Step 1

We are given the angular velocity of the concrete culvert is $\omega = 0.5\;{\rm{rad/s}}$, the radius of gyration is ${k_G} = 3.5\;{\rm{ft}}$, and the weight of the culvert and man is $W = 500\;{\rm{lb}}$.

We are asked to determine the angular acceleration of the culvert.

 
Step 2

The diagram of the system is shown as:

We have the radius of the culvert is $r = 4\;{\rm{ft}}$.

We have the distance between point O and G is $d = 0.5\;{\rm{ft}}$.

 
Step 3

The formula to calculate the horizontal component of acceleration of point G is,

\[{\left( {{a_G}} \right)_x} = {a_0} - d{\omega ^2}\]
 
Step 4

Substitute the values in the above expression.

\[\begin{array}{c} {\left( {{a_G}} \right)_x} = {a_0} - \left( {0.5} \right){\left( {0.5} \right)^2}\\ {\left( {{a_G}} \right)_x} = {a_0} - 0.125 \end{array}\] ... (1)
 
Step 5

The formula to calculate the vertical component of acceleration of point G is,

\[{\left( {{a_G}} \right)_y} = d\alpha \]
 
Step 6

Substitute the values in the above expression.

\[\begin{array}{l} {\left( {{a_G}} \right)_y} = \left( {0.5} \right)\alpha \\ {\left( {{a_G}} \right)_y} = 0.5\alpha \end{array}\] ... (2)
 
Step 7

For condition of no slipping, the acceleration of point A is zero.

The formula to calculate the acceleration at point A is,

\[{a_A} = {a_0} - r\alpha \]
 
Step 8

Substitute the values in the above expression.

\[\begin{array}{c} 0 = {a_0} - \left( 4 \right)\alpha \\ {a_0} = 4\alpha \end{array}\] .. (3)
 
Step 9

The formula to calculate the moment about the point A is,

\[\begin{array}{l} Wd = {I_G}\alpha + \left[ {m{{\left( {{a_G}} \right)}_x} \times r} \right] + \left[ {m{{\left( {{a_G}} \right)}_y} \times d} \right]\\ Wd = {I_G}\alpha + \left[ {\left( {\frac{W}{g}} \right){{\left( {{a_G}} \right)}_x} \times r} \right] + \left[ {\left( {\frac{W}{g}} \right){{\left( {{a_G}} \right)}_y} \times d} \right] \end{array}\] ... (4)
 
Step 10

The formula to calculate the moment of inertia about the point G is,

\[\begin{array}{l} {I_G} = m{\left( {{k_G}} \right)^2}\\ {I_G} = \left( {\frac{W}{g}} \right){\left( {{k_G}} \right)^2} \end{array}\]

Here, g is the gravitational acceleration having a standard value of $32.2\;{\rm{ft/}}{{\rm{s}}^2}$.

 
Step 11

Substitute the values in the above expression.

\[\begin{array}{l} {I_G} = \left( {\frac{{500\;{\rm{lb}}}}{{32.2\;{\rm{ft/}}{{\rm{s}}^2}}}} \right){\left( {3.5\;{\rm{ft}}} \right)^2}\\ {I_G} = 190.22\;{\rm{lb}} \cdot {\rm{ft}} \cdot {{\rm{s}}^2} \end{array}\]
 
Step 12

On solving equations (1), (2) and (3), we get:

\[\begin{array}{l} {\left( {{a_G}} \right)_x} = 4\alpha - 0.125\\ {\left( {{a_G}} \right)_y} = 0.5\alpha \end{array}\]
 
Step 13

Substitute the values in the equation (4).

\[\begin{array}{c} \left( {500} \right)\left( {0.5} \right) = \left( {190.22} \right)\alpha + \left[ {\left( {\frac{{500}}{{32.2}}} \right)\left( {4\alpha - 0.125} \right) \times \left( 4 \right)} \right] + \left[ {\left( {\frac{{500}}{{32.2}}} \right)\left( {0.5\alpha } \right) \times \left( {0.5} \right)} \right]\\ 257.764 = 442.54\alpha \\ \alpha = 0.58\;{\rm{rad/}}{{\rm{s}}^2} \end{array}\]