Crane Problem, Finding Angular Velocity

AI Thread Summary
The discussion focuses on calculating the angular velocity components (wx, wy, wz) of a crane's boom in a fixed reference frame. The crane's boom is positioned at a 33-degree angle, rotating with specific angular velocities and accelerations. The user attempts to solve for the angular velocity using the equation w = w1 * w2 and incorporates the acceleration into their calculations. Despite their efforts, the final answer is incorrect, prompting a request for clarification on how to properly account for the crane's acceleration in the equations. The conversation emphasizes the complexities of integrating angular motion and acceleration in dynamic systems.
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Homework Statement



At the instant shown in the figure , the crane's boom has an angle = 33.0 and rotates around the vertical axis with angular velocity = 8.00×10^−2 radians/s that increases at the rate = 7.00×10^−2 radians/s^2. At the same instant, the crane's boom is rotating upward with angular velocity = 3.00×10^−2 that increases at the rate = 5.00×10^−2 . Determine wx, wy, and wz, the i, j, and k components of w, the boom's angular velocity in the fixed reference frame X, Y, Z. The translating-rotating reference frame, x, y, z, lines up with the fixed reference frame at the instant shown.

> The figure shows a crane with the z-axis vertical and the y-axis along the cranes boom. The angular velocity are all rotating counterclockwise. The origin is at the center of the crane.

Homework Equations



vp = wp X rp

vp is the velocity at a point
w is the angular velocity
rp is the position of the vector at a point

The Attempt at a Solution



w = w1 * w2 = (.03i + .08k)rad/s
\Omega = .08k rad/s
w = w_{12} + \Omega X w = (.05i + .07k) + .08k X (.03i + .08k)

=0.05i + 0.0024j + 0.07k

That was my final answer but it is wrong.
 
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How are you taking the acceleration of the crane into account?
 
I'm taking into account the acceleration in my equation but the answer is still not correct:

w = + \Omega X w = (.05i + .07k) + .08k X (.03i + .08k)
 
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