Loading web-font TeX/Math/Italic

6.8 HW 8

PDF (letter size)

PDF (legal size)

6.8.1 Problem 1

The before and after impact diagram is

Along the $y$ direction

$$\begin{align*} m_{A}v_{0}\cos \beta & =m_{A}v_{A_{y}}^{+}+m_{B}v_{B_{y}}^{+}\\ -e & =-1=\frac{v_{A_{y}}^{+}-v_{B_{y}}^{+}}{v_{A_{y}}^{-}-v_{B_{y}}^{-}}=\frac{v_{A_{y}}^{+}-v_{B_{y}}^{+}}{v_{0}\cos \beta } \end{align*}$$

These are 2 equations with 2 unknowns $v_{A_{y}}^{+},v_{B_{y}}^{+}$. From the second equation

$$\begin{equation} -v_{0}\cos \beta =v_{A_{y}}^{+}-v_{B_{y}}^{+}\tag{1} \end{equation}$$ Substituting this in the first equation (and canceling the mass since they are the same), gives $$\begin{align*} -v_{A_{y}}^{+}+v_{B_{y}}^{+} & =v_{A_{y}}^{+}+v_{B_{y}}^{+}\\ v_{A_{y}}^{+} & =0 \end{align*}$$

Therefore from (1)

$$\begin{align*} v_{B_{y}}^{+} & =v_{0}\cos \beta \\ & =9\cos \left ( 45\left ( \frac{\pi }{180}\right ) \right ) \\ & =6.364\text{ m/s} \end{align*}$$

Along the $x$ direction, since this is perpendicular to the line of impact then we know that

$$\begin{align*} v_{A_{x}}^{+} & =v_{A_{x}}^{-}=v_{0}\sin \beta =9\sin \left ( 45\left ( \frac{\pi }{180}\right ) \right ) =6.364\text{ m/s}\\ v_{B_{x}}^{+} & =v_{B_{x}}^{-}=0 \end{align*}$$

Hence velocity of $B$ is

$$\bar{v}_{B}=0\hat{\imath }+6.364\hat{\jmath }$$ And velocity of $A$ is $$\bar{v}_{A}=6.364\hat{\imath }+0\hat{\jmath }$$

6.8.2 Problem 2

The before and after impact diagram is

Along the $x$ axis, the conservation of linear momentum gives

$$\begin{align} m_{A}v_{A}^{-}\cos \alpha -m_{B}v_{B}^{-}\cos \beta & =m_{A}v_{A_{x}}^{+}+m_{B}v_{B_{x}}^{+}\nonumber \\ \left ( 1.48\right ) \left ( 26.7\right ) \cos \left ( 45\left ( \frac{\pi }{180}\right ) \right ) -\left ( 2.75\right ) \left ( 22.6\right ) \cos \left ( 15\left ( \frac{\pi }{180}\right ) \right ) & =\left ( 1.48\right ) v_{A_{x}}^{+}+\left ( 2.75\right ) v_{B_{x}}^{+}\nonumber \\ -32.09 & =\left ( 1.48\right ) v_{A_{x}}^{+}+\left ( 2.75\right ) v_{B_{x}}^{+}\tag{1} \end{align}$$

And

$$\begin{align} -e & =\frac{v_{A_{x}}^{+}-v_{B_{x}}^{+}}{v_{A_{x}}^{-}-v_{B_{x}}^{-}}\nonumber \\ -0.58 & =\frac{v_{A_{x}}^{+}-v_{B_{x}}^{+}}{v_{A}^{-}\cos \alpha +v_{B}^{-}\cos \beta }\nonumber \\ -0.58 & =\frac{v_{A_{x}}^{+}-v_{B_{x}}^{+}}{\left ( 26.7\right ) \cos \left ( 45\left ( \frac{\pi }{180}\right ) \right ) +\left ( 22.6\right ) \cos \left ( 15\left ( \frac{\pi }{180}\right ) \right ) }\nonumber \\ -0.58 & =\frac{v_{A_{x}}^{+}-v_{B_{x}}^{+}}{40.71}\nonumber \\ -23.612 & =v_{A_{x}}^{+}-v_{B_{x}}^{+}\tag{2} \end{align}$$

Now $v_{A_{x}}^{+},v_{B_{x}}^{+}$ is solved for using (1),(2). From (2) $v_{A_{x}}^{+}=-23.612+v_{B_{x}}^{+}$, substituting this in (1) gives

$$\begin{align*} -32.09 & =\left ( 1.48\right ) \left ( -23.612+v_{B_{x}}^{+}\right ) +\left ( 2.75\right ) v_{B_{x}}^{+}\\ -32.09 & =-34.945+4.23v_{B_{x}}^{+}\\ v_{B_{x}}^{+} & =\frac{-32.09+34.945\,}{4.23}\\ & =0.675\,\text{\ m/s} \end{align*}$$

From (2)

$$\begin{align*} v_{A_{x}}^{+} & =-23.612+0.675\\ & =-22.937\text{ m/s} \end{align*}$$

Now we do the same for the $y$ direction. But along this direction we know that

$$\begin{align*} v_{A_{y}}^{+} & =v_{A_{y}}^{-}\\ & =v_{A}^{-}\sin \alpha \\ & =\left ( 26.7\right ) \sin \left ( 45\left ( \frac{\pi }{180}\right ) \right ) \\ & =18.88\text{ m/s} \end{align*}$$

And

$$\begin{align*} v_{B_{y}}^{+} & =v_{B_{y}}^{-}\\ & =-v_{B}^{-}\sin \beta \\ & =\left ( -22.6\right ) \sin \left ( 15\left ( \frac{\pi }{180}\right ) \right ) \\ & =-5.849\,\text{ m/s} \end{align*}$$

Therefore, after impact

$$\begin{align*} \bar{v}_{A} & =-22.938\text{ }\hat{\imath }+18.879\,\hat{\jmath }\\ \bar{v}_{B} & =0.675\hat{\imath }-5.849\,\hat{\jmath } \end{align*}$$

6.8.3 Problem 3

Using

$$\tau =4I\ddot{\theta }$$ Where $\tau$ is applied torque and $I$ is mass moment of inertia around the spin axis of one blade (we have 4). But $I=m\left ( \frac{L}{2}\right ) ^{2}=\frac{mL^{2}}{4}$, since blade is modeled as point mass. Therefore $$\ddot{\theta }=\frac{\tau }{\frac{4mL^{2}}{4}}=\frac{\beta t}{mL^{2}}$$ But $\ddot{\theta }=\frac{d}{dt}\dot{\theta }$, then the above becomes $$\begin{align*} \frac{d}{dt}\dot{\theta } & =\frac{\beta t}{mL^{2}}\\ d\dot{\theta } & =\frac{\beta t}{mL^{2}}dt\\ \int _{0}^{\dot{\theta }_{f}}d\dot{\theta } & =\frac{\beta t}{mL^{2}}\int _{0}^{10}tdt\\ \dot{\theta }_{f} & =\frac{\beta }{mL^{2}}\left ( \frac{t^{2}}{2}\right ) _{0}^{10}\\ & =\frac{\beta }{2mL^{2}}100\\ & =\frac{\left ( 63\right ) }{2\left ( 89\right ) \left ( 4.5\right ) ^{2}}100\\ & =1.748\,\text{\ rad/sec} \end{align*}$$

6.8.4 Problem 4

The angular momentum $\bar{h}$ is the moment of the linear momentum. The linear momentum is $m\bar{v}$. Using radial and tangential coordinates, then

$$m\bar{v}=m\left ( L\dot{\theta }\hat{u}_{\theta }+0\hat{u}_{r}\right )$$ Therefore $$\begin{align} \bar{h} & =\bar{r}\times m\bar{v}\nonumber \\ & =L\hat{u}_{r}\times mL\dot{\theta }\hat{u}_{\theta }\nonumber \\ & =\begin{vmatrix} \hat{u}_{r} & \hat{u}_{\theta } & \hat{k}\\ L & 0 & 0\\ 0 & mL\dot{\theta } & 0 \end{vmatrix} \nonumber \\ & =\hat{k}mL^{2}\dot{\theta }\tag{1} \end{align}$$

The above is what we want. But we need to find $\dot{\theta }$. Taking time derivative of $\bar{h}$ gives

$$\frac{d}{dt}\bar{h}=\hat{k}mL^{2}\ddot{\theta }$$ But $\frac{d}{dt}\bar{h}$ is the torque $\tau$, which we can see to be $$\tau =-mgL\sin \theta$$ The minus sign, since clockwise. Using the above 2 equations, then we write $$\begin{align} -mgL\sin \theta & =mL^{2}\ddot{\theta }\nonumber \\ \ddot{\theta } & =-\frac{g}{L}\sin \theta \tag{2} \end{align}$$

To integrate this, we need a trick. Since

$$\begin{align*} \ddot{\theta } & =\frac{d}{dt}\dot{\theta }\\ & =\left ( \frac{d}{d\theta }\frac{d\theta }{dt}\right ) \dot{\theta }\\ & =\left ( \frac{d}{d\theta }\dot{\theta }\right ) \dot{\theta }\\ & =\dot{\theta }\frac{d\dot{\theta }}{d\theta } \end{align*}$$

Then (2) becomes

$$\dot{\theta }\frac{d\dot{\theta }}{d\theta }=-\frac{g}{L}\sin \theta$$ Now it is separable. $$\begin{align*} \dot{\theta }d\dot{\theta } & =-\frac{g}{L}\sin \theta d\theta \\ \int _{0}^{\dot{\theta }}\dot{\theta }d\dot{\theta } & =-\frac{g}{L}\int _{33^{0}}^{\theta }\sin \theta d\theta \\ \frac{\dot{\theta }^{2}}{2} & =-\frac{g}{L}\left ( -\cos \theta \right ) _{33^{0}}^{\theta }\\ \frac{\dot{\theta }^{2}}{2} & =\frac{g}{L}\left ( \cos \theta -\cos 33^{0}\right ) \\ \dot{\theta } & =\pm \sqrt{\frac{2g}{L}\left ( \cos \theta -\cos 33^{0}\right ) } \end{align*}$$

All this work was to find $\dot{\theta }$. Now we go back to (1) and find the angular momentum

$$\begin{align*} \bar{h} & =\hat{k}mL^{2}\dot{\theta }\\ & =\pm \hat{k}\sqrt{\frac{2g}{L}\left ( \cos \theta -\cos 33^{0}\right ) }mL^{2}\\ & =\pm \hat{k}\sqrt{2gL^{3}\left ( \cos \theta -\cos 33^{0}\right ) }m\\ & =\pm \hat{k}\sqrt{\frac{2L^{3}}{g}\left ( \cos \theta -\cos 33^{0}\right ) }W \end{align*}$$

Substituting numerical values

$$\begin{align*} \bar{h} & =\pm \hat{k}1.8\sqrt{\frac{2\left ( 5.3\right ) ^{3}}{\left ( 32.2\right ) }\left ( \cos \theta -\cos \left ( 33\left ( \frac{\pi }{180}\right ) \right ) \right ) }\\ & =\pm \hat{k}1.8\sqrt{9.247\left ( \cos \theta -0.839\right ) }\\ & =\pm \hat{k}1.8\sqrt{9.247}\sqrt{\left ( \cos \theta -0.839\right ) }\\ & =\pm 5.474\sqrt{\left ( \cos \theta -0.839\right ) }\hat{k} \end{align*}$$

6.8.5 Problem 5

There is no external torque, hence angular momentum is conserved. Let $\bar{h}_{1}$ be the angular momentum initially and let $\bar{h}_{2}$ be angular momentum be at some instance of time later on.  Therefore

$$\begin{align*} \bar{h}_{1} & =\bar{r}_{1}\times m\bar{v}_{1}\\ & =r_{0}\hat{u}_{r}\times m\left ( r_{0}\omega _{0}\hat{u}_{\theta }\right ) \\ & =\begin{vmatrix} \hat{u}_{r} & \hat{u}_{\theta } & \hat{k}\\ r_{0} & 0 & 0\\ 0 & mr_{0}\omega _{0} & 0 \end{vmatrix} \\ & =mr_{0}^{2}\omega _{0}\hat{k} \end{align*}$$

And at some later instance

$$\begin{align*} \bar{h}_{2} & =\bar{r}_{2}\times m\bar{v}_{2}\\ & =r\hat{u}_{r}\times m\left ( \dot{r}\hat{u}_{r}+r\omega \hat{u}_{\theta }\right ) \\ & =\begin{vmatrix} \hat{u}_{r} & \hat{u}_{\theta } & \hat{k}\\ r & 0 & 0\\ m\dot{r} & mr\omega & 0 \end{vmatrix} \\ & =mr^{2}\omega \hat{k} \end{align*}$$

Equating the last two results gives

$$\begin{align} mr_{0}^{2}\omega _{0} & =mr^{2}\omega \nonumber \\ \omega & =\left ( \frac{r_{0}}{r}\right ) ^{2}\omega _{0}\tag{1} \end{align}$$

Now the equation of motion in radial direction is $F=ma_{r}$, but $F=0$, since there is no force on the collar. Therefore

$$\begin{align*} ma_{r} & =0\\ m\left ( \ddot{r}-r\omega ^{2}\right ) & =0\\ \ddot{r} & =r\omega ^{2} \end{align*}$$

Using (1) in the above

$$\begin{align*} \ddot{r} & =r\left [ \left ( \frac{r_{0}}{r}\right ) ^{2}\right ] ^{2}\omega _{0}^{2}\\ \ddot{r} & =\frac{r_{0}^{4}}{r^{3}}\omega _{0}^{2} \end{align*}$$

But $\ddot{r}=\dot{r}\frac{d\dot{r}}{dr}$, hence the above becomes

$$\dot{r}d\dot{r}=\frac{r_{0}^{4}}{r^{3}}\omega _{0}^{2}dr$$ Now we can integrate $$\begin{align*} \int _{0}^{\dot{r}}\dot{r}d\dot{r} & =\int _{r_{0}}^{r}\frac{r_{0}^{4}}{r^{3}}\omega _{0}^{2}dr\\ \frac{\dot{r}^{2}}{2} & =\frac{1}{2}\omega _{0}^{2}r_{0}^{4}\left ( \frac{-1}{r^{2}}\right ) _{r_{0}}^{r}\\ & =\frac{1}{2}\omega _{0}^{2}r_{0}^{4}\left ( \frac{1}{r_{0}^{2}}-\frac{1}{r^{2}}\right ) \end{align*}$$

Therefore

$$\dot{r}=\omega _{0}r_{0}^{2}\sqrt{\left ( \frac{1}{r_{0}^{2}}-\frac{1}{r^{2}}\right ) }$$ To find $\dot{r}$ when it hits the end, we just need to replace $r$ by $r_{0}+d$ in the above $$\dot{r}_{end}=\omega _{0}r_{0}^{2}\sqrt{\frac{1}{r_{0}^{2}}-\frac{1}{\left ( r_{0}+d\right ) ^{2}}}$$ Numerically the above is $$\begin{align*} \dot{r}_{end} & =\left ( 1.6\right ) \left ( 0.5\right ) ^{2}\sqrt{\left ( \frac{1}{\left ( 0.5\right ) ^{2}}-\frac{1}{\left ( 0.5+1.9\right ) ^{2}}\right ) }\\ & =0.782\,\text{ m/s} \end{align*}$$

6.8.6 Problem 6

Using

$$\bar{h}_{1}+\int _{0}^{t}\tau dt=\bar{h}_{2}$$ Where $\bar{h}_{1}$ is initial angular momentum which is zero, and $\bar{h}_{2}$ is final angular momentum which is $I\omega _{f}$ where $I=2MR^{2}$ where $M$ is mass of large ball and $I$ is the mass moment of inertial of the large ball about the spin axis.

But torque $\tau =I_{2}\ddot{\theta }$ where $I_{2}=2\left ( mr^{2}\right )$ where $m$ is mass of each small ball and $I_{2}$ is the mass moment of inertial of the small ball about the spin axis. Hence the above becomes

$$\begin{align*} \int _{0}^{t}\tau dt & =\bar{h}_{2}\\ 2\left ( mr^{2}\right ) \ddot{\theta }\int _{0}^{t}dt & =\bar{h}_{2} \end{align*}$$

Since $\ddot{\theta }$ is constant. Hence

$$2\left ( mr^{2}\right ) \ddot{\theta }t=2MR^{2}\omega _{f}$$ Solving for final angular velocity $$\begin{align*} \omega _{f} & =\frac{2\left ( mr^{2}\right ) \ddot{\theta }t}{2MR^{2}}\\ & =\frac{2\left ( \frac{4.3}{32.2}\right ) \left ( 0.75\right ) ^{2}\left ( 4.7\right ) \left ( 12\right ) }{2\left ( \frac{172}{32.2}\right ) \left ( 3.7\right ) ^{2}}\\ & =0.05793\text{ rad/sec} \end{align*}$$