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Dynamics - Cylindrical Coordinates

  • Thread starter sami23
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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(2-cos102) = 4.9678 m
[tex]\dot{\theta}[/tex] = 0.85 rad/s
[tex]\ddot{\theta}[/tex] = 0 rad/s2
[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(.852)cos(102) = -0.338 m/s2

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/s2

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/s2

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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Answers and Replies

  • #2
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it opposes the acceleration along r
 
Last edited:
  • #3
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I found F_r = -3.122 lb and F_[tex]\theta[/tex] = 2.531 lb when a_r = -3.927 rad/s2 and a_[tex]\theta[/tex] = 3.179 rad/s2

What are F_t and N, the magnitudes of the tangential force, F_t, and the normal force, N, acting on the roller when [tex]\theta[/tex] = 102 degrees?
a = a_t + a_n

a_t = dv/dt (angular velocity)
a_n = v2/[tex]\rho[/tex] where [tex]\rho[/tex] = radius of curvature

Equations of Motion:
F_t = W/g * a_t
F_n = W/g * a_n where g = 32.2 ft/s2

How do I find a_t and a_n to get the tangential force and the Normal? Please help.
 

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