Work and Energy with centripedal acceleration and springs.

[CaptainSFS]

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 = 15 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?

I found the answer to be 13.5m. I had to find velocity in the problem, which I found to be 16.275m/s.

b) For this part, we launch the cart 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" (as defined above) negotiate the loop?

This is the part I'm having trouble with.

Homework Equations

The Attempt at a Solution

I figured I would just set it up like I did for the first part. in the first part I used (PE) mgh = (KE) .5mv². So instead I used the PE for springs, and set it up... .5kx² = .5mv². I end up dividing mv² by k (10000), and then I take the sqrt of that to get 3.64m. This is not the correct answer, I'm doing something completely wrong. any help? thanks. :)

 

[anathema]

Thank you for your post. I am a scientist and I would be happy to help you with this problem.

For part b), you are on the right track by using the conservation of energy equation. However, there are a few mistakes in your approach. Let me guide you through the correct solution step by step.

1. First, let's draw a free-body diagram of the cart at the bottom of the loop, where the normal force is equal to 0.8 times the weight of the cart:

[insert image of free-body diagram]

From this diagram, we can see that the forces acting on the cart are the weight (mg) and the normal force (N). Since the normal force is 0.8 times the weight, we can write N = 0.8mg.

2. Now, let's write the conservation of energy equation for the cart at the bottom of the loop:

PE (spring) + PE (gravitational) = KE

We can ignore the kinetic energy (KE) term since the cart is released from rest. The potential energy (PE) for the spring is given by 0.5kx^2, where k is the spring constant and x is the amount the spring is compressed. The gravitational potential energy (PE) is given by mgh, where m is the mass of the cart, g is the acceleration due to gravity, and h is the height of the cart at the bottom of the loop.

3. Now, we can substitute in the values for the variables in the equation:

0.5kx^2 + mgh = 0

Remember, the cart is released from rest, so its initial velocity is 0. Also, the potential energy at the bottom of the loop is 0 since the cart is at ground level.

4. We can rearrange the equation to solve for x:

x = √(2mgh/k)

5. Now, we can substitute in the values for the variables in the equation:

x = √(2*500*9.8*h/10000)

Simplifying, we get:

x = √(98h/10000)

6. Finally, we can substitute in the value for h (the minimum height above the top of the loop) that we found in part a) of the problem:

x = √(98*13.5/10000) = 0.41 m

 

[baba89]

Hello! It looks like you are on the right track with using the conservation of energy principle to solve these problems. For the first part, you correctly used the equation PE = mgh and KE = 1/2mv^2 to find the minimum height h above the top of the loop.

For the second part, you are correct in using the equation PE (spring) = 1/2kx^2 and KE = 1/2mv^2. However, the key difference in this problem is that the cart is being launched horizontally, so you will need to take into account the horizontal component of the velocity.

To solve this, you can use the equation for centripetal acceleration, a = v^2/r, to find the velocity at the top of the loop. Then, you can use the equation for kinetic energy, KE = 1/2mv^2, to find the velocity at the bottom of the loop. From there, you can set the two equations for kinetic energy equal to each other and solve for the minimum amount X that the spring must be compressed.

I hope this helps! Keep up the good work with using the principles of work and energy in your problem solving. Good luck with your studies!