Chapter 5.8, Problem 185RP

To determine

The instantaneous velocity of point $A$.

The instantaneous acceleration of point $A$.

Answer to Problem 185RP

The instantaneous velocity of point $A$ is $r\omega \left(i+\frac{1}{3}j\right)$.

The instantaneous acceleration of point $A$ is $r\left[\left(\frac{{\omega }^2}{3}-\alpha \right)i-\left({\omega }^2+\frac{\alpha }{3}\right)j\right]$.

Explanation of Solution

Write the expression for position vector point $A$.

    $r=-\overline{OC}i+\overline{CA}j$        (I)

Here, unit vector in x-direction is $i$, unit vector in y-direction is $j$, the horizontal distance between the point $O$ and centre point of circular disk is $C$ is $\overline{OC}$ and vertical distance between point centre point $C$ and point $A$ is $\overline{CA}$

Substitute $\frac{r}{3}$ for $\overline{OC}$ and $r$ for $\overline{CA}$ in Equation (I).

    $r=-\frac{r}{3}i+rj$        (II)

Here, radius of circular disk is $r$.

Write the expression for instantaneous velocity of point $A$.

    $v_A=\omega \times r$        (III)

Here, angular velocity vector is $\omega$.

Substitute $-\omega k$ for $\omega$ and $-\frac{r}{3}i+rj$ for $r$ in Equation (III).

    $\begin{array}{c}{v}_A=-\omega k\times \left(-\frac{r}{3}i+rj\right)\\ =-\frac{r}{3}\omega \left(k\times i\right)-r\omega \left(k\times j\right)\\ =-\frac{r}{3}\omega \left(-j\right)-r\omega \left(-i\right)\\ =r\omega \left(i+\frac{1}{3}j\right)\end{array}$

Thus, the instantaneous velocity of point $A$ is $r\omega \left(i+\frac{1}{3}j\right)$.

Write the expression for the instantaneous acceleration of point $A$.

    $a_{\text{A}}=\alpha \times r-{\omega }^{2}r$        (IV)

Here, angular acceleration vector is $\alpha$ and position vector point $A$ is $r$.

Substitute $\alpha k$ for $\alpha$ and $-\frac{r}{3}i+rj$ for $r$ in Equation (IV).

    $\begin{array}{c}{a}_{\text{A}}=\left(\alpha k\right)\times \left(-\frac{r}{3}i+rj\right)-{\omega }^2\left(-\frac{r}{3}i+rj\right)\\ =-\frac{r\alpha }{3}\left(k\times i\right)+r\alpha \left(k\times j\right)+\frac{r{\omega }^2}{3}i-r{\omega }^{2}j\\ =-\frac{r\alpha }{3}\left(j\right)+r\alpha \left(-i\right)+\frac{r{\omega }^2}{3}i-r{\omega }^{2}j\\ =r\left[\left(\frac{{\omega }^2}{3}-\alpha \right)i-\left({\omega }^2+\frac{\alpha }{3}\right)j\right]\end{array}$

Thus, the instantaneous acceleration of point $A$ is $r\left[\left(\frac{{\omega }^2}{3}-\alpha \right)i-\left({\omega }^2+\frac{\alpha }{3}\right)j\right]$.