{align*} \mathbf{\tau }_{0} & =\frac{d}{dt}\mathbf{H}_{0}\\ & =\left ( \frac{d}{dt}\mathbf{H}_{0}\right ) _{0xyz}+\mathbf{\omega }\times \mathbf{H}_{0}\\ & =\begin{bmatrix} \frac{MR}{2}L\omega _{0}^{2}\\ 0\\ 0 \end{bmatrix} +\det \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ 0 & 0 & \omega _{0}\\ 0 & -\frac{MR^{2}}{2}\omega _{disk} & M\left ( \frac{R^{2}}{4}+\frac{h^{2}}{12}+L^{2}\right ) \omega _{0}\end{vmatrix} \\ & =\begin{bmatrix} \frac{MR}{2}L\omega _{0}^{2}\\ 0\\ 0 \end{bmatrix} +\begin{bmatrix} \omega _{0}\frac{MR^{2}}{2}\omega _{disk}\\ 0\\ 0 \end{bmatrix} \end{align*}

Using the no-slip condition \(\omega _{disk}=\frac{L}{R}\omega _{0}\), the above becomes\begin{align*} \mathbf{\tau }_{0} & =\begin{bmatrix} \frac{MR}{2}L\omega _{0}^{2}\\ 0\\ 0 \end{bmatrix} +\begin{bmatrix} \omega _{0}\frac{MR^{2}}{2}\frac{L}{R}\omega _{0}\\ 0\\ 0 \end{bmatrix} \\ & =\begin{bmatrix} \frac{MR}{2}L\omega _{0}^{2}\\ 0\\ 0 \end{bmatrix} +\begin{bmatrix} \frac{MR}{2}L\omega _{0}^{2}\\ 0\\ 0 \end{bmatrix} \\ & =\begin{bmatrix} MRL\omega _{0}^{2}\\ 0\\ 0 \end{bmatrix} \end{align*}

Or in vector form\[ \mathbf{\tau }_{0}=MRL\omega _{0}^{2}\ \mathbf{i+}0\mathbf{j+}0\mathbf{k}\] And the magnitude of the torque is \[ \tau =MRL\omega _{0}^{2}\] Which is the same as was found above. This completes problem 4.