Roller coaster minimum speed around loop and effective weight

[and9]

Homework Statement

A roller coaster goes around a vertical loop of 20.0m. a) What is the minimum speed the cars must be moving so that the passengers do not fall out at the top? b) Assuming that the car maintains the same constant speed what is th effective weight of the passengers at the bottom? c) Again assuming car maintains this same constant speed, what is the net force (magnitutde and direction) acting on a 25kg passenger at point half way down the loop?

2. My attempt at solution

a) At top, N -> 0
N + mg = m(v²/r)
mg=m(v²/r)
g=v²/r
v_min=sqrt(rg) = 14 m/s

b) At bottom, N counters mg
N - mg = m(v²/r)
N = m((v²/r) + g)
N = m (9.8 + g) = 19.6m or 2mg

c) I had no idea what to do here?

 

[tiny-tim]

Welcome to PF!

Hi and9! Welcome to PF!

c) Again assuming car maintains this same constant speed, what is the net force (magnitutde and direction) acting on a 25kg passenger at point half way down the loop?
…
c) I had no idea what to do here?

Same as a) and b) … find N, and add it (as a vector) to the weight.

 

[tomcue]

Hello, thank you for your question and attempt at solving it. I would like to provide some additional information and clarification to your solution.

a) Your solution for part a) is correct. The minimum speed required for the roller coaster to go around the loop without the passengers falling out is 14 m/s.

b) Your solution for part b) is also correct. At the bottom of the loop, the normal force (N) must be equal to the sum of the weight (mg) and the centripetal force (m(v^2/r)). Therefore, the effective weight of the passengers at the bottom is 2 times their actual weight, or 19.6 m or 2mg.

c) For part c), we can use Newton's second law, F=ma, to calculate the net force acting on the passenger at the halfway point of the loop. The passenger's mass is given as 25 kg, and we know that the net force is equal to the sum of the forces acting on the object. In this case, the forces are the normal force (N) and the weight (mg), which are in opposite directions.

Therefore, the net force acting on the passenger is:

F = N - mg

At the halfway point, the normal force is equal to the centripetal force, which is equal to the weight (mg). So we can rewrite the equation as:

F = mg - mg = 0

This means that at the halfway point, the net force acting on the passenger is zero. This makes sense, as the passenger is not accelerating and is moving at a constant speed. The direction of the net force is also zero, as it is the sum of two forces in opposite directions.

I hope this helps to clarify your solution. Keep up the good work in your scientific studies!