How can I safely negotiate a loop-the-loop with a cart of mass 500 kg?

[rgalvan2]

The two problems below are related to a cart of mass M = 500 kg going around a circular loop-the-loop of radius R = 11 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 0.4 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?

h = m
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?

X = m
c) When the car is descending vertically (ie at a height R above the ground) in the loop, what is its speed |v|?

|v| = m/s
d) At the bottom of the loop, on the flat part of the track, the cart must be stopped in a distance of d = 15 m. What retarding acceleration |a| is required?

|a| = m/s2


I don't even know how to start this problem. Help please!

 

[anathema]

Hello! Let's break down each part of the problem and go through the steps to solve them.

a) For this part, we need to find the minimum height above the top of the loop that the cart can be released from in order to safely negotiate the loop. To do this, we can use the conservation of energy principle. At the top of the loop, the cart has both potential and kinetic energy. At the minimum height, the cart will have only kinetic energy, since it will have just enough energy to make it through the loop. So, we can set the potential energy at the top of the loop equal to the kinetic energy at the minimum height.

The potential energy at the top of the loop is given by mgh, where m is the mass of the cart, g is the acceleration due to gravity, and h is the height above the top of the loop. The kinetic energy at the minimum height is given by (1/2)mv^2, where v is the speed of the cart at the minimum height. Setting these two equal, we get:

mgh = (1/2)mv^2

We can cancel out the mass of the cart, and solve for h:

h = (1/2)v^2/g

Now, we know that the normal force at the top of the loop must be at least 0.4 times the weight of the cart. The weight of the cart is given by mg, so we can write:

N = 0.4mg

We also know that the normal force is equal to the centripetal force at the top of the loop, which is given by mv^2/R. So, we can set these two equal and solve for v:

0.4mg = mv^2/R

v = √(0.4gR)

Now, we can substitute this value for v into our equation for h:

h = (1/2)(0.4gR)^2/g = 0.08R

So, the minimum height above the top of the loop that the cart can be released from is 0.08 times the radius of the loop.

b) For this part, we are launching the cart horizontally from a compressed spring. We can use the conservation of energy principle again to find the minimum amount that the spring must be compressed. At the start, the cart has only potential energy from the compressed spring. At the minimum height, it

 

[baba89]

I would first recommend reviewing the principles of circular motion and understanding the forces involved in this scenario. This will help in solving the problems and ensuring the cart can safely negotiate the loop-the-loop.

a) To determine the minimum height h above the top of the loop, we need to consider the forces acting on the cart at the top of the loop. These forces are the normal force (N) exerted by the track, the weight (mg) of the cart, and the centripetal force (Fc) required for circular motion. The normal force must be at least 0.4 times the weight of the cart, so we can set up the following equation:

N ≥ 0.4mg

At the top of the loop, the normal force is equal to the sum of the weight and the centripetal force (N = mg + Fc). Substituting this into the previous equation, we get:

mg + Fc ≥ 0.4mg

Solving for the centripetal force (Fc) and substituting in the equation for centripetal force (Fc = mv²/R), we get:

mg + mv²/R ≥ 0.4mg

Simplifying, we get:

v² ≥ 0.6gR

To safely negotiate the loop, the speed of the cart at the top of the loop must be at least equal to this value. Since the cart is initially at rest, we can use the equation for conservation of energy to solve for the minimum height h:

mgh = 0.5mv² + 0.5kX²

Where X is the compression of the spring, which is zero in this case since the cart is at rest. Solving for h, we get:

h = v²/2g = (0.6gR)/2g = 0.3R

Therefore, the minimum height above the top of the loop that the cart can be released from rest is 0.3 times the radius of the loop.

b) In this scenario, the cart is launched horizontally from a compressed spring. To determine the minimum amount X that the spring must be compressed, we can use the same equation for conservation of energy:

mgh = 0.5mv² + 0.5kX²

Since the cart is launched horizontally, the initial height (h) is zero. Solving for X, we get: