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

Question

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

Accepted Answer

r = (40[latex]{e}^{0.5 heta}[/latex])

[latex]\dot {r}[/latex] = 2 [latex]{e}^{0.05 heta}\dot { heta}[/latex]

[latex]\ddot {r}[/latex] = 0.1 [latex]{e}^{0.05 heta}{(\dot { heta})}^{2}[/latex] + 2[latex]{e}^{0.05 heta}\ddot { heta}[/latex]

[latex] heta[/latex] = [latex]\frac { \pi }{ 6 }[/latex]

[latex]\dot { heta}[/latex] = -4

[latex]\ddot { heta}[/latex] = 0

r = 40[latex]{e}^{0.5(\frac { \pi }{ 6 })}[/latex] = 41.0610

[latex]\dot {r}[/latex] = 2[latex]{e}^{0.5(\frac { \pi }{ 6 })}[/latex](-4) = -8.2122

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

v = [latex]\dot {r}[/latex] = -8.2122 = 8.21 mm/s

a = [latex]\ddot {r}[/latex] - r[latex]{\dot { heta}}^{2}[/latex]= 1.64244 - 41.0610[latex]{(-4)}^{2}[/latex] = -665.33 = -665 mm/[latex]{s}^{2}[/latex]

Question : The partial surface of the cam is that of a logarithmic spir...

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