# Why does the speed change at the bottom of the loop-the-loop?

• Vigorous
In summary, the track of a roller coaster cannot pull it upwards at the top of the loop-the-loop, so it relies on the weight of the roller coaster to act towards the center. The normal force (N) is equal to 3 times the weight (Mg) and the minimum speed (vmin) is equal to the square root of the radius multiplied by gravity (Rg). At the bottom of the circle, the forces in the radial direction are analyzed and the normal force is equal to 5 times the weight. The speed at the bottom is determined using the law of conservation of energy. This problem can also be solved using the small angle approximation for simple harmonic motion.
Vigorous
Homework Statement
A roller coaster of mass M is at the top of the Loop-the-loop of radius R at twice the
minimum speed possible. What force does the track exert on it? What force does it
exert when it is at the bottom of the circle? (Use conservation of energy if needed.)
Relevant Equations
ac=mv^2/R
1 At the top of the loop-the-loop, the track cannot pull the roller coaster upward so it acts along with the weight of the roller coaster towards the center
N+Mg=Ma=Mv^2/R
N=M(v^2/R-g)=M(4Rg/R-g)=3Mg
N cannot act upwards, therefore vmin=(Rg)^1/2. Since its stated in the problem that the roller coaster is moving at twice the minimum speed, its speed is 2(Rg)^1/2.
At the bottom of the circle
If we also analyze the forces in the radial direction at the bottom of the circle
N-Mg=Mv^2/R=4Mg N=5Mg
My answer to the second part of the problem is incorrect but I don't get why the speed should change if the acceleration and change in velocity change are perpendicular as is the case at all times in circular motion.

Vigorous said:
My answer to the second part of the problem is incorrect but I don't get why the speed should change if the acceleration and change in velocity change are perpendicular as is the case at all times in circular motion.
Careful. Don't confuse this situation with uniform circular motion, where the speed is constant.

Car at top of loop has enough velocity as to keep moving horizontally (until gravity takes over and it starts describing a parabolic trajectory); therefore, it needs some force from the track to keep it describing a circular trajectory.
Car at bottom: same situation, only that the car has higher velocity and requires greater corrective force.

1) Thank you for the clarification, I drew the free body diagram again and considering the points above if the weight force is resolved along the tangential and radial direction it will provide tangential acceleration. But how can we determine the changes in the normal force and speed as it goes from top to bottom given the initial speed without resorting to the law of conservation of energy. I thought about writing the displacement velocity and acceleration in the radial and tangential directions but I haven't still been exposed to that material or may someone send me any resources explaining how to rewrite these vectors in terms of polar axes.

Lnewqban
Vigorous said:
But how can we determine the changes in the normal force and speed as it goes from top to bottom given the initial speed without resorting to the law of conservation of energy.
What do you have against using conservation of energy?

There's a reason they only ask you to analyze forces at the top and bottom of the circular path: Those points are much easier to analyze since the acceleration is purely centripetal. And since the acceleration depends upon the speed, conservation of energy is the easy way to find the speed at the bottom.

Lnewqban
To add to what @Doc Al posted, this problem is no different from a pendulum bob at the end of a massless rod that is allowed to go all the way around from zero to 2π. You can write down the equation of motion, but you cannot solve it analytically to get the velocity as a function of time. That is why one resorts to the small angle approximation to describe the oscillatory motion of a simple pendulum.

Mechanical energy conservation is equivalent to doing the so called first integral which allows you to find the velocity as a function of position. It involves the standard transformation $$a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}.$$

## 1. Why does the speed decrease at the bottom of the loop-the-loop?

The speed decreases at the bottom of the loop-the-loop due to the force of gravity acting against the motion of the object. As the object reaches the top of the loop, it has potential energy. As it moves down the loop, this potential energy is converted into kinetic energy, causing the object to speed up. However, as it reaches the bottom of the loop, the force of gravity is pulling the object downwards, causing it to slow down.

## 2. Why is the speed the highest at the top of the loop-the-loop?

The speed is highest at the top of the loop-the-loop because the object has the most potential energy at this point. As the object moves up the loop, it gains potential energy due to its increased height. This potential energy is then converted into kinetic energy, causing the object to speed up. Therefore, the speed is highest at the top of the loop.

## 3. What factors affect the speed at the bottom of the loop-the-loop?

The speed at the bottom of the loop-the-loop is affected by several factors, including the mass of the object, the radius of the loop, and the initial speed of the object. A heavier object will have more inertia and will therefore maintain a higher speed at the bottom of the loop. A smaller radius of the loop will result in a tighter turn, causing the object to have a higher speed. And a higher initial speed will also result in a higher speed at the bottom of the loop.

## 4. Can the speed at the bottom of the loop-the-loop be controlled?

Yes, the speed at the bottom of the loop-the-loop can be controlled by adjusting the initial speed of the object and the radius of the loop. By increasing the initial speed, the object will have more kinetic energy and will maintain a higher speed at the bottom of the loop. Similarly, by increasing the radius of the loop, the object will have a wider turn and will therefore have a lower speed at the bottom.

## 5. What happens if the speed is too low at the bottom of the loop-the-loop?

If the speed is too low at the bottom of the loop-the-loop, the object may not have enough kinetic energy to complete the loop and will fall off the track. This is because the force of gravity will be greater than the centripetal force needed to keep the object moving in a circular motion. This is why it is important to have enough initial speed and the correct radius of the loop for the object to successfully complete the loop-the-loop.

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