## Question:

The partial surface of the cam is that of a logarithmic spiral r = (40${e}^{0.5\theta}$) mm , where $\theta$ is in radians. If the cam is rotating at a constant angular rate of $\dot {\theta}$ = 4 rad/s , determine the magnitudes of the velocity and acceleration of the follower rod at the instant $\theta$ = 30° .

## Step-by-step

r = (40${e}^{0.5\theta}$)

$\dot {r}$ = 2 ${e}^{0.05\theta}\dot {\theta}$

$\ddot {r}$ = 0.1 ${e}^{0.05\theta}{(\dot {\theta})}^{2}$ + 2${e}^{0.05\theta}\ddot {\theta}$

$\theta$ = $\frac { \pi }{ 6 }$

$\dot {\theta}$ = -4

$\ddot {\theta}$ = 0

r = 40${e}^{0.5(\frac { \pi }{ 6 })}$ = 41.0610

$\dot {r}$ = 2${e}^{0.5(\frac { \pi }{ 6 })}$(-4) = -8.2122

$\ddot {r}$ = 0.1${e}^{0.5(\frac { \pi }{ 6 })}{(-4)}^{2}$ + 0 = 1.64244

v = $\dot {r}$ = -8.2122 = 8.21 mm/s

a = $\ddot {r}$ – r${\dot {\theta}}^{2}$= 1.64244 – 41.0610${(-4)}^{2}$ = -665.33 = -665 mm/${s}^{2}$