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6.10 HW 10

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6.10.1 Problem 1

Since the wheel rolls without slip with angular velocity $\omega _{disk}=15$ rad/sec and its radius is $r=\frac{5}{12}$ ft, then the center of the wheel moves to the left (since disk is rolling with counter clock wise) with velocity $$\begin{align*} V_{B} & =r\omega _{disk}\\ & =\left ( \frac{5}{12}\right ) \left ( 15\right ) \\ & =6.25\text{ ft/sec} \end{align*}$$

In vector format $$\vec{V}_{B}=-6.25\hat{\imath }+0\hat{\jmath }$$ For the point $A$ $$\begin{align} \vec{V}_{A} & =\vec{V}_{B}+\vec{\omega }_{AB}\times \vec{r}_{A/B}\nonumber \\ & =\left ( -6.25\hat{\imath }+0\hat{\jmath }\right ) +\omega _{AB}\hat{k}\times \left ( -L\cos \theta \hat{\imath }+L\sin \theta \hat{\jmath }\right ) \nonumber \\ & =-6.25\hat{\imath }-\omega _{AB}L\cos \theta \hat{\jmath }-\omega _{AB}L\sin \theta \hat{\imath }\nonumber \\ & =\hat{\imath }\left ( -6.25-\omega _{AB}L\sin \theta \right ) +\hat{\jmath }\left ( -\omega _{AB}L\cos \theta \right ) \tag{1} \end{align}$$

Since point $A$ can only move in vertical direction, then its $\hat{\imath }$ component above must be zero. Therefore $$\begin{align*} -6.25-\omega _{AB}L\sin \theta & =0\\ \omega _{AB} & =\frac{-6.25}{L\sin \theta } \end{align*}$$

Numerically $\omega _{AB}=\frac{-6.25}{\left ( \frac{33}{12}\right ) \sin \left ( 48\left ( \frac{\pi }{180}\right ) \right ) }=-3.058$ rad/sec$.$

Now from (1) we find $\vec{V}_{A}$ since now we know $\omega _{AB}$ $$\begin{align*} \vec{V}_{A} & =\hat{\jmath }\left ( -\omega _{AB}L\cos \theta \right ) \\ & =\hat{\jmath }\left ( \frac{6.25}{L\sin \theta }L\cos \theta \right ) \\ & =\hat{\jmath }\left ( \frac{6.25}{\tan \theta }\right ) \end{align*}$$

Since $\theta =48^{0}$ then the above becomes $$\begin{align*} \vec{V}_{A} & =\frac{6.25}{\tan \left ( 48\left ( \frac{\pi }{180}\right ) \right ) }\hat{\jmath }\\ & =5.627525\hat{\jmath }\\ & =5.628\hat{\jmath }\text{ ft/sec} \end{align*}$$

6.10.2 Problem 2

The first step is to find the vector velocities of point $B$ and $C$ and then resolve them along the $x,y$ directions as follows

Now we look at point $C$. We see that its $x$ component of the velocity is $$Vc_{x}=L\omega _{CB}\cos \phi -R\omega \sin \theta$$ This is just read from the diagram. In other words, the $x$ component of the velocity of $B$ is added.  Since $C$ can only move in the vertical direction, then $Vc_{x}=0$. We use this to solve for $\omega _{CB}$ $$\begin{equation} \omega _{CB}=\frac{R\omega \sin \theta }{L\cos \phi } \tag{1} \end{equation}$$ Everything on the right above is known. We find $\phi$ using $R\cos \theta =L\sin \phi$, hence $$\begin{align*} \phi & =\arcsin \left ( \frac{R\cos \theta }{L}\right ) \\ & =\arcsin \left ( \frac{\left ( 2.3\right ) \cos \left ( 26\frac{\pi }{180}\right ) }{5.3}\right ) \\ & =22.957^{0} \end{align*}$$

And $\omega =4890\left ( \frac{2\pi }{60}\right ) =512.0796$ rad/sec. Hence from (1) $$\begin{align*} \omega _{CB} & =\frac{\left ( 2.3\right ) \left ( 512.0796\right ) \sin \left ( 26\frac{\pi }{180}\right ) }{\left ( 5.3\right ) \cos \left ( 22.957\left ( \frac{\pi }{180}\right ) \right ) }\\ & =105.7955\text{ rad/sec} \end{align*}$$

In vector form $$\vec{\omega }_{CB}=105.7955\hat{k}$$ From the diagram, we see that the vertical component of the velocity of point $C$ is $$\begin{align*} Vc_{y} & =L\omega _{CB}\sin \phi +R\omega \cos \theta \\ & =\left ( 5.3\right ) \left ( 105.7955\right ) \sin \left ( 22.957\left ( \frac{\pi }{180}\right ) \right ) +\left ( 2.3\right ) \left ( 512.0796\right ) \cos \left ( 26\frac{\pi }{180}\right ) \\ & =1277.286\text{ in/sec}\\ & =106.441\text{ ft/sec} \end{align*}$$

In vector form $$\vec{V}c=106.441\hat{\jmath }$$

6.10.3 Problem 3

The first step is to find the vector velocities of point $A$ and $B$ and then resolve them along the $x,y$ directions as follows

Point $A$ will have velocity in $x$ direction of $$V_{A,x}=L\omega \sin \theta -V_{Bx}$$ But $\sin \theta =\frac{d}{L}=\frac{1.8}{6}=0.3$, hence $\theta =\arcsin \left ( 0.3\right ) =17.458^{0}$. Since $A$ can only move in vertical direction, then the above is zero. We use this to find $\omega$ $$\begin{align*} L\omega \sin \theta -V_{Bx} & =0\\ \omega & =\frac{V_{Bx}}{L\sin \theta }\\ & =\frac{6}{6\sin \left ( 17.458\left ( \frac{\pi }{180}\right ) \right ) }\\ & =3.333\text{ rad/sec} \end{align*}$$

In vector format $\vec{\omega }=3.333\hat{k}$ rad/sec. Hence the velocity of $A$ in vertical direction is $$\begin{align*} V_{Ay} & =-L\omega \cos \theta \\ & =-6\left ( 3.333\right ) \cos 17.458\left ( \frac{\pi }{180}\right ) \\ & =-19.076\,83\text{ ft/sec} \end{align*}$$

In vector format $$\vec{V}_{A}=-19.077\hat{\jmath }$$ This is the same velocity as weight $C$ but $C$ will be going up. Hence $$\vec{V}_{C}=19.077\hat{\jmath }$$

6.10.4 Problem 4

The first step is to find the acceleration vectors of point $A$ and $B$ and then resolve them along the $x,y$ directions as follows

To find $\omega _{AB}$ we need to resolve velocity vectors and set the $x$ component of the velocity of $A$ to zero to solve for $\omega _{AB}$. If we do that as before, we get $$\begin{equation} V_{B_{x}}-L\omega _{AB}\sin \theta =0 \tag{1} \end{equation}$$ The above is just the $x$ component of $\vec{V}_{A}$. We know $V_{B}$ which is velocity of center of wheel. It is $$\begin{align*} V_{B_{x}} & =R\omega _{disk}\\ & =1.58\left ( 2.1\right ) \\ & =3.318\text{ m/s} \end{align*}$$

And to the right. Hence $\vec{V}_{B}=3.\,\allowbreak 318\hat{\imath }$. Now we use (1) to solve for $\omega _{AB}$ $$\begin{align*} \omega _{AB} & =\frac{V_{Bx}}{L\sin \theta }=\frac{3.318}{\left ( 3.43\right ) \sin \left ( 69\frac{\pi }{180}\right ) }\\ & =1.0362\text{ rad/sec} \end{align*}$$

Hence $\vec{\omega }_{AB}=1.0362\hat{k}$. Now we have all the information to solve for $\alpha _{AB}$. The $x$ component of $\vec{a}_{A}$ is zero, since $A$ does not move in $x$ direction. Hence from the figure, we see that $$L\omega _{AB}^{2}\cos \theta -L\alpha _{AB}\sin \theta =0$$ There is no acceleration to transfer from point $B$ since $B$ is not accelerating. Solving the above for $\alpha _{AB}$ gives $$\begin{align*} \alpha _{AB} & =\frac{L\omega _{AB}^{2}\cos \theta }{L\sin \theta }\\ & =\frac{\omega _{AB}^{2}}{\tan \theta }\\ & =\frac{1.0362^{2}}{\tan \left ( 69\frac{\pi }{180}\right ) }\\ & =0.41216\text{ rad/sec}^{2} \end{align*}$$

In vector format $\vec{\alpha }_{AB}=0.41216\hat{k}.$ Hence the vertical component of the acceleration $\vec{a}_{A}$ is (from the diagram) $$\begin{align*} a_{Ay} & =-L\omega _{AB}^{2}\sin \theta -L\alpha _{AB}\cos \theta \\ & =-\left ( 3.43\right ) \left ( 1.0362^{2}\right ) \sin \left ( 69\frac{\pi }{180}\right ) -\left ( 3.43\right ) \left ( 0.41216\right ) \cos \left ( 69\frac{\pi }{180}\right ) \\ & =-3.945\text{ m/s}^{2} \end{align*}$$

In vector format $$\vec{a}_{A}=0\hat{\imath }-3.945\hat{\jmath }$$

6.10.5 Problem 5

$$\vec{\alpha }_{BC}=7.507\hat{k}\text{ rad/sec}^{2}$$

Need to type the solution. This uses constraints method.

6.10.6 Problem 6

We need to first find $\omega _{BC}$. This follows similar approach to problem 2. The first step is to find the vector velocities of point $B$ and $C$ and then resolve them along the $x,y$ directions as follows

Now we look at point $C$. We see that its $x$ component of the velocity is $$Vc_{x}=L\omega _{CB}\cos \phi -R\omega \sin \theta$$ This is just read from the diagram. In other words, the $x$ component of the velocity of $B$ is added.  Since $C$ can only move in the vertical direction, then $Vc_{x}=0$. We use this to solve for $\omega _{CB}$ $$\begin{equation} \omega _{CB}=\frac{R\omega \sin \theta }{L\cos \phi } \tag{1} \end{equation}$$ Everything on the right above is known. We find $\phi$ using $R\cos \theta =L\sin \phi$, hence $$\begin{align*} \phi & =\arcsin \left ( \frac{R\cos \theta }{L}\right ) \\ & =\arcsin \left ( \frac{\left ( 2.1\right ) \cos \left ( 28\frac{\pi }{180}\right ) }{5.8}\right ) \\ & =0.325\,4\text{ radians}\\ & =18.6441^{0} \end{align*}$$

And $\omega =5030\left ( \frac{2\pi }{60}\right ) =526.740\,4$ rad/sec. Hence from (1) $$\begin{align*} \omega _{CB} & =\frac{\left ( 2.1\right ) \left ( 526.740\,4\right ) \sin \left ( 28\frac{\pi }{180}\right ) }{\left ( 5.8\right ) \cos \left ( 0.325\,4\right ) }\\ & =94.495\text{ rad/sec} \end{align*}$$

In vector form $$\vec{\omega }_{CB}=94.495\hat{k}$$ Now we draw the acceleration vectors and resolve them

The $x$ component of the acceleration of point $C$ is zero. Hence from the diagram $$L\alpha _{CB}\cos \phi +L\omega _{CB}^{2}\sin \phi -R\alpha _{disk}\sin \theta -R\omega ^{2}\cos \theta =0$$ Solving for $\alpha _{CB}$ $$\alpha _{CB}=\frac{R\alpha _{disk}\sin \theta +R\omega ^{2}\cos \theta -L\omega _{CB}^{2}\sin \phi }{L\cos \phi }$$ Since $\alpha _{disk}=0$ since we are told $\omega$ is constant, then the above simplifies to $$\alpha _{CB}=\frac{R\omega ^{2}\cos \theta -L\omega _{CB}^{2}\sin \phi }{L\cos \phi }$$ Using numerical values gives $$\begin{align*} \alpha _{CB} & =\frac{\left ( 2.1\right ) \left ( 526.740\,4\right ) ^{2}\cos \left ( 28\frac{\pi }{180}\right ) -\left ( 5.8\right ) \left ( 94.\,\allowbreak 494\,71\right ) ^{2}\sin \left ( 0.325\,4\right ) }{\left ( 5.8\right ) \cos \left ( 0.325\,4\right ) }\\ & =90598.94\text{ rad/sec}^{2} \end{align*}$$

In vector form $$\vec{\alpha }_{CB}=-90598.94\hat{k}$$ The acceleration of point $C$ is only in vertical direction. From diagram $$\begin{align*} a_{C,y} & =L\alpha _{CB}\sin \phi -L\omega _{CB}^{2}\cos \phi -R\omega ^{2}\sin \theta \\ & =\left ( 5.8\right ) \left ( 90598.95\right ) \sin \left ( 0.325\,4\right ) -\left ( 5.8\right ) \left ( 94.495\right ) ^{2}\cos \left ( 0.325\,4\right ) -\left ( 2.1\right ) \left ( 526.740\,4\right ) ^{2}\sin \left ( 28\frac{\pi }{180}\right ) \\ & =-154624.9\text{ in/sec}^{2}\\ & =-12885.41\text{ ft/sec}^{2} \end{align*}$$

Hence in vector form $$\vec{a}_{C}=-12885.\,37\hat{\jmath }\text{ ft/sec}^{2}$$