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Homework Statement
A cam has a shape that is described by the function r = r_0(2  cos [tex]\theta[/tex]), where r_0 = 2.25 ft. A slotted bar is attached to the origin and rotates in the horizontal plane with a constant angular velocity ([tex]\dot{\theta}[/tex] dot) of 0.85 radians/s. The bar moves a roller weighing 25.6 lb along the cam's perimeter. A spring holds the roller in place; the spring's spring constant is 1.75 lb/ft. The friction in the system is negligible. When [tex]\theta[/tex] = 102 degrees, what are F_r and F_[tex]\theta[/tex], the magnitudes of the cylindrical components of the total force acting on the roller?
Homework Equations
r = r_0(2  cos [tex]\theta[/tex]) = 2.25(2cos102) = 4.9678 m
[tex]\dot{\theta}[/tex] = 0.85 rad/s
[tex]\ddot{\theta}[/tex] = 0 rad/s^{2}
[tex]\dot{r}[/tex] = (r_0sin[tex]\theta[/tex])([tex]\dot{\theta}[/tex]) = 2.25(.85)sin(102) = 1.87 m/s
[tex]\ddot{r}[/tex] = (r_0cos[tex]\theta[/tex])([tex]\dot{\theta}[/tex])^2 + (r_0sin[tex]\theta[/tex])([tex]\ddot{\theta}[/tex]) = 2.25(.85^{2})cos(102) = 0.338 m/s^{2}
a_r = [tex]\ddot{r}[/tex]  r([tex]\dot{\theta}[/tex])^2
a_[tex]\theta[/tex] = r[tex]\theta[/tex] + 2[tex]\dot{r}[/tex][tex]\dot{\theta}[/tex]
F_r = W/g * a_r
F_[tex]\theta[/tex] = W/g * a_[tex]\theta[/tex]
F_s = ks (spring)
The Attempt at a Solution
a_r = [tex]\ddot{r}[/tex]  r([tex]\dot{\theta}[/tex])^2 = 0.338  4.9678(.85^2) = 3.927 m/s^{2}
a_[tex]\theta[/tex] = r[tex]\theta[/tex] + 2[tex]\dot{r}[/tex][tex]\dot{\theta}[/tex] = 4.9678*0 + 2*1.87*.85 = 3.179 rad/s^{2}
F_r = W/g * a_r = (25.6/32.2)*(3.927) = 3.12 lb
F_[tex]\theta[/tex] = W/g * a_[tex]\theta[/tex] = (25.6/32.2)*(3.179) = 2.53 lb
But it's wrong and I didn't consider the spring force. How do I incorporate Fs? Please help. I'm stuck
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