Quick rollercoaster AP physics question...?

Question

I know that there are pretty simple ways to solve these problems (there always are with these type of assignments) but I really can't figure them out. I have the following image: http://www.eggloaf.com/coaster.PNG
And I am supposed to calculate the minimum value of h such that the cart can stay on the top of the loop without falling off of the track (friction is negligible). I've tried just about every formula I can think of, but I can't come up with a numeric answer...
Also, it asks to calculate the normal force as a function of theta from 0 to 180 degrees, where theta is the angle formed between the bottom of the loop and the position of the cart. I don't even know where to begin with this one...

Honestly, any help at all would be really appreciated.

Answer 1

At the bottom of the track where theta is zero, the normal force is made up of the centrifugal force plus the weight of the mass. At the top, given the condition that the cart will stay on the top of the loop without falling off of the track, the normal force is just the centrifugal force minus the weight and at this point they must be equal and opposite. The normal force starts from 0 degrees and goes to 90 degrees when the mass is 1/4 the way up the right side. It follows the radius vector central angle. So it goes from a max at 0 degrees to 0 at 90 degrees then to a min at 180. Looks like a cosine function to me.

Answer 2

Your diagram doesn't match the question.

I will assume it is an inside loop where the cart starts at some height,h, and enters the loop of diameter d at the bottom. Find the h that will keep the cart from dropping at the top

using energy
m*g*(h-d)=.5*m*v^2

when the cart is at the top of the loop there is a centripetal force of
m*v^2/r
or
m*2*v^2/d

the minimum v is when the force is balanced with m*g
therefore
m*g=m*2*v^2/d
solve for v^2
v^2=g*d/2
plug into the energy equation
m*g*(h-d)=.5*m*g*d/2
solve for h
h=5*d/4

For the normal force as a funtion of the angle:
Let's look at a fbd of the cart for 0<θ<90

The speed of the the cart is related to the angle as
.5*m*v^2=m*g*(h-d*sin(θ)/2)
v^2=2*g*(h-d*sin(θ)/2)

the sum of forces radial to the center of the loop are the normal force, N
N=m*g*(4*(h-d*sin(θ)/2)/d+cos(θ))

Let's look at a fbd of the cart for 90<θ<180

The speed of the the cart is related to the angle as
.5*m*v^2=m*g*(h-d*(1-cos(θ))/2)
v^2=2*g*(h-d*(1-cos(θ))/2)

the sum of forces radial to the center of the loop are the normal force, N
N=m*g*(4*(h-d*(1-cos(θ))/2)/d+cos(θ))

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Answer 3

I in simple terms took Honors Physics and it grew to become into killer! Hated the class cuz it grew to become into troublesome, clever of you to income over the summer season. i don't comprehend what e book, yet good success with the class! attempting to study the textbooks is like attempting to study yet another language, lol.