Since it is experiencing a constant torque and constant angular acceleration, the angular displacement can be calculated using:

$\displaystyle \Delta \theta =\omega_ot+\frac{1}{2}\alpha t^2$

The angular acceleration is easily calculated using the angular velocity and the time:

$\displaystyle \alpha=\frac{\Delta\omega}{\Delta t}=\frac{12\frac{rad}{s}-0\frac{rad}{s}}{3s}=4\frac{rad}{s^2}$

Using this value, we can find the angular displacement:

$\displaystyle \Delta \theta =(0\frac{rad}{s})(3s)+\frac{1}{2}(4\frac{rad}{s^2})(3s)^2$

$\displaystyle \Delta \theta=18rad$

Convert the angular displacement to revolutions by diving by $\displaystyle 2\pi$:

$\displaystyle \frac{18rad}{2\pi\frac{rad}{rev}}\approx2.9rev$

We can find the angular acceleration using the rotaional motion equivalent of Newton's second law. In rotational motion, torque is the product of moment of inertia and angular acceleration: 

$\displaystyle \tau = I\alpha$

The moment of inertia for a circular disk is:

$\displaystyle I=\frac{1}{2}mr^2$

The tourque is the product of force and distance (in this case, the radius):

$\displaystyle \tau=F\cdot r$

We can plug these into our first equation:

$\displaystyle (F)(r)=(\frac{1}{2}mr^2)(\alpha)$

Simplify and rearrange to derive an equation for angular acceleration:

$\displaystyle \alpha=\frac{Fr}{\frac{1}{2}mr^2}=\frac{2F}{mr}$

Use our given values to solve:

$\displaystyle \alpha=\frac{2(25N)}{(3kg)(0.5m)}=33.3\frac{rad}{s^2}$

Torque is given by,

$\displaystyle \tau =rFsin\theta$

Since all of the forces are equal in magnitude, the magnitude of the torque is then influenced by the radius r and the angle theta between the radius and the force.

For $\displaystyle F_1$,

$\displaystyle \tau_{1} =\left (\frac{1}{2}L \right )F\sin\, 45^{\circ}=0.354\cdot L\cdot F$

For $\displaystyle F_2$,

$\displaystyle \tau_{2} =\left (\frac{1}{2}L \right )F\sin\, 90^{\circ}=0.5\cdot L \cdot F$

For $\displaystyle F_3$,

$\displaystyle \tau_{3} =LF\sin\, 30^{\circ}=0.5 \cdot L \cdot F$

For $\displaystyle F_4$,

$\displaystyle \tau_{4} =LF\sin\, 90^{\circ}=L\cdot F$

Combining this information yields the relationship,

$\displaystyle \tau _{1} < \tau _{2} = \tau _{3} < \tau _{4}$

The net torque on the beam is given by addition of the torques caused by the weight of the man and the weight of the beam itself, each at its respective distance from the end of the beam:

$\displaystyle \tau _{net}=\tau _{man}+\tau _{beam}$

Let's assign the direction of positive torque in the direction of the torques of the man's and the beam's weights, noting that they will add together since they both point in the same direction.

$\displaystyle \tau _{net}=L\cdot W_{man}\cdot\sin 90^{\circ}+\left ( \frac{1}{2} L\right )\cdot W_{beam}\cdot\sin\, 90^{\circ}$

$\displaystyle \tau _{net}=L\cdot m_{man}\cdot g+\left ( \frac{1}{2} L\right )\cdot m_{beam}\cdot g$

We can further simplify by combining like terms:

$\displaystyle \tau _{net}=L\cdot g \cdot (m_{man}+\frac{1}{2} m_{beam})$

Using the given numerical values,

$\displaystyle \tau _{net}=(3.05\: m)\cdot(9.8\: m/s^{2})\cdot(85\: kg+\frac{1}{2} 40\: kg)=3140\: N\cdot m$

A torque analysis is appropriate in this situation due to the inclusion of distances from a given pivot point. Generally,

$\displaystyle \tau_{net}=I\alpha$

This is a static situation. There are two torques about the pivot caused by the weights of two children. We will note that these weights cause torques in opposite directions about the pivot, such that

$\displaystyle \tau _{net}=0=\tau _{1}-\tau _{2}$

Consequently,

$\displaystyle \tau _{1}=\tau _{2}$

$\displaystyle r_{1}\cdot m_{1} \cdot g \cdot \sin\: 90^{\circ}=r_{2}\cdot m_{2}\cdot g \cdot \sin\: 90^{\circ}$

Or more simply,

$\displaystyle r_{1}\cdot m_{1}=r_{2} \cdot m_{2}$

Solving for $\displaystyle m_2$,

$\displaystyle m_{2}=\frac{m_{1}r_{1}}{r_{2}}=\frac{36\: kg\cdot1.8\: m}{2.4\: m}}=27\: kg$

A torque analysis is appropriate in this situation due to the inclusion of distances from a given pivot point. Generally,

$\displaystyle \tau_{net}=I\alpha$

This is a static situation. As such, any pivot point can be chosen about which to do a torque analysis. The quickest way to the unknown force asked for in the question is to do the torque analysis about the left end of the plank. There are three torques about this pivot: two clockwise caused by the weights of the gymnast and the plank itself, and one counterclockwise caused by the force from the right support. Designating clockwise as positive,

$\displaystyle \tau _{net}=0=\tau _{gymnast}+\tau _{plank}-\tau _{right}$

Consequently,

$\displaystyle \tau _{gymnast}+\tau _{plank}=\tau _{right}$

$\displaystyle r _{gymnast}\cdot W_{gymnast}\cdot\sin\: 90^{\circ}+r _{plank}\cdot W_{plank}\cdot \sin\: 90^{\circ}=r _{right}\cdot F_{right}\cdot \sin\: 90^{\circ}$

This simplifies to

$\displaystyle r _{gymnast}\cdot m_{gymnast}\cdot g+r _{plank}\cdot m_{plank}\cdot g=r _{right}\cdot F_{right}$

Solving for $\displaystyle F_{right}$,

$\displaystyle F_{right} = \frac{(r _{gymnast}\cdot m_{gymnast}+r _{plank}\cdot m_{plank})\cdot g}{r _{right}}$

Solving with numerical values,

$\displaystyle F_{right} = \frac{(2\: m\cdot 55\: kg+3\: m\cdot 23\: kg)\cdot 9.8\: m/s^{2}}{6\: m}=290\: N$

The moments of inertia for both axes $\displaystyle A$ and $\displaystyle B$ are equal because both of these axes are equivalently passing through the center of the mass.

By the parallel axis theorem for moments of inertia ($\displaystyle I = I_{cm}+ MD^2$), the moment of inertia for axis $\displaystyle C$ is larger than $\displaystyle A$ or $\displaystyle B$ 

because it is located a distance

$\displaystyle \frac{L}{2}$ away from the center of mass.

The moment of inertia for a point mass is $\displaystyle I = mR^2$.

To calculate the total moment of inertia, we add the moment of inertia for each part of the object, such that

$\displaystyle I_{total}=I_{1}+I_{2}+I_{3}+I_{4}$

$\displaystyle I_{total}=m_{1}R_{1}^{2}+m_{2}R_{2}^{2}+m_{3}R_{3}^{2}+m_{4}R_{4}^{2}$

The masses of the bowls are all equal in this problem, so this simplifies to

$\displaystyle I_{total}=m (R_{1}^{2}+R_{2}^{2}+R_{3}^{2}+R_{4}^{2})$

Plugging in and solving with numerical values,

$\displaystyle I_{total}=(0.6\: kg) [(0.8\: m)^{2}+(0.6\: m)^{2}+(0.4\: m)^{2}+(0.2\: m)^{2}]=0.72\: kg\cdot m^{2}}$

For a long thin rod about its center of mass,

$\displaystyle I_{cm}=\frac{1}{12}mL^{2}$

According to the parallel axis theorem,

$\displaystyle I_{//}=I_{cm}+md^{2}$

where $\displaystyle d$ is defined to be the distance between the center of mass of the object and the location of the axis parallel to one through the center of mass. For this problem, $\displaystyle d$ is the difference between the given distance and half the length of the rod.

Combining the above,

$\displaystyle I=\frac{1}{12}mL^{2}+m(\frac{1}{2}L-l)^{2}$

Inserting numerical values,

$\displaystyle I=\frac{1}{12}(12)(1.82)^{2}+(12)(\frac{1.82}{2}-.036)^{2}=6.9\: kg\cdot m^{2}$

The moment of inertia for a long uniform thin rod about its end is given by

$\displaystyle I_{rod,end}=\frac{1}{3}mL^{2}$

Reducing the mass and the length by a factor of four introduces the following factors into the equation,

$\displaystyle I_{new}=\frac{1}{3}(\frac{1}{4}m)(\frac{1}{4}L)^{2}=\frac{1}{3}(\frac{1}{4})m(\frac{1}{64})L^{2}$

Simplifying,

$\displaystyle I_{new}=(\frac{1}{64})\frac{1}{3}mL^{2}=\frac{1}{64}I_{0}$