Solving Two Problems: 500 kg Cart on a 9 m Loop-the-Loop

[jromega3]

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 = 9 m, as shown in the figures. Assume that friction can be ignored. Also assume that, 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. (Note: This is different from the conditions needed to "just negotiate" the loop.) You may treat the cart as a point particle.

Homework Equations

The Attempt at a Solution

 

[djeitnstine]

What are you solving for and where is your attempt?

 

[nlightner]

I would approach this problem by first analyzing the given conditions and assumptions. The fact that friction can be ignored simplifies the problem and allows us to focus on the forces acting on the cart. The normal force exerted by the track on the cart at the top of the loop is an important factor to consider, as it determines the safety of the cart's movement.

Next, I would use the equation F = ma to calculate the centripetal force needed for the cart to safely negotiate the loop. This force must be equal to the sum of the weight of the cart (mg) and the normal force (N) at the top of the loop. Therefore, the equation can be written as F = mg + N.

Using the given condition that N ≥ 0.4mg, we can rearrange the equation to solve for the minimum speed (v) needed for the cart to safely negotiate the loop:

v ≥ √(gR(1+0.4))

This equation gives us the minimum speed needed for the cart to safely complete the loop-the-loop. However, we also need to consider the maximum speed that the cart can have without losing contact with the track. This can be determined using the equation v = √(gR), which gives us the speed at which the normal force is equal to zero.

Therefore, the range of speeds that the cart can safely travel at is √(gR) ≤ v ≤ √(gR(1+0.4)). This range can be used to determine the maximum and minimum kinetic energies of the cart as it travels around the loop.

In conclusion, by analyzing the forces acting on the cart and using equations to calculate the minimum and maximum speeds, we can ensure that the cart safely negotiates the loop-the-loop while also considering the maximum kinetic energy that it can have without losing contact with the track. This approach can also be applied to similar problems involving circular motion and safety considerations.