A 50 lb bar rotates clockwise 2 rad/s. The spring attached to its end always remains vertical due to the roller guide at C. If the spring has an unstretched length of 2 ft and a stiffness of k=6lb/ft, determine the angular velovity of the bar the instant it has rotated 30 clockwise
image is a bar horizontal at end spring is attached and is vertical, at other end of spring is a roller(C)
To determine the angular velocity of the bar when it has rotated 30 degrees clockwise (not radians), we can use the concept of conservation of mechanical energy.
First, let's convert the given information into the appropriate units:
- The unstretched length of the spring is 2 ft.
- The stiffness of the spring is k = 6 lb/ft.
- The rotation speed is given as 2 rad/s.
Now, let's calculate the potential energy stored in the stretched spring when the bar has rotated 30 degrees. We'll then equate this to the kinetic energy of the rotating bar.
1. Find the change in length of the spring:
The angular displacement (θ) is given as 30 degrees. Convert it to radians: θ = 30° * (π/180) = π/6 radians.
The change in the length of the spring can be calculated using the formula:
ΔL = L - L0, where L is the stretched length of the spring and L0 is the unstretched length.
ΔL = 2 ft * (1 - cos(π/6)) ≈ 2 ft * (1 - 0.866) = 0.27 ft.
2. Calculate the potential energy stored in the spring:
Using the formula for the potential energy stored in a spring:
PE = (1/2) * k * ΔL^2, where k is the spring stiffness and ΔL is the change in length of the spring.
PE = (1/2) * 6 lb/ft * (0.27 ft)^2 ≈ 0.22 lb*ft.
3. Equate potential energy to kinetic energy:
Since the system is conserving mechanical energy, we can write:
PE = KE, where KE is the kinetic energy of the rotating bar.
KE = (1/2) * I * ω^2, where I is the moment of inertia and ω is the angular velocity.
4. Calculate the moment of inertia:
The moment of inertia of the bar can be calculated using the formula:
I = (1/3) * m * L^2, where m is the mass of the bar and L is the length of the bar.
Given that the bar weighs 50 lb, and its length is not specified, we'll assume a length of 1 ft for simplicity.
I = (1/3) * 50 lb * (1 ft)^2 = 16.67 lb*ft^2.
5. Solve for the angular velocity:
Equating potential energy to kinetic energy and substituting the values, we get:
0.22 lb*ft = (1/2) * 16.67 lb*ft^2 * ω^2.
Simplifying, we find that:
ω^2 = (0.22 lb*ft * 2) / (16.67 lb*ft^2) ≈ 0.0264.
Taking the square root of both sides, we find:
ω = sqrt(0.0264) ≈ 0.163 rad/s.
Therefore, the angular velocity of the bar when it has rotated 30 degrees clockwise is approximately 0.163 rad/s.