To start off with, it is useful to consider the energy of the system. Initially, the block is at a certain unknown height, and will thus have gravitational potential energy.

\(\displaystyle E_{i}=mgh\)

Since we are assuming that there is no friction in this case, then we know that total mechanical energy will be conserved. Therefore, all of the gravitational potential energy contained in the block will become kinetic energy when it slides down to the bottom of the loop. But, once the block begins to slide up the loop, it will lose kinetic energy and will regain some gravitational potential energy. Therefore, at the top of the loop, the block will have a combination of kinetic and gravitational potential energy whose sum is equal to the initial energy of the system.

\(\displaystyle E_{f}=mg(2r)+\frac{1}{2}mv^{2}\)

Due to conservation of energy, we can equate the two.

\(\displaystyle mgh=mg(2r)+\frac{1}{2}mv^{2}\)

In addition to considering energy, it is also necessary to consider the forces acting on the block in this scenario. When at the top of the loop, the block will experience a downward force due to its weight, and another downward force due to the normal force of the loop on the block. Furthermore, because the block is traveling along a circular path while in the loop, it will experience a centripetal force. At the top of the loop, the centripetal force will be due to a combination of the weight of the block as well as the normal force.

\(\displaystyle F_{c}=\frac{mv^{2}}{r}=N+mg\)

We're looking for a starting height that will just allow the block to travel around the loop. The minimum height will be the height such that the block will just start to fall off. When falling off, the block will no longer be touching the loop and therefore, the normal force will be equal to zero. This is the situation we are looking for, and since the normal force is zero, only the block's weight will contribute to the centripetal force at the top of the loop.

\(\displaystyle \frac{mv^{2}}{r}=mg\)

\(\displaystyle v=\sqrt{rg}\)

The preceding expression gives us the value of velocity that will allow the block to have enough kinetic energy while at the top of the loop to not fall off. We can plug this expression into the previous energy equation.

\(\displaystyle mgh=mg(2r)+\frac{1}{2}m(\sqrt{rg})^{2}\)

\(\displaystyle mgh=2mgr+\frac{1}{2}mgr=\frac{5}{2}mgr\)

\(\displaystyle h=\frac{5}{2}r\)

\(\displaystyle h=\frac{5}{2}(12 m)=30m\)

Like all force problems, this one starts with a clear free body diagram:

The tension \(\displaystyle (T)\) points along the vine (tensions can only pull), so it goes straight up. The force of gravity \(\displaystyle (mg)\) points straight down, as it always does. The two do not add to zero, however, since the person is undergoing circular motion. Instead, they add to a net force pointing towards the center of the circle that the person is making, which is up at the place where the vine is attached to the tree. Draw a vector diagram:

Then write the equation about the lengths of the vectors: the length of the gravity vector plus the length of the net force vector equals the length of the tension vector:

\(\displaystyle T=mg+ F_{net}\)

\(\displaystyle T=90kg*10\frac{N}{kg}+ \frac{mv^{2}}{r}\)

The net force for an object undergoing circular motion is mass times speed squared divided by the radius of the circle. \(\displaystyle 10\frac{N}{kg}\) is the gravity force constant. Some use \(\displaystyle 9.81\frac{N}{kg}\), but the AP physics 1 test allows you to use \(\displaystyle 10\frac{N}{kg}\), which makes it a lot easier. Plug in the numbers, and solve for the tension:

\(\displaystyle T=90kg* 10\frac{N}{kg}+ \frac{90kg* (10\frac{m}{s^2})}{15m}\)

\(\displaystyle T=1500N\)

This answer is reasonable since the vine has to both hold the person up and provide a centripetal force; that is why the tension is is greater than his weight alone.

The formula for centripetal force is:

\(\displaystyle F_{centripetal}=m*\frac{v^2}{r}\)

Where \(\displaystyle m\) is the mass of the object, \(\displaystyle v\) is its velocity, and \(\displaystyle r\) is the radius of the circle made by the motion of the object around the center. We can model an atom like a circle, with its nucleus being its center. Since we are given the diameter of an atom, its radius will be

\(\displaystyle r=\frac{1}{2}*diameter= \frac{1}{2}*4*10^{-10}m= 2*10^{-10}m\)

\(\displaystyle v^2=(2200\frac{km}{s})^2=(2.2*10^6 \frac{m}{s^2})^2=4.84*10^{12}\frac{m^2}{s^2}\)

\(\displaystyle m=9*10^{-31}kg\) 

\(\displaystyle F_{centripetal}=(9*10^{-31}kg)\frac{4.84*10^{12}\frac{m^2}{s^2}}{2*10^{-10}m}=2.2*10^{-8}N\)

If we imagine the hill as a semi-circle, it appears that the car is moving along a circle. At the apex of the hill, the car's acceleration points downwards as this points towards the center of the circle. If an object travels in a circular fashion, at a constant speed, the direction of acceleration is always towards the center of the circle. This type of acceleration arises do to the change in velocity. Although the speed is constant, the direction changes. 

We can solve this problem by looking at all of the forces acting on the car at the top of the circular path.

\(\displaystyle mg-N=m\frac{v^2}{r}\)

The weight is positive, because it points towards the center of the circle, while the normal force \(\displaystyle N\) is negative because it points away from the circle.

Since we are asked to look at the special case where the occupants of the car feel weightless, this tells us that the normal force experienced by the occupants must be zero.

\(\displaystyle mg=m\frac{v^2}{r}\Rightarrow r=\frac{v^2}{g}\)

\(\displaystyle r=\frac{(15\: \textup{m/s})^2}{9.8\:\textup{m/s}^2}\)

\(\displaystyle r=\frac{225}{9.8}\:\textup{m}\)

\(\displaystyle r\approx 22.9\:\textup{m}\)

For this problem, we recognize that the force keeping the mass m suspended is the tension in the string. This tension is balanced by the gravitational force on m such that m does not accelerate. So how do we relate the speed of M to the tension? We must ask ourselves why is there tension in the string at all? Any object that travels in a circle has some centripital acceleration acting on it that point towards the center of the circle (for constant speed). This force is equal to the mass time the centripial acceleration. 

\(\displaystyle F=Ma=M\frac{v^2}{R}\)

I have stated previously that this force must be equal to the gravititational force acting on mass m. So,

\(\displaystyle F=Ma=M\frac{v^2}{R}=mg\)

Solving for v,

\(\displaystyle v=\sqrt{\frac{mgR}{M}}\)

Plugging in values yields a speed of 1.6 m/s

Let's start with what we know, and what we want to know. We are given the mass of an object, and we're told that it's traveling in a circular orbit. Also, we have been given the diameter of the orbit. What we are trying to find is the velocity of the object.

In this problem, we'll need to consider the centripetal force of the object as it travels in a circular orbit. The force that is contributing to the centripetal force is the weight of the object. Therefore, we can write the following expression:

\(\displaystyle F_{Centripetal}=F_{Weight}\)

\(\displaystyle \frac{mv^{2}}{r}=mg\)

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

\(\displaystyle v=\pm \sqrt{rg}\)

The above expression tells us that the object's velocity is dependent on the acceleration due to gravity, as well as the radius of the orbit. Since we're told that the diameter of the orbit is \(\displaystyle 5.1\:meters\), we can find the radius by taking half of this value, which is \(\displaystyle 2.55\:meters\). Moreover, the plus-or-minus sign indicates that the object can either be traveling in a clockwise or counter-clockwise orbit. In either case, the object will have the same magnitude of velocity.

Solving for velocity, we can plug in values to obtain:

\(\displaystyle v=\sqrt{rg}=\sqrt{(2.55\: meters)(9.8\: \frac{meters}{second^{2}})}=5\:\frac{m}{s}\)

Because the hammer is moving at a constant angular velocity, the centripetal force along the hammer rope can be expressed as \(\displaystyle F = m\cdot \left ( v^{2}/r\right )\). By plugging in the numbers given by the problem we get \(\displaystyle F = (7.257kg) \cdot ((25m/s)^{2}/1.215m)\) which, when simplified gives us approximately \(\displaystyle 3733 kg\cdot m / s^{2}\) or \(\displaystyle 3733N\)

In order to find the radius of the circle that corresponds to the maximum tension the string can experience without breaking, we must rearrange the centripetal force equation for objects moving in a circle:

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

\(\displaystyle F_c = T_{max}\Rightarrow r= m \frac{v^2}{T_{max}}\)\(\displaystyle =.25kg\left ( \frac{(15\frac{m}{s})^2}{500N} \right ) = 11.25cm\)

To solve this problem, it is important to remember the signs of each force acting on the coaster.  Since we are inside the loop at the top, both the weight and the normal force point toward to center of the circular path, and are thus positive.

\(\displaystyle mg + N = m \frac{v^2}{r}\)

Next, we note that the normal force is what a rider "feels" during their ride on the rollercoaster.  We want the speed at which a rider would feel three times as heavy, or three times their weight.

\(\displaystyle mg+N=mg+3mg=m\frac{v^2}{r}\Rightarrow v=\sqrt{4gr}=\sqrt{4(9.81\frac{m}{s^2})50m}=44.3\frac{m}{s}\)