A slotted bar is attached to the origin and rotates in the horizontal plane with a constant angular velocity of 0.665 radians/s . The bar moves a roller weighing 33.6 lb along the cam's perimeter. A spring holds the roller in place; the spring's spring constant is 1.34 . The friction in the system is negligible. When theta = 101 degrees, what are Fr and F(theta), the magnitudes of the cylindrical components of the total force acting on the roller?

Sorry, but I don't get the picture.

If the bar is rotating in a horizontal plane with constant angular velocity, the magnitudes of the forces do not change with time nor position (θ).

Is the spring installed in a radial direction keeping the roller, or otherwise how?

Is the roller rolling in the radial direction?

A diagram would certainly help.

To find the magnitudes of the cylindrical components of the total force acting on the roller, we can break down the force into two components: the radial force (Fr) and the tangential force (Fθ).

Given:
Angular velocity, ω = 0.665 radians/s
Weight of the roller, W = 33.6 lb
Spring constant, k = 1.34

To find Fr, we'll start by determining the centripetal force acting on the roller.

The centripetal force (Fc) is given by the equation:
Fc = mrω²

Where:
m = mass of the roller
r = radius of motion
ω = angular velocity

Since we are given the weight of the roller (W), we can convert it to mass using the equation:
W = mg

Where:
g = acceleration due to gravity

Let's first calculate the mass of the roller:
m = W/g
= 33.6 lb / 32.2 ft/s² (acceleration due to gravity)
≈ 1.045 lb•s²/ft

Note: We convert lb to a mass unit, lb•s²/ft, to maintain consistency with other units used in the calculations.

Now, let's find the radius of motion (r). In a slotted cam mechanism, the roller moves along the perimeter of the cam, so the radius of motion can be calculated using the length of the slotted bar.

Given the position angle, θ = 101 degrees

Let's calculate the arc length of the slotted bar:
arc length (s) = θ * radius of motion (r)

Since we are given the position angle in degrees, let's convert it to radians before using it in the equation:
θ (in radians) = θ (in degrees) * π/180

Substituting the given values into the equation:
s = (101 degrees * π/180) * r

Now, rearranging the equation to solve for r:
r = s / (101 degrees * π/180)

Next, we'll determine the centripetal force (Fc) using the mass (m) and the angular velocity (ω):
Fc = m * r * ω²

Finally, we have determined the radial force (Fr), which is equal to the centripetal force:
Fr = Fc

Next, to find Fθ, we'll calculate the force exerted by the spring using Hooke's Law.

Hooke's Law states:
F = -k * δ

Where:
F = force exerted by the spring
k = spring constant
δ = displacement from the equilibrium position

Let's assume the equilibrium position (θ = 0) corresponds to the roller being at the lowest point of the cam. The displacement from the equilibrium position (δ) can be calculated as:
δ = s - (r * θ)

The negative sign in Hooke's Law indicates that the force exerted by the spring is always opposite to the displacement from the equilibrium position.

Now, we can calculate Fθ using Hooke's Law:
Fθ = -k * δ

After obtaining Fr and Fθ, we can calculate their magnitudes:
Fr = |Fr|
Fθ = |Fθ|

I will now perform the calculations for you.

To find the magnitudes of the cylindrical components of the total force acting on the roller, we need to consider the different forces acting on it.

First, we need to find the acceleration of the roller along the cam's perimeter. Since the angular velocity is constant, the acceleration can be calculated using the centripetal acceleration formula:

a = r * ω^2

where:
a = acceleration
r = radius of rotation = distance of roller from the origin
ω = angular velocity

Note that the roller is at a fixed distance from the origin, which is the radius of the slotted bar. However, the distance of the roller from the origin changes as the bar rotates.

To find the distance of the roller from the origin, we can use trigonometry. Considering the right-angled triangle formed by the radius of the bar, the perpendicular distance from the origin to the roller, and the angle θ, we can use the sine function:

sin(θ) = (distance of roller from origin) / (radius of bar)

Rearranging the equation, we get:

(distance of roller from origin) = sin(θ) * (radius of bar)

Substituting this value in the centripetal acceleration formula, we have:

a = (sin(θ) * radius of bar) * ω^2

Next, we need to find the force acting on the roller due to the spring. The force exerted by the spring is given by Hooke's law:

F = -k * x

where:
F = force exerted by the spring
k = spring constant
x = displacement from the equilibrium position

In this case, the displacement from the equilibrium position is the distance of the roller from the origin (same as before), so we can write:

F = -k * (sin(θ) * radius of bar)

Finally, the total force acting on the roller can be calculated by summing the forces in the radial and tangential directions:

Fr = m * a (radial force)
Fθ = m * aω (force in the tangential direction)

where:
m = mass of the roller

Substituting the values we have:

Fr = m * ((sin(θ) * radius of bar) * ω^2)
Fθ = m * ((sin(θ) * radius of bar) * ω)

Now you can plug in the given values (m = 33.6 lb, ω = 0.665 radians/s, radius of bar, and θ = 101 degrees) into the equations to find the magnitudes of Fr and Fθ.