Known

$\displaystyle r = 12cm$ which should be converted to meters ($\displaystyle 0.12m$)

$\displaystyle v = 40 \: revolutions \: per \: minute$ which should be converted to $\displaystyle m/s$

$\displaystyle \frac{40rev}{min} * \frac{2\pi (0.12m)}{rev} * \frac{1min}{60s} = 0.5m/s$

We are dealing with circular motion so the first thing to do is to identify the force that is causing the centripetal motion.  In this case it is the force of friction which is holding the penny in place.  Therefore we know that the centripetal force is equal to the force of friction.

$\displaystyle F_c = F_f$

The centripetal force is equal to the mass times the centripetal acceleration.

$\displaystyle F_c = ma_c$

To calculate the centripetal acceleration you will need to know the velocity of the penny and the radius that the penny sits at.

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

Therefore $\displaystyle F_c = \frac{mv^2}{r}$ 

We also know that the frictional force is related to the coefficient of friction and the normal force.

$\displaystyle F_f = \mu F_N$

In this case the penny is sitting on the surface and the gravitational force is equal in size to the normal force.  Therefore 

$\displaystyle F_f = \mu F_g$

The force of gravity is equal to the mass times the gravitational acceleration.

$\displaystyle F_f = \mu mg$

We can put all this together to a larger equation that says that

$\displaystyle \frac{mv^2}{r} = \mu mg$

Notice that mass falls out of both sides of the equation

$\displaystyle \frac{v^2}{r} = \mu g$

We can now rearrange and solve for the coefficient of friction

$\displaystyle \mu = \frac{v^2}{rg}$

Now we can plug in our known values

$\displaystyle \mu = \frac{0.5^2}{(0.12)(9.8)}$

In circular motion, velocity is tangential to the circular path. Since the object is moving counterclockwise, at the top of the circle this tangent line points to the left. It may help to draw a diagram to better visualize this motion.

Velocity is given by distance per unit time.

$\displaystyle v=\frac{d}{t}$

When moving in a circle, the distance is the circumference, and each rotation takes exactly one period. We can substitute into the velocity formula.

$\displaystyle v=\frac{C}{T}=\frac{2\pi r}{T}$

If the children are in a straight line, that means that their periods (how long it takes to make one revolution) will be the same. The only thing that changes is r, the distance from the center. Since radius is in the numerator, we can conclude that increasing the distance from the center will increase the velocity.

$\displaystyle v_1 = \frac{2 \pi(0.5m)}{T}$

$\displaystyle v_2=\frac{2 \pi(3m)}{T}$

Velocity is given by distance per unit time.

$\displaystyle v=\frac{d}{t}$

When moving in a circle, the distance is the circumference, and each rotation takes exactly one period. We can substitute into the velocity formula.

$\displaystyle v=\frac{C}{T}=\frac{2\pi r}{T}$

If the children are in a straight line, that means that their periods (how long it takes to make one revolution) will be the same. The only thing that changes is $\displaystyle r$, the distance from the center. Since radius is in the numerator, we can conclude that increasing the distance from the center will increase the velocity.

$\displaystyle v_1=\frac{2\pi (0.5m)}{T}=(0.5m)\frac{2\pi}{T}$

$\displaystyle v_2=\frac{2\pi (3m)}{T}=(3m)\frac{2\pi}{T}$

The centripetal force is what is acting on the rider.  At the top of the Ferris wheel, the normal force is pointing up, and the gravitational force is pointing down.  The sum of these two forces must equal the centripetal force pointing downward toward the center of the circle.  Therefore the normal force must be smaller than the gravitational force.  At the bottom of the Ferris wheel, the same forces are present.  However, the sum of these forces must equal the centripetal force point upward toward the center of the circle.  Therefore the normal force must be greater than the gravitational force.  Since the normal force must be greater than the gravitational force at the bottom and less than the gravitational force at the top, the force at the bottom must be greater than the force on the top.

Centripetal acceleration is the acceleration towards the center when an object is moving in a circle. Though the speed may be constant, the change in direction results in a non-zero acceleration.

The formula for this is $\displaystyle a_c=\frac{v^2}{r}$, where $\displaystyle v$ is the perceived tangential velocity and $\displaystyle r$ is the radius of the circle.

Plug in the given values and solve for the acceleration.

$\displaystyle a_c=\frac{(3\frac{m}{s})^2}{(1m)}$

$\displaystyle a_c=\frac{9\frac{m^2}{s^2}}{1m}$

$\displaystyle a_c=9\frac{m}{s^2}$

Centripetal force is the force that constantly moves the object towards the center; it is what keeps the object moving in a circle rather than flying off tangentially to the circle.

The formula for force is $\displaystyle F_c=ma_c$.

To find the centripetal force, we need to find the centripetal acceleration. We do this with the formula $\displaystyle a_c=\frac{v^2}{r}$, where $\displaystyle v$ is the perceived tangential velocity and $\displaystyle r$ is the radius of the circle.

Plug in the given values and solve for the acceleration.

$\displaystyle a_c=\frac{(3\frac{m}{s})^2}{(1m)}$

$\displaystyle a_c=\frac{9\frac{m^2}{s^2}}{1m}$

$\displaystyle a_c=9\frac{m}{s^2}$

Plug that into the first equation to solve for the force.

$\displaystyle F_c=(2kg)(9\frac{m}{s^2})$

$\displaystyle F_c=18N$

The period, $\displaystyle T$, of an object moving in circular motion is the amount of time it takes for the object to make one complete loop of the circle.

If we start with the linear understanding of velocity,$\displaystyle v=\frac{\Delta x}{\Delta t}$, we can apply the same concept here. Our velocity should be the change in distance over the change in time. In this case, we don't have a definite time, $\displaystyle t$, but we do have a period in terms of one complete loop.

We can set up an equation for the period using the circumference of the circle as our distance: $\displaystyle v=\frac{2\pi r}{T}$.

Plug in the given values to solve for the period.

$\displaystyle v=\frac{2\pi r}{T}$

$\displaystyle 3\frac{m}{s}=\frac{2\pi (1m)}{T}$

$\displaystyle T=\frac{2\pi (1m)}{3\frac{m}{s}}$

$\displaystyle T=\frac{2\pi}{3}s$

Centripetal acceleration is the acceleration towards the center when an object is moving in a circle. Though the speed may be constant, the change in direction results in a non-zero acceleration.

The formula for this is $\displaystyle a_c=\frac{v^2}{r}$, where $\displaystyle v$ is the perceived tangential velocity and $\displaystyle r$ is the radius of the circle.

Plug in the given values and solve for the acceleration.

$\displaystyle a_c=\frac{(8\frac{m}{s})^2}{(0.5m)}$

$\displaystyle a_c=\frac{64\frac{m^2}{s^2}}{0.5m}$

$\displaystyle a_c=128\frac{m}{s^2}$

Centripetal force is the force that constantly moves the object towards the center; it is what keeps the object moving in a circle rather than flying off tangentially to the circle.

The formula for force is $\displaystyle F_c=ma_c$.

To find the centripetal force, we need to find the centripetal acceleration. We do this with the formula $\displaystyle a_c=\frac{v^2}{r}$, where $\displaystyle v$ is the perceived tangential velocity and $\displaystyle r$ is the radius of the circle.

Plug in the given values and solve for the acceleration.

$\displaystyle a_c=\frac{(8\frac{m}{s})^2}{(0.5m)}$

$\displaystyle a_c=\frac{64\frac{m^2}{s^2}}{0.5m}$

$\displaystyle a_c=128\frac{m}{s^2}$

Plug the acceleration and given mass into the first equation to solve for force.

$\displaystyle F_c=(2kg)(128\frac{m}{s^2})$

$\displaystyle F_c=256N$