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Angular velocity and acceleration of a flywheel system

  1. Aug 17, 2004 #1
    Ok, did I do this correctly?

    The question goes:

    Determine the angular velocity and angular acceleration of the disk and the link AB at the instant A rotates through 90 degrees.

    See attachment for the figure.

    What I did--first the disk:

    Assuming [itex]\alpha_{disk}[/itex] is constant
    [tex]
    \alpha_{disk}=\alpha_0=\alpha_{90}=6\frac{rad}{s^2}
    [/tex]

    [tex]
    \omega_{90}^2=\omega_{0}^2+2\alpha(\theta_{90}-\theta_{0})
    [/tex]

    [tex]
    \omega_{90}=\sqrt{(5\frac{rad}{s})^2+2(6\frac{rad}{s^2})(\frac{\pi}{2})}=6.622\frac{rad}{s}[/tex]

    So, angular velocity of the disk is 6.622 rad/s, and angular acceleration of the disk is 5 rad/s^2.

    Now for the link:

    [tex]
    \vec{v}_A=\vec{ \omega }\times \vec{r}_A
    [/tex]

    [tex]
    \vec{v}_A=\omega(r_A)i=6.622\frac{rad}{s}(0.5\frac{ft}{rad})=3.311\frac{ft}{s}i
    [/tex]

    [tex]
    \vec{v}_B=\vec{v}_A+\vec{ \omega }\times \vec{r}_{B/A}
    [/tex]

    Used trig to find this:
    [tex]
    \vec{r}_{B/A}=2\cos 19i+2\sin 19j
    [/tex]

    [tex]\vec{v}_B=3.311i+\omega(0.652i-1.891j)[/tex]

    [tex]
    v_bi=3.311i+0.652\omega i
    [/tex]
    [tex]
    v_bj=-1.891\omega j
    [/tex]

    solve simultaneously and I get:

    [tex]
    \omega=-1.302\frac{rad}{s}
    [/tex]

    Is this right thus far?

    [edit]forgot picture
     

    Attached Files:

    Last edited: Aug 18, 2004
  2. jcsd
  3. Aug 18, 2004 #2
    Interesting problem. How are you calculating the angular acceleration of the disk? And shouldn't the angular acceleration/velocity of the disk and the link be the same?
     
  4. Aug 18, 2004 #3
    I'm sorry, [itex]\alpha_{Disk}[/itex] was given. In regards to your second question the answer is no. The angular acceleration of the disk is converted to an angular and translational acceration of the link. The link moves to the right as well as rotates clockwise.
     
    Last edited: Aug 18, 2004
  5. Aug 18, 2004 #4
    I figured that in the reference frame of the center of mass of the link, the link doesn't move. Therefore, if you take any particle on the link and look at its motion, it is in pure rotation. For example, take the point on the link attached to the wheel. This point is in pure rotation.

    [edit]I just realized that what I said doesn't make sense. I was trying to say that the link doesn't rotate, i.e. the angle it makes with the horizontal is constant. You said this wasn't the case so I retract what I said.[/edit]
     
    Last edited: Aug 18, 2004
  6. Aug 19, 2004 #5
    Anybody???
     
  7. Aug 19, 2004 #6
    Can you explain this expression:

    [tex]
    \vec{v}_B=\vec{v}_A+\vec{ \omega }\times \vec{r}_{B/A}
    [/tex]

    I'm not sure what you are doing exactly.
     
  8. Aug 19, 2004 #7
    Nevermind. I see what your are doing. How did you find out that [itex]\vec{r}_{B/A}[/itex] makes an angle of 19 degrees with the horizontal?
     
  9. Aug 19, 2004 #8
    I found the 19'ish degrees using the law of cosines and some vector addition.

    First I found vector [itex]\vec{r}_{C/A}[/itex] by using [itex]\vec{r}_{C/A}=\vec{r}_{B/A}+\vec{r}_{C/B}[/itex]

    [tex]
    \vec{r}_{C/A}=(2\cos 30i+2\sin 30j)+(0i+1.5j)
    [/tex]

    [tex]
    \vec{r}_{C/A}=1.732i+2.5j
    [/tex]

    Next I rotated A to A' and said [itex]\vec{r}_{C/A}=\vec{r}_{A^\prime/A}+\vec{r}_{C/A^\prime}[/itex] thus [itex]\vec{r}_{C/A^\prime}=\vec{r}_{C/A}+(-\vec{r}_{A^\prime/A})[/itex]

    [tex]
    \vec{r}_{C/A^\prime}=(1.732i+2.5j)+(-0.5i-0.5j)
    [/tex]

    [tex]
    \vec{r}_{C/A^\prime}=1.232i+2j
    [/tex]

    Next I found the magnitude of [itex]\vec{r}_{C/A^\prime}[/itex]

    [tex]
    r_{C/A^\prime}=\parallel\vec{r}_{C/A^\prime}\parallel=2.349
    [/tex]


    Now, from here I just used the law of cosines to find the angles of triangle A'B'C where A'B'=2 and B'C=1.5 and A'C=2.349.

    Next, I found angle [itex]\beta[/itex] which is the angle between A'C and the horizontal axis. To do this I used vector trig to find the angle.

    Finally, I subtracted angle CA'B' from angle [itex]\beta[/itex] to get 19'ish degrees.

    I'm sure there's a easier way to do this, but this is the route I chose.

    Hope I was clear enough.
     
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