To calculate torque, this equation is needed:

$\displaystyle \tau=rFsin\Theta$

Next, identify the given information:

$\displaystyle r= 0.5 m$

$\displaystyle F=1.3N$

$\displaystyle \Theta=45^{o}$

Plug these numbers into the equation to determine the torque:

$\displaystyle \tau=rFsin\Theta = (0.5m)(1.5N)sin(45^{o})=0.46N\cdot m$

We can determine the vertical components by using some geometry.

Since this problem boils down to being a problem of conservation of energy, we can state that

$\displaystyle E_{i}=E_{f}\rightarrow PE_{i}+KE_{i}=PE_{f}+KE_{f}$

Since the mass is released from rest and we can state at the bottom of its arc it will be at height 0 m, this equation can be simplified:

$\displaystyle PE_{i}=KE_{f} \rightarrow mgh=\frac{1}{2}mv^{2} \rightarrow gh=\frac{1}{2}v^{2}$

$\displaystyle h$ is equal to the height of the mass above 0 m; therefore, by referring to the diagram, we can determine that $\displaystyle h=R-Rcos\Theta$. By plugging this in and rearranging the equation, we can solve for the speed of the mass when the string is vertically aligned:

$\displaystyle v=\sqrt{2gR(1-cos\Theta)}=\sqrt{2(9.81\frac{m}{s^2})(8\cdot10^{-2}m)(1-cos30^{o})}=0.46\frac{m}{s}$

When the object is released, it has no kinetic energy (it isn't moving) and a potential energy of

$\displaystyle P.E.=\frac{1}{2}kx^2=\frac{1}{2}k(10m)^2=50m^2\cdot k$

When the object passes through the equilibrium point, it has no potential energy ($\displaystyle x=0$) and a kinetic energy of

$\displaystyle K.E.=\frac{1}{2}mv^2=\frac{1}{2}(5kg)(7\frac{m}{s})^2=122.5\frac{kg\cdot m^2}{s^2}$

Due to the conservation of energy, these two quantities must be equal to each other:

$\displaystyle 50m^2\cdot k=122.5\frac{kg*m^2}{s^2}$

$\displaystyle k = \frac{122.5\frac{kg\cdot m^2}{s^2}}{50m^2}=2.45\frac{kg}{s^2}$

First, we should convert 80 centimeters to 0.8 meters.

We know the force applied to the weight by gravity is

$\displaystyle F_g=mg$

$\displaystyle 294N=(30kg)(9.81\frac{m}{s^2})$

The force applied by the spring in the opposite direction must be equal to this:

$\displaystyle -kx=-294N$

$\displaystyle k(0.8m) = 294N$

$\displaystyle k=368\frac{N}{m}=368\frac{kg}{s^2}$

The angular velocity of the harmonic system is equal to the square root of the spring constant over the mass:

$\displaystyle \omega=\sqrt{\frac{k}{m}}$ 

Since we need the angular velocity to be $\displaystyle \frac{2\pi}{60}$ and we are given the spring constant as $\displaystyle 0.01\frac{kg}{s^2}$, then we can set up an equation:

$\displaystyle \frac{2\pi}{60}=\sqrt{\frac{0.01}{m}}$

$\displaystyle \frac{4\pi^2}{60^2}=\frac{0.01}{m}$

$\displaystyle m=\frac{60^2*0.01}{4\pi^2}=\frac{9}{\pi^2}kg$

The velocity equation can be found by differentiating the position equation.

$\displaystyle v(t)=x'(t)=\frac{d}{dt}(2sin(4\pi t)+4)=2*4\pi cos(4\pi t)$

$\displaystyle =8\pi cos(4\pi t)$

Here, we made use of the chain rule in taking the derivative.

You may recall the kinematic equation that relates final velocity, initial velocity, acceleration, and distance, respectively:

$\displaystyle v_{f}^{2}=v_{i}^{2}+2ad$

Well, for rotational motion (such as in this problem), there is a similar equation, except it relates final angular velocity, intial angular velocity, angular acceleration, and angular distance, respectively:

$\displaystyle \omega _{f}^{2}=\omega _{i}^{2}+2\alpha \theta$

The wheel starts at rest, so the initial angular velocity, $\displaystyle \omega _{i}$, is zero. The total number of revolutions of the wheel is given to be 5 revolutions. Each revolution is equivalent to an angular distance of $\displaystyle 2\pi$ radians. So, we can convert the total revolutions to an angular distance to get:

$\displaystyle \theta=5\:rev\cdot \frac{2\pi }{1\:rev}=10\pi$

The final angular velocity was given as $\displaystyle \omega$ in the text of the question. So, we should use the above equation to solve for the angular acceleration, $\displaystyle \alpha$.

$\displaystyle \omega ^{2}=0+2\alpha\cdot 10\pi$

$\displaystyle \alpha=\frac{\omega ^{2}}{20\pi }$

For a rotating object, or an object moving in a circular path, the relationship between angular acceleration and linear acceleration is

$\displaystyle a=\alpha r$

Linear acceleration is given by $\displaystyle a$, angular acceleration is $\displaystyle \alpha$, and the radius of the circular path is $\displaystyle r$.

For circular/centripetal motion, the linear acceleration is related to the object's linear velocity by

$\displaystyle a_{centripetal}=\frac{v^{2}}{r}$

We know the linear velocity is $\displaystyle 9.0 \:\frac{m}{s}$, and the radius is 1.5 m, so we can find the linear acceleration...

$\displaystyle a_{c}=\frac{(9.0\:\frac{m}{s})^{2}}{1.5\:m}=54\:\frac{m}{s^2}$

Now that we have the linear acceleration, we can use this in the equation at the top to find the angular acceleration...

$\displaystyle \alpha=\frac{a}{r}=\frac{54\:\frac{m}{s^2}}{1.5\Lm}=36\:\frac{rad}{s^2}$

The definition of angular velocity is $\displaystyle \omega=\frac{\Delta \Theta }{\Delta t}$. 

By identifying the given information to be $\displaystyle \Delta \Theta =2\pi$ and $\displaystyle \Delta t =3s$, we can plug this into the equation to calculate the angular velocity:

$\displaystyle \omega=\frac{2\pi}{3 s}=2.09s^-1$

The angular velocity of the second hand of a clock can be found by dividing the number of radians the second hand will travel over a known period of time. Thankfully for a clock, we know that the second hand will make one revolution, i.e. covering $\displaystyle 2\pi rad$ in one minute, or 60s. The formula for angular velocity is:

$\displaystyle \omega = \frac{\theta}{t}$

So the angular velocity, $\displaystyle \omega$ is $\displaystyle \frac{2\pi rad}{60s}$, which simplifies to our answer, $\displaystyle \frac{\pi rad}{30s}$