Math Help - verticle circle radius

I believe this is another Verticle circular motion problem, they still confuse me, why is it i must add the net forces for the answer, and do i use an equation (m * v^2) / r more than once?

heres the problem:

A roller coaster at an amusement park has a dip that bottoms out in a vertical circle of radius r.

A passenger feels the seat of the car pushing upward on her with a force equal to twice her weight as she goes through the dip.

If r = 16.5 m, how fast is the roller coaster traveling at the bottom of the dip?

in m/s

2. Originally Posted by rcmango
I believe this is another Verticle circular motion problem, they still confuse me, why is it i must add the net forces for the answer, and do i use an equation (m * v^2) / r more than once?

heres the problem:

A roller coaster at an amusement park has a dip that bottoms out in a vertical circle of radius r.

A passenger feels the seat of the car pushing upward on her with a force equal to twice her weight as she goes through the dip.

If r = 16.5 m, how fast is the roller coaster traveling at the bottom of the dip?

in m/s
One would suspect either that you don't read my posts or you don't tell me if you are confused by the answers/hints that I give you.

As always, start with a Free-Body Diagram. I will choose a +y direction upward. There is a weight on the coaster, w, straight down. There is an upward (presumably a normal) force equal to 2w acting straight up. That's it for the vertical direction.

So what is the net force on any object moving in a circular path? Right! It's the centripetal force. Thus
$\sum F = -w + 2w = F_c$

I will leave you with the rest.

-Dan

3. Okay then i reached the answer using this equation: v = sqrt(r * g)

however, i don't believe i am supposed to solve this problem using that.

also, i am confused with the formula you refer to: F = -w +2w = Fc

can you break this down in an explanation of each variable, or what it really means.

Thankyou.

4. Originally Posted by rcmango
Okay then i reached the answer using this equation: v = sqrt(r * g)

however, i don't believe i am supposed to solve this problem using that.

also, i am confused with the formula you refer to: F = -w +2w = Fc

can you break this down in an explanation of each variable, or what it really means.

Thankyou.
Where did the $v = \sqrt{rg}$ come from?

We are looking at the car at the bottom of the circular dip. The net force on any object moving in a circular path is equal to the centripetal force. The Free-Body Diagram for the girl at this point contains two forces: the weight w acting straight down, and the normal force N acting straight up. So Newton's 2nd says:
$\sum F = -w + N = F_c$
(where I have chosen the positive direction as upward.)

The problem states that the car is acting on the girl with a force magnitude of twice her weight. Thus N = 2w. Thus
$\sum F = -w + 2w = \frac{mv^2}{r}$

$w = \frac{mv^2}{r}$

$mg = \frac{mv^2}{r}$

$v^2 = rg$

$v = \sqrt{rg}$
(Now you can use this equation since you've just derived it.)

-Dan