A bucket crane consists of a uniform boom of mass
M = 201 kg and length L= 59.05 ft that pivots at a point on
the bed of a fixed truck. The truck supports an elevated bucket
with a worker inside at the other end of the boom, as shown in
the figure. The bucket and the worker together can be modeled
as a point mass of weight 201 lb located at the end point of
the boom.
Suppose that when the boom makes an angle of 68.7° with the
horizontal truck bed, the bucket crane suddenly loses power,
causing the bucket and boom to rotate freely toward the
ground. Find the magnitude of the angular acceleration la of
the system just after the crane loses power. Take the rotation
axis to be at the point where the boom pivots on the truck bed.
Use g = 9.81 m/s2 for the acceleration due to gravity. For unit
conversions, assume that 1 m = 3.28 ft and 1 lb = 4.45 N.
Express your answer to at least two decimal places.

Newton's second law for rotational motion allows finding the results for the angular acceleration of the system at the moment of losing power is:

              α =0.616 rad / s²

Newton's second law for rotational motion relates the torque to the moment of inertia and the angular acceleration, in the special case where the angular acceleration is zero, it is called the rotational equilibrium condition.

           [tex]\sum \tau = I \alpha[/tex]

           [tex]\tau = F \ r \ sin \theta[/tex]

where τ is the torque, F is the force, r the distance to the pivot point and θ the angle between the force, the distance, I is the moment of inertia and α is the angular acceleration.

In this case they indicate the values ​​in several different units, let's reduce to the international system of measurements (SI).

We place our reference system at the base of the beam on the truck and counterclockwise turns as positive, and in the attached we see a free-body diagram of the system.

          [tex]W_m \ L \ sin 68.7 + Mg \ \frac{L}{2} \ sin 68.7 = I \alpha[/tex]

The moment of inertia is an additive scalar quantity, the total moment of inertia is the moment of inertia of the pen plus the moment of inertia of the man.

Moment of inertia of the beam with respect to one end.

           [tex]I_[/tex] = ⅓ ML²

moment of inertia of man

           [tex]I_m = M_m L^2[/tex]

Let's substitute

           [tex](W_m + \frac{Mg}{2} ) \ L \ sin 68.7 = (\frac{1}{3} M + m_m ) L^2 \alpha \\\alpha = \frac{(W_m + \frac{Mg}{2} ) sin 68.7 }{ (\frac{1}{3} M + m_m ) L }[/tex]

Let's calculate        

The mass of man is

           m = [tex]\frac{W_m}{g}[/tex]

           m = 894.1 / 9.8

           m = 91.2 kg

          [tex]\alpha = \frac{(894.1 +2.1 \ 9.8 ) \ sin 68.7 }{ 18.0 ( \frac{201}{3} + 91 }[/tex]

          α = 0.616 rad / s²

In conclusion using Newton's second law for rotational motion we can find the results for the angular acceleration of the system at the moment of losing power is 0.616 rad / s²

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