Solving for Centripetal Force using a Free Body Diagram

Question

A 70 kg student is riding a roller coaster and is at the top of the vertical loop. The loop has a radius of 16 m, and the car's velocity at the top is 12 m/s. What is the student's apparent weight at that moment?

 

SHOW WORK

Answer 1

I know it is way too late to matter to Chantelle, but...

 

When you stand on a scale, what the scale is actually reading is its force of pushing back on you.  In other words, it reads the normal force it applies to you.  In a situation of being stationary at equilibrium, this normal force must equal your weight, so that the two cancel out [(F_N-F_g) = 0].

 

In other situations where you are not in equilibrium, though, this changes.  And being in circular motion (as is described in this problem) is decidedly NOT in equilibrium (since the object in motion is always being accelerated by the centripetal force).

 

The key is that the net force on an object in uniform circular motion MUST be the centripetal force (and we are assuming we are looking at a snippet of uniform circular motion, since the student is said to be moving at a constant 12 m/s over the interval of interest to us).  The centripetal force, in other words, must the be the resultant of all the other forces put together.  Otherwise, the object would not stay on its circular path.

 

So let's look at the forces acting on the student at the top of the vertical loop.  There are two:

 

1.  the force of gravity (F_g), down

2.  the normal force of the seat (F_N), up

 

These two forces must add up to the centripetal force needed to hold the student in that circular path:

 

F_c = mv²/r

 

and that centripetal force has to be directed down at the indicated position, since the student is on top of the loop, and the center (toward which the centripetal force always points) is directly below.

 

Hence, if we set up Newton's 2nd law of motion in the vertical direction at the indicated point, we have:

 

F_net = F_N - F_g = -F_c   (assuming down is negative)

 

Even though the student is not said to be sitting on a scale, the student's perception of weight is determined by the normal force of whatever pushes up against him or her.  So the normal force is the student's "apparent weight," (a needless term if there ever was one), and we just have to solve for it.

 

F_N - (70 kg)(9.8 m/s²) = -(70 kg)(12 m/s)²/(16 m) --> F_N - 686 N = -630 N --> F_N = 56 N

 

This is the normal force exerted by the seat in the student at this position.  Notice it is much less than the student's weight (686 N), which is why the student feels lighter going over the top of a hill like that (which you may have experienced going over the top of a hill fast in a car).  

Answer 2

F_c = m * a = (m * v²) / r = m * ω2 * r      (v = ω * r)

 

Since you are given a tangential velocity and not an angular velocity, you want to use the force equation using v.

 

F_c = (m * v²) / r = (70kg * (12m/s)²) / 16m = 630 kgm/s²

 

F_g = m * g = 70kg * 9.81m/s² = 686.7 kgm/s²

 

The net force is equal to mass times acceleration, which would be the centripetal force. The two downward forces that add up to the net downward force would be the normal force and the gravitational force. This means that the normal force would be the difference between the centripetal force (net force) and the gravitational force. This difference would be the relative weight.

 

F_net = F_n + F_g  -->  ma = F_n + F_g  -->  F_c = W + F_g  -->  W = F_c - F_g

 

W = F_g - F = 686.7kgm/s² + 630 kgm/s² = 56.7 kgm/s² = 56.7 N

Answer 3

The centripetal force is F_c = mv²/r = (70 kg)(12 m/s²)²/(16 m) = 630 N.

 

The student's weight on the ground is F_g = mg = (70 kg)(9.8 m/s²) = 686 N.

 

The student's apparent weight is then the difference, F_a = F_g - F_c = 56 N.

 

Note that this apparent weight is towards the ground, away from the roller coaster cart.