Dynamics - Cylindrical Coordinates

In summary, the problem involves a cam with a slotted bar attached to it, a roller with weight 25.6 lb, and a spring with spring constant of 1.75 lb/ft. The system has negligible friction and the roller moves along the cam's perimeter with a constant angular velocity of 0.85 radians/s. The goal is to find the magnitudes of the cylindrical components of the total force acting on the roller when \theta = 102 degrees. Using the equations of motion, we can find that F_r = -3.122 lb and F_\theta = 2.531 lb. To find the tangential force, F_t, and the normal force, N, we need to first calculate the
  • #1
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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it opposes the acceleration along r
 
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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.
 

Related to Dynamics - Cylindrical Coordinates

1. What are cylindrical coordinates?

Cylindrical coordinates are a type of coordinate system used to describe the location of a point in three-dimensional space. They consist of three components: the distance from the origin (ρ), the angle from the positive x-axis (ϕ), and the height above the xy-plane (z).

2. How are cylindrical coordinates related to Cartesian coordinates?

Cylindrical coordinates are related to Cartesian coordinates through the following equations: x = ρcos(ϕ), y = ρsin(ϕ), and z = z. These equations allow for the conversion of coordinates between the two systems.

3. What is the significance of using cylindrical coordinates in dynamics?

Cylindrical coordinates are useful in dynamics because they allow for the description of motion in three-dimensional space, which is often the case in real-world scenarios. They also simplify certain equations and make it easier to solve problems involving cylindrical symmetry.

4. How do you calculate the velocity and acceleration of a point using cylindrical coordinates?

The velocity of a point in cylindrical coordinates can be calculated using the equations: vx = ρcos(ϕ)uρ - ρsin(ϕ)uϕ + wz, vy = ρsin(ϕ)uρ + ρcos(ϕ)uϕ + wz, and vz = wz, where uρ, uϕ, and uz are the unit vectors in the ρ, ϕ, and z directions, respectively. The acceleration can be calculated using the same equations, but with the addition of angular acceleration (α) terms.

5. What are some common applications of cylindrical coordinates in dynamics?

Cylindrical coordinates are commonly used in dynamics for problems involving circular motion, such as the motion of objects on a turntable or a rotating wheel. They are also useful in problems involving cylindrical objects, such as a spinning top or a rolling cylinder. Additionally, they are used in the study of fluid mechanics and electromagnetics.

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