How to Calculate F_t and N for a Roller on a Cam Mechanism?

[sami23]

Homework Statement

A cam has a shape that is described by the function r = r_0(2 - cos \theta), 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 (\dot{\theta} 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 \theta = 102 degrees, what are F_r and F_\theta, the magnitudes of the cylindrical components of the total force acting on the roller?

Homework Equations

r = r_0(2 - cos \theta) = 2.25(2-cos102) = 4.9678 m
\dot{\theta} = 0.85 rad/s
\ddot{\theta} = 0 rad/s²
\dot{r} = (r_0sin\theta)(\dot{\theta}) = 2.25(.85)sin(102) = 1.87 m/s
\ddot{r} = (r_0cos\theta)(\dot{\theta})^2 + (r_0sin\theta)(\ddot{\theta}) = 2.25(.85²)cos(102) = -0.338 m/s²

a_r = \ddot{r} - r(\dot{\theta})^2
a_\theta = r\theta + 2\dot{r}\dot{\theta}

F_r = W/g * a_r
F_\theta = W/g * a_\theta

F_s = ks (spring)

The Attempt at a Solution

a_r = \ddot{r} - r(\dot{\theta})^2 = -0.338 - 4.9678(.85^2) = -3.927 m/s²

a_\theta = r\theta + 2\dot{r}\dot{\theta} = 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_\theta = W/g * a_\theta = (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

 

[sgd37]

it opposes the acceleration along r

 

[sami23]

I found F_r = -3.122 lb and F_\theta = 2.531 lb when a_r = -3.927 rad/s² and a_\theta = 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 \theta = 102 degrees?
a = a_t + a_n

a_t = dv/dt (angular velocity)
a_n = v²/\rho where \rho = 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.