A roller-coaster car has a mass of 1230 kg when fully loaded with passengers?

Question

As the car passes over the top of a circular hill of radius 20 m, its speed is not changing.
(a) What are the magnitude and direction of the force of the track on the car at the top of the hill if the car's speed is 11 m/s?
N

upward
downward
no force
(b) What are the magnitude and direction of the force of the track on the car at the top of the hill if the car's speed is 17 m/s?
N

upward
downward
no force

Answer 1

a)
F=m(g-ac)
ac=V^2/R
F=1230(8.81 - (11)^2/20)=
F=4625 N (downward)

b)
F=m(g-ac)
F=1230(8.81 - (17)^2/20)=
F=-5707N (upward) The car be flying

Answer 2

[Edit: In my analysis, I'm assuming that the track can either push up or pull down on the car as necessary; this is the way real roller coasters are built.]

Whenever _anything_ goes in a circle at a constant speed, its acceleration is toward the center of the circle, and has this amount:

a = v²/r

We also know that the direction of the acceleration is toward the center of the circle, which in this case is straight downward.

At this point, we have to be careful about positive and negative signs. When you're dealing with "ups" and "downs," you have to make a decision about whether you want "up" quantities to be positive (and "downs" to be negative); or the other way around. Doesn't matter which choice you make, but you have to stick with it throughout the problem.

So, here's an idea: We said before,

a = v²/r

That formula makes "a" come out positive; also we know that "a" points down. So let's use the convention: "postive is down; negative is up."

Now, from Newton's 2nd Law:

F_net = ma

or:
F_net = (m)(v²/r)

Now, that is the _net_ force; which means it's a combination of the individual forces acting on the car. What are those individual forces? There are only two I can think of:

1) The force of gravity, mg (we would have called it "–mg" if we'd chosen "up" to be positive)
2) The force of the track on the car, F_t (don't know yet whether this points down (negative) or up (positive))

So the net force, F_net, is a combination of those:

F_net = mg + F_t

So, combining the two equations for F_net:

mg + F_t = (m)(v²/r)

Next, just use algebra to solve for F_t. They tell you all the other numbers (m, v and r).

> upward
> downward
> no force

Solve for F_t and then see if it's positive (down), negative (up), or zero.

> if the car's speed is 17 m/s?

Same equation, different numbers. Solve again for the (new) F_t, and see whether it's positive, negative, or zero.

Answer 3

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