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6.13 HW 13

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6.13.1 Problem 1

Free body diagram is

Method one, using work-energy

Applying work energy

$$\begin{equation} T_{1}+U_{1}+\int _{0}^{\theta _{final}}Md\theta =T_{2}+U_{2} \tag{1} \end{equation}$$ But $T_{1}=0$ and $U_{1}=0$ (using initial position as datum). $$\int _{0}^{\theta _{final}}Md\theta =M\theta _{final}=M\frac{d}{R}$$ Where $d$ is distance travelled (since no slip, we use $d=R\theta$). $$\begin{align*} T_{2} & =\frac{1}{2}mv_{cg}^{2}+\frac{1}{2}I_{cg}\omega ^{2}\\ & =\frac{1}{2}m\left ( R\omega \right ) ^{2}+\frac{1}{2}\left ( mk_{G}^{2}\right ) \omega ^{2} \end{align*}$$

And

$$U_{2}=\frac{1}{2}kd^{2}-Wd\sin \theta$$ Hence (1) becomes $$\begin{align*} M\frac{d}{R} & =\frac{1}{2}m\left ( R\omega \right ) ^{2}+\frac{1}{2}\left ( mk_{G}^{2}\right ) \omega ^{2}+\frac{1}{2}kd^{2}-Wd\sin \theta \\ M\frac{d}{R}-\frac{1}{2}kd^{2}+Wd\sin \theta & =\omega ^{2}\left ( \frac{1}{2}mR^{2}+\frac{1}{2}mk_{G}^{2}\right ) \\ \omega ^{2} & =\frac{M\frac{d}{R}-\frac{1}{2}kd^{2}+Wd\sin \theta }{\frac{1}{2}mR^{2}+\frac{1}{2}mk_{G}^{2}}\\ & =\frac{2\left ( M\frac{d}{R}-\frac{1}{2}kd^{2}+Wd\sin \theta \right ) }{\frac{W}{g}\left ( R^{2}+k_{G}^{2}\right ) }\\ & =\frac{g}{W\left ( R^{2}+k_{G}^{2}\right ) }\left ( 2M\frac{d}{R}-kd^{2}+2Wd\sin \theta \right ) \end{align*}$$

Or

$$\begin{equation} \omega =\sqrt{\frac{g}{W\left ( R^{2}+k_{G}^{2}\right ) }}\sqrt{d\left ( 2\frac{M}{R}+2W\sin \theta -kd\right ) } \tag{2} \end{equation}$$ Hence choice $B$. Plug-in numerical values gives $k=4,R=1.4,W=10,\theta =28^{0}$, and since $I_{disk}=\frac{1}{2}mR^{2}=mk_{G}^{2}$ then $k_{G}^{2}=\frac{R^{2}}{2}=\frac{1.4^{2}}{2}=0.98$, then (2) becomes for $\omega =0$ $$\begin{align*} 0 & =\sqrt{\frac{32.2}{10\left ( 1.4^{2}+0.98\right ) }}\sqrt{\left ( 4\right ) \left ( 2\frac{M}{1.4}+2\left ( 10\right ) \sin \left ( 28\left ( \frac{\pi }{180}\right ) \right ) -\left ( 4\right ) \left ( 4\right ) \right ) }\\ 0 & =1.047\sqrt{5.714\,M-26.442} \end{align*}$$

Solving for moment $M$ gives

$$M=4.627\,\ \text{ft-lb}$$ Method two, using Newton methods

$\sum F_{x}$ gives (where positive $x$ is as shown in diagram, going down the slope).

$$\begin{equation} W\sin \theta -kx-F=m\ddot{x}=mR\ddot{\theta } \tag{1} \end{equation}$$ Taking moment about CG of disk. But note that now anti-clock wise is negative and not positive, due to right-hand rule) $$\begin{equation} -M-FR=-I_{cg}\ddot{\theta } \tag{2} \end{equation}$$ From (2) we solve for $F$ and use (1) to find $\ddot{\theta }$. From (2) $$F=\frac{I_{cg}\ddot{\theta }-M}{R}$$ Plug the above into (1) $$\begin{align*} W\sin \theta -kx-\frac{I_{cg}\ddot{\theta }-M}{R} & =mR\ddot{\theta }\\ W\sin \theta -kx & =mR\ddot{\theta }+\left ( \frac{I_{cg}\ddot{\theta }-M}{R}\right ) \\ W\sin \theta -kx & =mR\ddot{\theta }+\frac{I_{cg}\ddot{\theta }}{R}-\frac{M}{R}\\ W\sin \theta -kx & =\ddot{\theta }\left ( mR+\frac{I_{cg}}{R}\right ) -\frac{M}{R}\\ \frac{M}{R}+mg\sin \theta -kx & =\ddot{\theta }\left ( mR+\frac{mk_{G}^{2}}{R}\right ) \end{align*}$$

Hence

$$\begin{align*} \ddot{\theta } & =\frac{\frac{M}{R}+W\sin \theta -kx}{mR+\frac{mk_{G}^{2}}{R}}\\ & =\frac{M+WR\sin \theta -kRx}{\frac{W}{g}\left ( R^{2}+k_{G}^{2}\right ) }\\ & =\frac{g}{W\left ( R^{2}+k_{G}^{2}\right ) }\left ( M+WR\sin \theta -kRx\right ) \end{align*}$$

The above shows that $\ddot{\theta }$ is not constant. To find $\omega$ we need to integrate both sides. Since $\ddot{\theta }=\frac{d\omega }{dt}=\frac{d\omega }{dx}\frac{dx}{dt}=\frac{d\omega }{dx}R\omega$ then the above can be written as

$$R\omega d\omega =\frac{g}{W\left ( R^{2}+k_{G}^{2}\right ) }\left ( M+WR\sin \theta -kRx\right ) dx$$

Integrating

$$\frac{R}{2}\omega ^{2}=\frac{g}{W\left ( R^{2}+k_{G}^{2}\right ) }\left ( Mx+WRx\sin \theta -kR\frac{x^{2}}{2}\right )$$

When $x=d$, the above becomes

$$\begin{equation} \frac{R}{2}\omega ^{2}=\frac{g}{W\left ( R^{2}+k_{G}^{2}\right ) }\left ( Md+WRd\sin \theta -kR\frac{d^{2}}{2}\right ) \tag{3} \end{equation}$$

Hence

$$\begin{align} \omega ^{2} & =2\left ( \frac{g}{W\left ( R^{2}+k_{G}^{2}\right ) }\left ( \frac{Md}{R}+Wd\sin \theta -k\frac{d^{2}}{2}\right ) \right ) \nonumber \\ & =\frac{g}{W\left ( R^{2}+k_{G}^{2}\right ) }d\left ( \frac{2M}{R}+2W\sin \theta -kd\right ) \tag{4} \end{align}$$

Compare (4) to (2) in first method, we see they are the same.

6.13.2 Problem 2

$$\begin{align*} h_{AB} & =I_{A}\omega _{AB}\\ & =\left ( \frac{1}{3}m_{AB}R^{2}\right ) \omega _{AB}\\ & =\frac{1}{3}\left ( 2.2\right ) \left ( 0.76\right ) ^{2}\left ( 5\right ) \\ & =2.\,\allowbreak 119\text{ kg m}^{2}/s \end{align*}$$

For bar $BC$, it has zero $\omega _{BC}$ at this instance. Therefore the only angular momentum comes from translation. WHich is

$$h_{BC}=m_{BC}v_{cg}R$$

But $v_{cg}$ for bar $BC$ is $R\omega _{AB}$, hence

$$\begin{align*} h_{BC} & =m_{BC}R^{2}\omega _{AB}\\ & =\left ( 3.4\right ) \left ( 0.76\right ) ^{2}\left ( 5\right ) \\ & =9.\,\allowbreak 819\,\text{\ kg m}^{2}/s \end{align*}$$

Finally, for bar $DC$, since it point $C$ moves with speed $v=R\omega _{AB}$, then

$$\begin{align*} R\omega _{AB} & =H\omega _{CD}\\ \omega _{CD} & =\frac{R}{H}\omega _{AB} \end{align*}$$

Therefore

$$\begin{align*} h_{CD} & =I_{CD}\omega _{CD}\\ & =\frac{1}{3}m_{CD}H^{2}\frac{R}{H}\omega _{AB}\\ & =\frac{1}{3}m_{CD}HR\omega _{AB}\\ & =\frac{1}{3}\left ( 5.2\right ) \left ( 1.54\right ) \left ( 0.76\right ) \left ( 5\right ) \\ & =10.\,\allowbreak 143\,\text{ kg m}^{2}/s \end{align*}$$

6.13.3 Problem 3

$$\begin{align} torque & =I_{cg}\ddot{\theta }\nonumber \\ -M_{0}\left ( 1+ct\right ) & =mk_{G}^{2}\ddot{\theta }\nonumber \\ \frac{d\dot{\theta }}{dt} & =-\frac{M_{0}}{mk_{G}^{2}}\left ( 1+ct\right ) \nonumber \\ \int _{\omega _{AB}}^{0}d\dot{\theta } & =-\frac{M_{0}}{mk_{G}^{2}}\int _{0}^{t_{s}}\left ( 1+ct\right ) dt\nonumber \\ -\omega _{AB} & =-\frac{M_{0}}{mk_{G}^{2}}\left ( t_{s}+\frac{c}{2}t_{s}^{2}\right ) \nonumber \\ \omega _{AB} & =\frac{M_{0}}{mk_{G}^{2}}\left ( t_{s}+\frac{c}{2}t_{s}^{2}\right ) \tag{1} \end{align}$$

Hence

$$\begin{align*} \left ( 1150\right ) \frac{2\pi }{60} & =\frac{3400}{\frac{3400}{32.2}\left ( 15.1\right ) ^{2}}\left ( t_{s}+\frac{0.012}{2}t_{s}^{2}\right ) \\ 120.\,428 & =0.141\left ( t_{s}+0.006t_{s}^{2}\right ) \end{align*}$$

Solving

$$t_{s}=302.763\text{ seconds}$$

Another way to solve this is to use conservation of angular momentum.

$$\begin{align*} h_{1}+\int \tau dt & =h_{2}\\ I_{cg}\omega _{B}+\int _{0}^{t_{s}}\left ( -M\right ) dt & =0\\ \left ( mk_{G}^{2}\right ) \omega _{B}-\int _{0}^{t_{s}}M_{0}\left ( 1+ct\right ) dt & =0\\ \left ( mk_{G}^{2}\right ) \omega _{B}-M_{0}\left ( t_{s}+\frac{c}{2}t_{s}^{2}\right ) & =0\\ \omega _{B} & =\frac{M_{0}}{mk_{G}^{2}}\left ( t_{s}+\frac{c}{2}t_{s}^{2}\right ) \end{align*}$$

Which is the same as (1).

6.13.4 Problem 4

It is easier to solve this using conservation of angular and linear momentum. There are two bodies in this problem. One has angular momentum and the second (cart) has linear momentum. So we need to apply

$$\begin{equation} p_{1}+\int _{0}^{t}fdt=p_{1}\tag{1} \end{equation}$$

Where $p=mv$, the linear momentum. The above is applied to the cart. And also apply

$$\begin{equation} h_{1}+\int _{0}^{t}\tau dt=h_{1}\tag{2} \end{equation}$$

Where $h=I\omega$, the angular momentum, and this is applied to the drum. Using the above two equations we will find final velocity of cart. We break the system to 2 bodies, using free body diagram. Let tension in cable be $T$. And since in state (1), $v_{A}=0$, then equation (1) becomes

$$\begin{align} \int _{0}^{t}\left ( T-W_{A}\right ) dt & =m_{A}v_{A}\nonumber \\ \int _{0}^{t}Tdt & =W_{A}t+\frac{W_{A}}{g}v_{A}\tag{3} \end{align}$$

In the above, $\int _{0}^{t}\left ( T-W_{A}\right ) dt$ is the impulse, and $v_{A}$ is the final speed we want to find.  We do not know the tension $T$.

Equation (2) becomes ($h_{1}=0$, since drum is not spining then)

$$\int _{0}^{t}Trdt=I_{cg}\omega _{D}$$

Where $Tr$ is the torque, caused by the tension $T$ in cable. But $v_{A}=-r\omega _{D}$, where the minus sign since it is moving downwards. Hence the above becomes

$$\begin{align} \int _{0}^{t}Tdt & =-\left ( \frac{W_{D}}{g}\frac{r^{2}}{2}\right ) \frac{v_{A}}{r^{2}}\nonumber \\ & =-\left ( \frac{W_{D}}{2g}\right ) v_{A}\tag{4} \end{align}$$

Comparing (3,4) we see that

$$\begin{align*} -\left ( \frac{W_{D}}{2g}\right ) v_{A} & =W_{A}t+\frac{W_{A}}{g}v_{A}\\ W_{A}t+\frac{W_{A}}{g}v_{A}+\left ( \frac{W_{D}}{2g}\right ) v_{A} & =0\\ v_{A}\left ( \frac{W_{A}}{g}+\frac{W_{D}}{2g}\right ) +W_{A}t & =0\\ v_{A} & =\frac{-2W_{A}gt}{2W_{A}+W_{D}} \end{align*}$$

Therefore

$$\begin{align*} v_{A} & =\frac{-2\left ( 325\right ) \left ( 32.2\right ) \left ( 2.5\right ) }{2\left ( 325\right ) +\left ( 117\right ) }\\ & =-68.\,\allowbreak 22\text{ ft/sec} \end{align*}$$

6.13.5 Problem 5