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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}

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