\(\displaystyle 1 radian=452m\)

\(\displaystyle 200\frac{km}{hr}*\frac{1000m}{1 km}*\frac{1 radian}{452m}*\frac{1 hr}{3600sec}=\frac{.123 rad}{sec}\)

Finding angular velocity:

\(\displaystyle 50\frac{m}{s}*\frac{1 rotation}{2*\pi*.25m}*\frac{2\pi radians}{rotation}=200\frac{radians}{second}\)

Using

\(\displaystyle I=.5MR^2\)

and

\(\displaystyle L=I\omega\)

Where \(\displaystyle L\) is angular momentum

Combining equations and plugging in values:

\(\displaystyle L=.5*25*.25^2*200\)

\(\displaystyle L=156.25\frac{kg*m^2}{s}\)

The expression for angular momentum is:

\(\displaystyle L=I\cdot w\)

Where I is the moment of inertia for a solid sphere:

\(\displaystyle I= \frac{2}{5}mr^2\)

And w is the angular velocity:

\(\displaystyle w= \frac{v}{r}\)

Plugging these in, we get:

\(\displaystyle L = \frac{2}{5}mvr\)

Plugging in our values, we get:

\(\displaystyle L = \frac{2}{5}(6kg)\left ( 4\frac{m}{s}\right )(0.15m)\)

\(\displaystyle 1.4\frac{kg\cdot m^2}{s}\)

Let's begin with the expression for conservation of energy:

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

Since the ball begins from rest, we can eliminate initial kinetic energy. Also, if we assume that the bottom of the slope has a height of 0, we can eliminate final potential energy to get:

\(\displaystyle U_i=K_f\)

Plugging in expressions, we get:

\(\displaystyle mgh_i=\frac{1}{2}mv_f^2+\frac{1}{2}Iw^2\)

Where I is the moment of inertia for a solid sphere:

\(\displaystyle I = \frac{2}{5}mr^2\)

And w is the angular velocity:

\(\displaystyle w = \frac{v}{r}\)

Plugging these in, we get:

\(\displaystyle mgh_i = \frac{1}{2}mv_f^2+\frac{1}{5}mv_f^2\)

Rearranging for final velocity, we get:

\(\displaystyle v_f=\sqrt{\left (\frac{10}{7}gh_i \right )}\)

Plugging in our values, we get:

\(\displaystyle v_f=\sqrt{\frac{10}{7}\left ( 10\frac{m}{s^2}\right )(12m)}\)

\(\displaystyle v_f = 13.1\frac{m}{s}\)

\(\displaystyle L=rmv\)

Where \(\displaystyle r\) is the radius of the circle

\(\displaystyle m\) is the mass of the object 

\(\displaystyle v\) is the linear velocity of the object

The car on the outside has a larger \(\displaystyle r\) and a larger \(\displaystyle v\), and the same \(\displaystyle m\), thus it has a higher angular momentum.

The first part of this problem is a transfer between angular and linear momentum. Conservation of momentum tells us that we can equate the angular momentum of the pendulum to the linear momentum of the brick, or \(\displaystyle L = m_{1}\omega r = p = m_{2}v\), where \(\displaystyle m_{1}\) and \(\displaystyle m_{2}\) are the masses of the pendulum and brick respectively, \(\displaystyle \omega\) is the angular velocity of the pendulum in \(\displaystyle rad/s\), \(\displaystyle r\) is the length of the pendulum rope, and \(\displaystyle v\) is the resultant velocity of the brick. We find that the resulting velocity of the brick can be written as \(\displaystyle v = \frac{m_{1}\omega r}{m_{2}} = 31.4 \frac{m}{s}\)

The next part of the problem is identifying that the kinematic equations can be used here to find the acceleration the brick undergoes as it comes to a stop. Here we can use \(\displaystyle v^2 = v_{0}^2 + 2a\Delta x\), where \(\displaystyle v_{0}\) is the end velocity of the brick (zero), \(\displaystyle a\) is the deceleration it undergoes, and \(\displaystyle \Delta x\) is the distance the brick travels before coming to a stop. Solving for \(\displaystyle a\) we have that \(\displaystyle a = \frac{v^2}{2\Delta x} = -49.4 \frac{m}{s^2}\).

Lastly, we know that the kinetic friction force can be written as \(\displaystyle f_{k} = -\mu_{k}N = -\mu_{k}m_{2}g\). Newton's second law says that this can be written as \(\displaystyle f_{k} = -\mu_{k}m_{2}g = m_{2}a\). Thus we find that \(\displaystyle \mu_{k} = \frac{a}{-g} = \frac{-49.4\frac{m}{s^2}}{-9.8\frac{m}{s^2}} = 5.0\).