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*}
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\[ \vec{V}c=106.441\hat{\jmath }\]
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\[ \vec{V}_{C}=19.077\hat{\jmath }\]
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\[ \vec{a}_{A}=0\hat{\imath }-3.945\hat{\jmath }\]
\[ \vec{\alpha }_{BC}=7.507\hat{k}\text{ rad/sec}^{2}\]
Need to type the solution. This uses constraints method.
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\[ \vec{a}_{C}=-12885.\,37\hat{\jmath }\text{ ft/sec}^{2}\]