What Is the Minimum Compression Distance X for Safe Loop Negotiation?

In summary, the problem involves a cart of mass 500 kg going around a 15 m radius circular loop-the-loop with no friction. In order for the cart to safely negotiate the loop, the normal force exerted by the track at the top of the loop must be at least 0.5 times the weight of the cart. The cart is launched horizontally from a compressed spring with k = 10000 N/m and the minimum amount X the spring must be compressed to safely negotiate the loop is being asked. The solution involves finding the total energy lost on the way up, which is 147150, and using it to find the kinetic
  • #1
TG3
66
0

Homework Statement


A cart of mass M = 500 kg going around a circular loop-the-loop of radius R = 15 m. Friction can be ignored. 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 0.5 times the weight of the cart. (Note: This is different from the conditions needed to "just negotiate" the loop.) You may treat the cart as a point particle.
(I have calculated the speed necessary to be 22. 52 m/s)
The cart is launched horizontally along a surface at the same height as the bottom of the loop by releasing it from rest from a compressed spring with spring constant k = 10000 N/m. What is the minimum amount X that the spring must be compressed in order that the cart "safely" negotiate the loop?


Homework Equations


Potential Energy = MGH
K=1/2 MV^2

The Attempt at a Solution



9.81 x 500 x 30 = 147150 = the amount of energy lost on the way up.

V=22.52
M= 500
22.52^2 x 500 x 1/2 = 126787.6

126787.6+147150= 273937.6
273937.6 =1/2 500 x v^2
1095.75 = v^2
33.1 = v
But this is wrong... where did I mess up?
 
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  • #2
TG3 said:

Homework Statement


A cart of mass M = 500 kg going around a circular loop-the-loop of radius R = 15 m. Friction can be ignored. 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 0.5 times the weight of the cart. (Note: This is different from the conditions needed to "just negotiate" the loop.) You may treat the cart as a point particle.
(I have calculated the speed necessary to be 22. 52 m/s)
The cart is launched horizontally along a surface at the same height as the bottom of the loop by releasing it from rest from a compressed spring with spring constant k = 10000 N/m. What is the minimum amount X that the spring must be compressed in order that the cart "safely" negotiate the loop?


Homework Equations


Potential Energy = MGH
K=1/2 MV^2

The Attempt at a Solution



9.81 x 500 x 30 = 147150 = the amount of energy lost on the way up.

V=22.52
M= 500
22.52^2 x 500 x 1/2 = 126787.6

126787.6+147150= 273937.6
273937.6 =1/2 500 x v^2
1095.75 = v^2
33.1 = v
But this is wrong... where did I mess up?

It appears that when you found the total energy you needed:

total energy = 126787.6+147150= 273937.6

That you turned around and substituted back into [tex] \frac{1}{2} m v^2 [/tex]

when what you wanted to do was to find the distance the spring was compressed for which you need to use

[tex] \frac{1}{2} k x^2 [/tex]
 
  • #3


Hello,

Your calculations for the speed needed to safely negotiate the loop seem to be correct. However, in order to calculate the minimum amount that the spring must be compressed, you need to consider the conservation of energy. At the bottom of the loop, the cart has both kinetic energy and potential energy due to the compressed spring. At the top of the loop, the cart has only kinetic energy. Therefore, the total energy at the bottom must be equal to the kinetic energy at the top. Using the equation for potential energy and kinetic energy, you can solve for the compression distance X. I hope this helps.
 

Related to What Is the Minimum Compression Distance X for Safe Loop Negotiation?

1. What is loop the loop kinetic energy?

Loop the loop kinetic energy is the energy possessed by an object that is moving in a circular path, such as a rollercoaster or a ferris wheel. It is a type of kinetic energy that is dependent on the object's mass, velocity, and the radius of the circular path.

2. How is loop the loop kinetic energy calculated?

The formula for calculating loop the loop kinetic energy is K = 1/2 * m * v^2, where K is the kinetic energy, m is the mass of the object, and v is the velocity of the object. In the case of loop the loop, the radius of the circular path is also a factor in the calculation.

3. What factors affect loop the loop kinetic energy?

The factors that affect loop the loop kinetic energy include the mass of the object, the velocity of the object, and the radius of the circular path. The greater the mass and velocity of the object, and the smaller the radius of the circular path, the higher the kinetic energy will be.

4. How is loop the loop kinetic energy related to potential energy?

In a loop the loop, as the object moves up the incline, it gains potential energy due to its increased height. As it moves down the incline, this potential energy is converted into kinetic energy. At the top of the loop, the object has the maximum potential energy, and at the bottom of the loop, it has the maximum kinetic energy.

5. What is the significance of loop the loop kinetic energy in amusement park rides?

Loop the loop kinetic energy is a crucial factor in designing safe and thrilling amusement park rides. The kinetic energy of the ride must be carefully calculated to ensure that it is enough to complete the loop, but not too much that it causes discomfort or danger to the riders. The calculation also helps in determining the necessary safety measures and restraints for the ride.

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