# Loop the Loop (circular motion)

• Trentonx
In summary, the cart must be released from rest at a height of 5.41 meters above the top of the loop in order to negotiate it safely.
Trentonx

## Homework Statement

The two problems below are related to a cart of mass M = 500 kg going around a circular loop-the-loop of radius R = 10 m, as shown in the figures. All surfaces are frictionless. In order for the cart to negotiate the loop safely, the normal force exerted by the track on the cart at the top of the loop must be at least equal to 0.8 times the weight of the cart. You may neglect the size of the cart. (Note: This is different from the conditions needed to "just negotiate" the loop.)
a) For this part, the cart slides down a frictionless track before encountering the loop. What is the minimum height h above the top of the loop that the cart can be released from rest in order that it safely negotiate the loop?

a_c = (V^2)/R
F = ma
V = sqrt(2gh)

## The Attempt at a Solution

Top of circle
a_c equals the sum of the other forces, so
4900+400=500((v^2)/10)
V = 10.295 m/s, a_c = 10.6 m/s/s
height to have given velocity - 5.41m
This is where I don't know. How does having centripital acceleration help in finding the height at all? Does it?

The potential energy due to the cart's initial position goes entirely into kinetic energy when it is at the bottom of the loop (I have assumed the reference level for potential energy to be PE = 0 at the bottom). From there that energy becomes part kinetic and part potential energy. At the top of the loop you have then mgh = K.E. + P.E. Knowing the centripetal acceleration gives you the V^2 that you need for the kinetic energy.
(The h in my equation above is the initial height where the cart is released.)

First of all at the top of the loop, you must have an additional contact force of .8*m*g.

That means your Centripetal acceleration needs to be 1.8*m*g

To determine the height ... that's the potential energy that you will draw from to give you the necessary kinetic energy.

So it's .8 the weight, not the mass. And because energy is never lost, then the sum of the potential and kinetic will always be the same, no matter where on the loop it is. That makes much more sense than the textbook. Thanks.

## 1. What is loop the loop (circular motion)?

Loop the loop, also known as circular motion, is a type of motion where an object moves in a circular path around a fixed point. This type of motion is commonly observed in rides at amusement parks, such as roller coasters.

## 2. What causes an object to follow a circular path in loop the loop?

An object follows a circular path in loop the loop due to the combination of its velocity and centripetal force. The velocity of the object provides the necessary momentum to keep it moving in a circular path, while the centripetal force acts as the inward force that keeps the object from flying off the path.

## 3. How is the centripetal force calculated in loop the loop?

The centripetal force in loop the loop can be calculated using the formula F = mv²/r, where F is the centripetal force, m is the mass of the object, v is its velocity, and r is the radius of the circular path.

## 4. What factors can affect the motion of an object in loop the loop?

Several factors can affect the motion of an object in loop the loop, such as its mass, velocity, and the radius of the circular path. Other factors include external forces, such as friction and air resistance, as well as the design and structure of the path itself.

## 5. How is energy conserved in loop the loop?

In loop the loop, energy is conserved as the object moves from one point to another. At the top of the loop, the object has maximum potential energy, which is then converted into kinetic energy as it moves down the path. At the bottom of the loop, the object has maximum kinetic energy, which is then converted back into potential energy as it moves up the next loop.

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