An object moving in constant circular motion must have centripetal acceleration, directed towards the center of its path, in order to maintain its circular motion. So \(\displaystyle \small a_{A}\neq0\) and is towards the center of car A's path. 

An object moving linearly at constant velocity has 0 acceleration, so \(\displaystyle \small a_{B}=0\).

In general, we can think of acceleration as a change in either the magnitude or direction of an object's velocity. Neither of these cars has a change in the magnitude of its velocity, but car A does experience a change in the direction of its velocity, while car B does not.

Centripetal acceleration is given by the equation:

\(\displaystyle a_c=\frac{v^2}{r}\)

We are given the angular velocity and the diameter. We can easily solve for the radius from the diameter, and use the angular velocity to find the linear velocity.

Convert the angular velocity to the proper units by changing revolutions to radians.

\(\displaystyle \omega=\frac{8rev}{s}*\frac{2\pi rad}{1rev}=16\pi\frac{rad}{s}\)

Now we can convert the angular velocity to the linear velocity using the radius.

\(\displaystyle v=\omega*r=16\pi\frac{rad}{s}*0.30m=4.8\pi\frac{m}{s}\)

 Use the radius and the linear velocity to calculate the centripetal acceleration from the original equation.

\(\displaystyle a_c=\frac{v^2}{r}=\frac{(4.8\pi\frac{m}{s})^2}{0.30m}\)

\(\displaystyle a_c=\frac{23.04\pi^2\frac{m^2}{s^2}}{0.30m}\)

\(\displaystyle a_c=76.8\pi^2\frac{m}{s^2}\)

Centripetal force is given by the equation:

\(\displaystyle F_c = ma_c=m\frac{v^2}{r}\)

The length of the string represents the radius of the circle being formed. Use the given string length as the radius, along with the mass of the ball and the velocity, to calculate the centripetal force.

\(\displaystyle F_c=(10kg)\frac{(5\frac{m}{s})^2}{1m}\)

\(\displaystyle F_c=250N\)

Note that we are asked the linear velocity of the pulley, not the angular velocity. With Newton’s second law, we know that F = ma. Additionally, in rotational motion, we know that \(\displaystyle F=\frac{mv^2}{r}\). The force (F) the instant the rope is cut is due to the platform with the 4 weights. Knowing the mass and radius of the pulley, we can solve for the linear velocity:

 \(\displaystyle F=\frac{mv^2}{r}\)

\(\displaystyle v=\sqrt{\frac{Fr}{m}}=\sqrt{\frac{(19.6N)(0.25m)}{0.01kg}}=22.1m/s\)

To calculate g's, we need to know the centripetal force the rider is experiencing. To calculate centripetal force, we need to know how fast the ride is traveling at the top of the loop. To do this, we will use the equation for conservation of energy:

\(\displaystyle E = U_i + K_i = U_f +K_f + F\)

If we set the height of the top of the loop to 10m, we can say that there is no initial potential energy. Since we are solving for final velocity, let's rearrange this for final kinetic energy:

\(\displaystyle K_f = K_i - U_f - F\)

Plugging in the expressions for each of these terms, we get:

\(\displaystyle \frac{1}{2}mv_f^2 = \frac{1}{2}mv_i^2-mgh_f - F_fd\)

Rearranging for final velocity:

\(\displaystyle v_f = \sqrt{v_i^2-2gh_f-\frac{2F_fd}{m}}\)

We have all of the values except for how far the coaster travels. It travels from the bottom to the top of the loop, which is half the circumference of the circle:

\(\displaystyle d = \pi r = 5\pi\)

Now we can plug in our values and solve:

\(\displaystyle v_f = \sqrt{(20\frac{m}{s})^2-(2\cdot10\frac{m}{s^2}\cdot10m)-\frac{2\cdot600N\cdot5\pi}{1075kg}}\)

\(\displaystyle v_f = \sqrt{400\frac{m^2}{s^2} - 200\frac{m^2}{s^2} - 5.58\pi\frac{m^2}{s^2}} = \sqrt{182.47\frac{m^2}{s^2}} = 13.51 \frac{m}{s}\)

Now that we know how fast the ride is traveling at the top of the loop, we can calculate the centripetal force on the man:

\(\displaystyle F_c = ma_c = m\frac{v^2}{r}\)

\(\displaystyle F_c = (75kg)\frac{(13.51\frac{m}{s})^2}{5m} = 2737 N\)

Now that we know the force in Newtons, we can convert this value to g's.

For this man:

\(\displaystyle 1'g' = mg = (75kg)(10\frac{m}{s^2}) = 750 N\)

Therefore,

\(\displaystyle G = \frac{2737N}{750\frac{N}{g}} = 3.65g\)