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6.12 HW 12

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6.12.1 Problem 1

A ball will roll with slip when the linear velocity $v$ of its center of mass is different from $r\omega$ where $r$ is the radius and $\omega$ is the spin angular velocity. Therefore, to find when the ball will roll without slipping, we need to find when $v=r\omega$. Let the initial state be such that $v_{1}=v_{0}$ (given) and $\omega _{1}=\omega _{0}$ (given). So we need to find the time $t$ to get to new state, such that $v_{2}=r\omega _{2}$

Using linear momentum

$$mv_{1}+\int _{0}^{t_{final}}F_{friction}dt=mv_{2}$$ But $F_{friction}=-\mu N=-\mu mg$ and the above becomes $$\begin{equation} mv_{1}-\mu mgt=mv_{2} \tag{1} \end{equation}$$ Using the angular momentum gives $$\begin{align} I\omega _{1}+\int _{0}^{t_{final}}F_{friction}rdt & =I\omega _{2}\nonumber \\ mr_{G}^{2}\omega _{1}-\mu mgrt & =-mr_{G}^{2}\left ( \frac{v_{2}}{r}\right ) \tag{2} \end{align}$$

Where in (2), $r_{G}$ is radius of gyration, and we replaced $\omega _{2}$ by $\frac{v_{2}}{r}$. Notice the sign in RHS of (2) is negative, since we assume $v_{2}$ is moving to the right, so in state 2, the ball will be spinning clock wise, which is negative,. Now we have two equations (1,2) with two unknowns $t$, which is the time to get to the state such that center of mass moves with same speed as $r\omega$ (i.e. no slip) and the second unknown is $v_{2}$ which is the speed at which the ball will be rolling at that time. We now solve (1,2) for $t,v_{2}$

(1) becomes

$$\begin{align} \left ( \frac{15}{32.2}\right ) \left ( 17\left ( \frac{5280}{3600}\right ) \right ) -\left ( 0.11\right ) \left ( \frac{15}{32.2}\right ) \left ( 32.2\right ) t & =\left ( \frac{15}{32.2}\right ) v_{2}\tag{1A}\\ \left ( \frac{15}{32.2}\right ) \left ( \frac{2.4}{12}\right ) ^{2}\left ( 9\right ) -\left ( 0.11\right ) \left ( \frac{15}{32.2}\right ) \left ( 32.2\right ) \left ( \frac{4.25}{12}\right ) t & =-\left ( \frac{15}{32.2}\right ) \left ( \frac{2.4}{12}\right ) ^{2}\left ( \frac{v_{2}}{\left ( \frac{4.25}{12}\right ) }\right ) \tag{2A} \end{align}$$

Or

$$\begin{align} 11.615-1.65t & =0.466v_{2}\tag{1A}\\ 0.168-0.584\,t & =-0.0526v_{2} \tag{2A} \end{align}$$

Solution is:

$$\begin{align*} t & =1.919\,6\text{ sec}\\ v_{2} & =18.134\,\text{ ft/sec} \end{align*}$$

Now that we know the time and the final velocity, we can find the acceleration of the ball

$$\begin{align*} v_{2} & =v_{1}+at\\ a & =\frac{v_{2}-v_{1}}{t}\\ & =\frac{18.134\,-17\left ( \frac{5280}{3600}\right ) }{1.9196}\\ & =-3.542\,\text{ ft/s}^{2} \end{align*}$$

Hence the distance travelled is

$$\begin{align*} s & =v_{0}t+\frac{1}{2}at^{2}\\ & =17\left ( \frac{5280}{3600}\right ) \left ( 1.9196\right ) +\frac{1}{2}\left ( -3.542\,\right ) \left ( 1.9196\right ) ^{2}\\ & =41.336\,1\text{ ft} \end{align*}$$

6.12.2 Problem 2

Using the following FBD

Notice that the Friction force $F$ is pointing downwards since the spool is spinning counter clockwise. Resolving forces along $x$ gives

$$\begin{equation} F-T+mg\sin \theta =m\ddot{x} \tag{1} \end{equation}$$ Taking moment about CG, using clockwise as positive now, since we changed $x$ positive direction from normal $$\begin{equation} FR-T\rho =I_{cg}\alpha \tag{2} \end{equation}$$ Where $\alpha$ is angular acceleration of spool. But $\ddot{x}=-\rho \alpha$ then (1) becomes $$\begin{equation} F-T+mg\sin \theta =-m\rho \alpha \tag{3} \end{equation}$$ But $$\begin{align*} F & =\mu _{k}N\\ & =\mu _{k}mg\cos \theta \end{align*}$$

Therefore (2) and (3) become

$$\begin{align} \mu _{k}mg\cos \theta R-T\rho & =I_{cg}\alpha \tag{2A}\\ \mu _{k}mg\cos \theta -T+mg\sin \theta & =-m\rho \alpha \tag{3A} \end{align}$$

In (2A) and (3A) there are 2 unknowns, $\alpha$ and $T$. Plugging numerical values gives

$$\begin{align*} \left ( 0.31\right ) \left ( 213\right ) \left ( 9.81\right ) \cos \left ( 27\left ( \frac{\pi }{180}\right ) \right ) \left ( 2.24\right ) -T\left ( 1.74\right ) & =\left ( 213\right ) \left ( 2\right ) ^{2}\alpha \\ \left ( 0.31\right ) \left ( 213\right ) \left ( 9.81\right ) \cos \left ( 27\left ( \frac{\pi }{180}\right ) \right ) -T+\left ( 213\right ) \left ( 9.81\right ) \sin \left ( 27\left ( \frac{\pi }{180}\right ) \right ) & =-\left ( 213\right ) \left ( 1.74\right ) \alpha \end{align*}$$

Or

$$\begin{align} 1292.823-1.74T & =852.0\alpha \tag{2A}\\ 1525.78-1.0T & =-370.62\alpha \tag{3A} \end{align}$$

Solution is:

$$\begin{align*} T & =1188.547\text{ N}\\ \alpha & =-0.9099\text{ rad/s}^{2} \end{align*}$$

Now since $\ddot{x}=-\rho \alpha$ then

$$\begin{align*} \ddot{x} & =-\left ( 1.74\right ) \left ( -0.9099\right ) \\ & =1.583\,\text{\ m/s}^{2} \end{align*}$$

6.12.3 Problem 3

The forces in play are

Resolving forces along $\hat{u}_{\phi }$

$$\begin{equation} -F-mg\sin \phi =m\left ( R-\rho \right ) \ddot{\phi } \tag{1} \end{equation}$$ Taking moment around C.G. of ball $$\begin{equation} -F\rho =I_{cg}\ddot{\theta } \tag{2} \end{equation}$$ The above are 2 equations in 3 unknowns ($F,\ddot{\theta },\ddot{\phi }$). So we need one more equation. Resolving along $\hat{u}_{r}$ will not give us an equation in any of these unknowns so it will not be useful for this. Here we must notice that acceleration of point $D$, where the ball touches the bottom of the bowl will be zero. This is because the ball rolls without slip. We can use this to come up with the third equation. The acceleration of this point in the $\hat{u}_{\phi }$ direction is zero, and given by $$\begin{equation} a_{D,\phi }=\left ( R-\rho \right ) \ddot{\phi }+\rho \ddot{\theta }=0 \tag{3} \end{equation}$$ Now we have three equations with three unknowns. Plug-in numerical values, using $I_{cg}=\frac{2}{5}m\rho ^{2}$ $$\begin{align} -F-\left ( 2.9\right ) \sin \left ( 40\left ( \frac{\pi }{180}\right ) \right ) & =\left ( \frac{2.9}{32.2}\right ) \left ( 4.2-1.2\right ) \ddot{\phi }\tag{1A}\\ -F\left ( 1.2\right ) & =\left ( \frac{2}{5}\left ( \frac{2.9}{32.2}\right ) \left ( 1.2^{2}\right ) \right ) \ddot{\theta }\tag{2A}\\ 0 & =\left ( 4.2-1.2\right ) \ddot{\phi }+\left ( 1.2\right ) \ddot{\theta } \tag{3A} \end{align}$$

Or

$$\begin{align} -F-1.864\, & =0.27\ddot{\phi }\tag{1A}\\ -1.2F & =0.0519\ddot{\theta }\tag{2A}\\ 0 & =1.2\ddot{\theta }+3\ddot{\phi } \tag{3A} \end{align}$$

Solving gives

$$\begin{align*} F & =-0.5326\text{ N}\\ \ddot{\theta } & =12.32\text{ rad/s}^{2}\\ \ddot{\phi } & =-4.\,928\text{ rad/s}^{2} \end{align*}$$

To find $N$, we resolve forces along $\hat{u}_{r}$

$$-N+mg\cos \phi =-m\left ( R-\rho \right ) \dot{\theta }^{2}$$ But $\dot{\theta }=\frac{v}{\left ( R-\rho \right ) }$, where $v=9$ ft/sec in this problem. Hence the above becomes $$\begin{align*} -N+mg\cos \phi & =-m\left ( \frac{v^{2}}{R-\rho }\right ) \\ N & =mg\cos \phi +m\left ( \frac{v^{2}}{R-\rho }\right ) \\ & =\left ( 2.9\right ) \cos \left ( 40\frac{\pi }{180}\right ) +\frac{2.9}{32.2}\left ( \frac{\left ( 9.1\right ) ^{2}}{\left ( 4.2-1.2\right ) }\right ) \\ & =4.708\text{ N} \end{align*}$$

Now to find $\vec{a}_{G}$. Since

$$\vec{a}_{G}=\left ( R-\rho \right ) \ddot{\phi }\hat{u}_{\phi }-\frac{v^{2}}{R-\rho }\hat{u}_{r}$$ Then $$\begin{align*} \vec{a}_{G} & =-\left ( 4.1-1.2\right ) 4.\,928\hat{u}_{\phi }-\frac{\left ( 9.1\right ) ^{2}}{\left ( 4.2-1.2\right ) }\hat{u}_{r}\\ & =-14.291\hat{u}_{\phi }-27.603\,\hat{u}_{r} \end{align*}$$

6.12.4 Problem 4

Let $r$ be radius of disk. Then, about joint $O$ at top,

$$\begin{align*} I_{disk} & =m_{disk}\frac{r^{2}}{2}+m_{disk}\left ( L+r\right ) ^{2}\\ & =\left ( 0.37\right ) \frac{\left ( 0.08\right ) ^{2}}{2}+0.37\left ( 0.75+0.08\right ) ^{2}\\ & =0.256\,077 \end{align*}$$

And

$$\begin{align*} I_{bar} & =m_{bar}\frac{L^{2}}{3}\\ & =\left ( 0.7\right ) \frac{\left ( 0.75\right ) ^{2}}{3}\\ & =0.131 \end{align*}$$

Hence overall

$$\begin{align*} I_{o} & =I_{disk}+I_{bar}\\ & =0.256\,+0.131\\ & =0.387\,\, \end{align*}$$

Therefore

$$\begin{align*} KE & =\frac{1}{2}I_{o}\omega ^{2}\\ & =\frac{1}{2}\left ( 0.387\right ) \left ( 0.23\right ) ^{2}\\ & =0.01024\,\text{J} \end{align*}$$

6.12.5 Problem 5

Since wheel rolls without spin, then friction on the ground against the wheel does no work. Therefore we can use work-energy to find $v_{final}$ since we do not need to find friction force and this gives us one equation with one unknown to solve for.

$$\begin{align} T_{1}+U_{1} & =T_{2}+U_{2}\nonumber \\ 0+mgh & =\frac{1}{2}mv_{cg}^{2}+\frac{1}{2}I_{cg}\omega ^{2}-mgh \tag{1} \end{align}$$

Where in the above, the datum is taken as horizontal line passing through the middle of the wheel. But

$$I_{cg}=mr_{G}^{2}$$ Where $r_{G}$ is radius of gyration. And $$v_{cg}=v_{o}\frac{\left ( R-h\right ) }{R}$$ And $\omega =\frac{v_{o}}{R}$ since rolls with no slip. Now we have all the terms needed to evaluate (1) and solve for $v_{o}$. Here $$m=\frac{260}{32.2}=8.075\text{ slug}$$ Hence (1) $$\begin{align*} mgh & =\frac{1}{2}m\left ( v_{o}\frac{\left ( R-h\right ) }{R}\right ) ^{2}+\frac{1}{2}mr_{G}^{2}\left ( \frac{v_{o}}{R}\right ) ^{2}-mgh\\ 260\left ( 0.6\right ) & =\frac{1}{2}\left ( \frac{260}{32.2}\right ) \left ( v_{o}\frac{\left ( 1.76-0.6\right ) }{1.76}\right ) ^{2}+\frac{1}{2}\left ( \frac{260}{32.2}\right ) \left ( 1.32\right ) ^{2}\left ( \frac{v_{o}}{1.76}\right ) ^{2}-260\left ( 0.6\right ) \\ 156 & =4.025v_{o}^{2}-156 \end{align*}$$

Therefore

$$v_{o}=8.805\text{ ft/s}$$ Where the positive root is used since it is moving to the right.

6.12.6 Problem 6

The velocities at each point are given by

$$\begin{align*} V_{B} & =R\omega _{AB}\\ & =4\left ( 3\right ) \\ & =12\text{ ft/s} \end{align*}$$

Looking at point $C$, we obtain two equations

$$\begin{align*} L\omega _{BC} & =-H\omega _{CD}\cos \phi \\ -V_{B} & =-H\omega _{CD}\sin \phi \end{align*}$$

Or

$$\begin{align*} \left ( 5.5\right ) \omega _{BC} & =-\left ( 6.5\right ) \omega _{CD}\cos \left ( 49\left ( \frac{\pi }{180}\right ) \right ) \\ -12 & =-\left ( 6.5\right ) \omega _{CD}\sin \left ( 49\left ( \frac{\pi }{180}\right ) \right ) \end{align*}$$

Solving gives

$$\begin{align*} \omega _{BC} & =-1.897\,\text{\ rad/sec}\\ \omega _{CD} & =2.446\,\text{ rad/sec} \end{align*}$$

We now need to find velocity of center of mass of bar $BC$. We see from diagram that it is given by

$$\begin{align*} \vec{v}_{CG} & =-V_{B}\hat{\imath }-\frac{L}{2}\omega _{BC}\hat{\jmath }\\ & =-12\hat{\imath }-\frac{5.5}{2}\left ( -1.\,897\right ) \hat{\jmath }\\ & =-12\hat{\imath }+5.\,217\hat{\jmath } \end{align*}$$

Hence

$$\begin{align*} \left \vert \vec{v}_{CG}\right \vert & =\sqrt{12^{2}+5.\,217^{2}}\\ & =13.085\ \text{ft/sec} \end{align*}$$

Now we have all the velocities needed. The K.E. of bar $AB$ is

$$\begin{align*} T_{AB} & =\frac{1}{2}I_{AB}\frac{1}{2}\omega _{AB}^{2}\\ & =\frac{1}{2}\left ( \frac{1}{3}m_{AB}R^{2}\right ) \omega _{AB}^{2}\\ & =\frac{1}{2}\left ( \frac{1}{3}\left ( \frac{3}{32.2}\right ) \left ( 4\right ) ^{2}\right ) \left ( 3\right ) ^{2}\\ & =2.236 \end{align*}$$

For bar $BC$ it has both translation and rotation KE

$$\begin{align*} T_{BC} & =\frac{1}{2}I_{BC}\frac{1}{2}\omega _{BC}^{2}+\frac{1}{2}m_{BC}v_{CG}^{2}\\ & =\frac{1}{2}\left ( \frac{1}{12}m_{BC}L^{2}\right ) \omega _{BC}^{2}+\frac{1}{2}m_{BC}v_{CG}^{2}\\ & =\frac{1}{2}\left ( \frac{1}{12}\left ( \frac{6.5}{32.2}\right ) \left ( 5.5\right ) ^{2}\right ) \left ( -1.\,\allowbreak 897\right ) ^{2}+\frac{1}{2}\left ( \frac{6.5}{32.2}\right ) \left ( \allowbreak 13.\,\allowbreak 085\right ) ^{2}\\ & =18.197 \end{align*}$$

And for bar $CD$ it has only rotation KE

$$\begin{align*} T_{CD} & =\frac{1}{2}I_{CD}\frac{1}{2}\omega _{CD}^{2}\\ & =\frac{1}{2}\left ( \frac{1}{3}m_{CD}H^{2}\right ) \omega _{CD}^{2}\\ & =\frac{1}{2}\left ( \frac{1}{3}\left ( \frac{11}{32.2}\right ) \left ( 6.5\right ) ^{2}\right ) \left ( 2.446\right ) ^{2}\\ & =14.392 \end{align*}$$

Therefore the total KE is

$$\begin{align*} KE & =T_{AB}+T_{BC}+T_{CD}\\ & =2.236+18.197+14.392\\ & =34.825\text{ J} \end{align*}$$