# Circular motion(Normal-Tangential)

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nrqed
Homework Helper
Gold Member
Heres the problem I was working on

http://i14.photobucket.com/albums/a322/guitaristx/train.jpg

Heres what I came up with so far...
http://i14.photobucket.com/albums/a322/guitaristx/trainwork.jpg

I feel like Im missing some numbers....any suggestions?
Consider the situation when the train is just about to "fly off" and leave the tracks. In that case, the normal force on the train is zero. This should alllow you to solve for rho.

EDIT: Question b does not make sense to me...What do they mean by "bottom of a hill"? Maybe we are to assume that we have now a track curved upward with the same radius?

ok i got

A = N-g

A = 0 - 9.8m/s^2

so

r = 106^2/-9.8 = -1148 m

does that make sense for the first question?

nrqed
Homework Helper
Gold Member
ok i got

A = N-g

A = 0 - 9.8m/s^2

so

r = 106^2/-9.8 = -1148 m

does that make sense for the first question?
Well a radius can't be negative.

Be careful, you are using

$$\sum F_y = m a_y$$

so your a_y should be $$- v^2/ r$$ not v^2/r.
Other than that it looks fine

how is Velocity negative when gravity is acting downward in the y direction?

when you solve A-sub-n you get -9.8m/s^2 however A-sub-n in the FBD is acting upward. So how do you know which sign to use?

nrqed
Homework Helper
Gold Member
how is Velocity negative when gravity is acting downward in the y direction?
I never said that a velocity is negative. I am talking about the y component of the acceleration. The radial (or centripetal) acceleration a_rad always points to ward the center of the circle. In your example, the center of the circle is below the train. Therefore

$$a_y = - a_{Rad} = - \frac{v^2}{R}$$

ok I see what your saying, than my FBD would be wrong? I have the A-rad pointing upward.

In terms of Normal and Tangential accerlation, when I draw the FBD if the normal Accerleration is upward, does that mean the Normal Force is also upwards?

nrqed