What Calculations Determine Roller Coaster Dynamics on a Vertical Track?

  • Thread starter Thread starter dr2112
  • Start date Start date
  • Tags Tags
    Circular Track
AI Thread Summary
The discussion focuses on calculating the dynamics of a roller coaster on a vertical circular track. Key calculations include determining the minimum speed required for the coaster to remain on the track at the top, which is derived from the equation v = square root of (rg). The speed at the bottom of the loop is linked to energy conservation principles, with the equation involving kinetic and potential energy. Additionally, the normal force experienced by passengers at both the bottom and top of the loop is analyzed using Newton's laws. Overall, the thread emphasizes applying physics equations to solve for speed and forces in roller coaster dynamics.
dr2112
Messages
9
Reaction score
0

Homework Statement



roller coaster traverses a vertical circular track.
a)what speed must car have to make it over the top w/o leaving the track? assume it is not attached to track.
b)what will speed be at bottom?
c)what will normal force on passenger be at the bottom of loop if radius is 14.5m?
d)at the top?

Homework Equations


a) EF=ma
n-mg=ma2
n=0 at top
b)Etop=1/2mv2(top)+mgh

The Attempt at a Solution


a) v=square root of (rg)
b)1/2mrg+mg(2*3.14) =2.5mgr
c)?
d)?
 
Physics news on Phys.org
Hi dr2112,

dr2112 said:

Homework Statement



roller coaster traverses a vertical circular track.
a)what speed must car have to make it over the top w/o leaving the track? assume it is not attached to track.
b)what will speed be at bottom?
c)what will normal force on passenger be at the bottom of loop if radius is 14.5m?
d)at the top?

Homework Equations


a) EF=ma
n-mg=ma2
n=0 at top
b)Etop=1/2mv2(top)+mgh

The Attempt at a Solution


a) v=square root of (rg)
b)1/2mrg+mg(2*3.14) =2.5mgr

The first term (1/2 mrg) looks right for the kinetic energy at the high point. What do the other two terms represent, and how did you get those numeric values?

c)?

Do the same thing for part c that you did in part a--use Newton's law for the forces.

d)?

You've already answered this in the working for part a.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top