The problem gives us a value in terms of revolutions per second, but we need radians per second.

Remember that $\displaystyle 1rev=2\pi rad$ so we can convert:

$\displaystyle 31\frac{rev}{s}*\frac{2\pi rad}{1rev}=62\pi\frac{rad}{s}=194.78\frac{rad}{s}$

The formula for angular momentum is:

$\displaystyle L=I\omega$

We need to find values for the moment of inertia and the angular velocity in order to find the angular momentum.

The tension in the string will be equal to the centripetal force on the ball.

$\displaystyle F_c=F_T$

Centripetal force is given by the formula:

$\displaystyle F_c=m\frac{v^2}{r}$

We are given the mass of the ball, its velocity, and the length of the string, which will determine the radius.

$\displaystyle F_c=(2.34kg)\frac{(5.2\frac{m}{s})^2}{3m}$

$\displaystyle F_c=(2.34kg)(9.01\frac{m}{s^2})$

$\displaystyle F_c=21.09N$

Since the centripetal force and force of tension will be equal, this is our answer.

$\displaystyle F_c=F_T=21.09N$

The relationship between linear and angular velocity is:

$\displaystyle \omega=\frac{v}{r}$

We know the radius, but we need to find the linear velocity. Fortunately, that's contained in the linear momentum. We know both the momentum and the mass, so we can find the linear velocity.

$\displaystyle p=m*v$

$\displaystyle 35\frac{kg*m}{s}=(2.2kg)v$

$\displaystyle \frac{35\frac{kg*m}{s}}{2.2kg}=v$

$\displaystyle 15.91\frac{m}{s}=v$

Plug that into our initial equation, along with the radius, to convert it to angular velocity.

$\displaystyle \omega=\frac{15.91\frac{m}{s}}{3.1m}$

$\displaystyle \omega=5.13\frac{rad}{s}$

Centripetal acceleration is equal to the tangential velocity squared over the radius:

$\displaystyle a_c=\frac{v^2}{r}$

We know the radius, but we need to find the linear velocity. Fortunately, that's contained in the linear momentum. We know both the momentum and the mass, so we can find the linear velocity.

$\displaystyle p=m*v$

$\displaystyle 35\frac{kg*m}{s}=(2.2kg)v$

$\displaystyle \frac{35\frac{kg*m}{s}}{2.2kg}=v$

$\displaystyle 15.91\frac{m}{s}=v$

Use the linear velocity and the radius in the initial equation to solve for the centripetal acceleration.

$\displaystyle a_c=\frac{v^2}{r}$

$\displaystyle a_c=\frac{(15.91\frac{m}{s})^2}{3.1m}$

$\displaystyle a_c=\frac{253.128\frac{m^2}{s^2}}{3.1m}$

$\displaystyle a_c=81.65\frac{m}{s^2}$

There is a direct relationship between angular momentum and linear momentum. Angular momentum is equal to the linear momentum times the radius:

$\displaystyle L=p*r$

We are given the value of both the linear momentum and the radius, allowing us to solve for the angular momentum.

$\displaystyle L=(35kg*\frac{m}{s})(3.1m)$

$\displaystyle L=108.5\,J*s$

Since the object is moving in a perfect circle and not rotating about some fixed point within itself (like spinning a ball or a frisbee), the equation for moment of inertia is:

$\displaystyle I=mr^2$

We are given the mass and radius, allowing us to calculate the moment of inertia from this equation.

$\displaystyle I=(2.2kg)(3.1m)^2$

$\displaystyle I=(2.2kg)(9.61m^2)$

$\displaystyle I=21.14kg*m^2$

The relationship between period and angular velocity is:

$\displaystyle \omega=\frac{2\pi}{T}$

Angular displacement is equal to the angular velocity times time:

$\displaystyle \theta=\omega*\Delta t$

We know the time, but we need to solve for the angular velocity. The relationship between linear and angular velocity is:

We can set up a proportion here:

$\displaystyle \frac{1rev}{T}=\frac{x\, rev}{14s}$.

In other words, the object can do one revolution per period ($\displaystyle T$), so it can do $\displaystyle x$ revolutions in $\displaystyle 14s$. First we need to find the period.