1.) The train on a roller coaster ride experiences a 42.0-kN normal force at the

Question

1.) The train on a roller coaster ride experiences a 42.0-kN normal force at the bottom of a smooth circular, vertical loop. Take the mass of the train and passengers to be 600kg.

a.) What normal force acts on the train at the top of the loop?

b.) Now assume that the radius of the loop is 10m with the same normal force acting at the bottom of the loop as before. What is the radius of the largest loop the train will just be able to completely pass through?

(NOTE: use g = 10.0m/s² for this problem)

Explanation / Answer

a) Applying Newton’s second law at the bottom,

Fnet = ma

-Fn – Fg = - Fc

Fn + Fg = Fc

42000 + 600*9.8 = Fc

Fc = 47880 N -----------(1)

Applying Newton’s second law at the top,

Fnet = ma

Fn – Fg = Fc

Fn = Fc + Fg

Fn = 47880 + 600*9.8

Fn = 53760 N

b)

From (1),

Fc = mv^2/r

47880 = (600*v^2)/10   => v= 28.25 m/s

the train just be able to completely pass through , Fn= 0 N

Thus.

Fn + Fg = Fc

Fg = Fc

mg= mv^2/R

g= v^2/R             

R= v^2/g = 28.25^2/10 = 79.8 m

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.