at the instant shown, the 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 degrees clockwise.

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To solve this problem, we need to use conservation of energy. At its initial state, the bar has rotational kinetic energy, and the spring has potential energy stored in it. When the bar has rotated 30 degrees clockwise, it still has rotational kinetic energy, and the spring will still have potential energy if it is stretched from its original position.

First, let's find the initial energy of the bar and the spring.

1. Rotational kinetic energy of the bar:

KE_rotational = 0.5 * I * ω^2
where I is the moment of inertia of the bar, and ω is the angular velocity.
Assuming the bar is a slender rod, its moment of inertia can be found using the formula:
I = (1/3) * m * L^2, where m is the mass of the bar and L is its length.

Mass of the bar, m = 50 lb * (1 slug / 32.2 lb) = 1.55 slug (approximately)
Length of the bar, L = 4 ft

Moment of inertia, I = (1/3) * 1.55 * 4^2 = 8.2 slug*ft^2

Initial angular velocity, ω1 = 2 rad/s

Now we can find the initial rotational kinetic energy of the bar:

KE_rotational1 = 0.5 * 8.2 * 2^2 = 16.4 ft*lb (approximately)

2. Potential energy stored in the spring:

PE_spring1 = 0.5 * k * Δx^2
where k is the stiffness of the spring and Δx is the initial stretch of the spring.

We can find the initial stretch of the spring using trigonometry:

2 ft (unstretched length) - (4 ft * cos(30)) = 0.268 ft (approximately)
So, the initial stretch is roughly 0.268 ft.

Now we can find the initial potential energy of the spring:

PE_spring1 = 0.5 * 6 * 0.268^2 = 0.214 ft*lb (approximately)

Now, let's find the angular velocity when the bar has rotated 30 degrees clockwise.

When the bar has rotated 30 degrees, the spring will still have potential energy, however, the displacement, and therefore the amount of energy stored, will be different. At this instant, the length of the spring can be calculated as follows:

New length of the spring = 4 * cos(30 + 30) = 2 * (1 - cos(60)) = 1.732 ft (approximately)

New stretch of the spring, Δx2 = 1.732 - 2 = -0.268 ft

PE_spring2 = 0.5 * 6 * (-0.268)^2 = 0.214 ft*lb (approximately)

It can be seen that the spring has stored the same amount of potential energy at both instants in our case.

Since the system is conservative (no external forces), by conservation of energy, we have:

(initial system energy) = (final system energy)
KE_rotational1 + PE_spring1 = KE_rotational2 + PE_spring2

Let ω2 be the angular velocity at 30 degrees. Then:

16.4 + 0.214 = 0.5 * 8.2 * ω2^2 + 0.214
16.4 = 4.1 * ω2^2
ω2^2 = 4 (approximately)
ω2 = ±2 rad/s

The angular velocity of the bar when it has rotated 30 degrees clockwise could be either 2 rad/s (if the bar is decelerating) or -2 rad/s (if the bar is accelerating).