Question

A 350-kg roller coaster car starts from rest at point A and slides down a frictionless loop-the-loop (Pig. P7.41). (a) How fast is this roller coaster car moving at point B? (b) How hard does it press against the track at point B?

Answer 1

Answer:

(a)v2 = 16m/s at point B

(b) F = 1.49×10⁴N.

Explanation:

Given m = 350kg

At A, h = 25m

At B h = 12m

(a) We are required to calculate the speed at this point, B.

This problem involves the application of the concept of conservation of mechanical energy for its solution since gravity is the only force doing work on the system.

So for this system,

ΔKE = – ΔPE

1/2×m(v2²– v1²) =– mg(h2 – h1)

From the diagram, g

h2 = 12m, h1 = 25m. From the question it was stated that the car starts from rest so v1 = 0m/s

In the above equation m is common to both sides so they cancel out. So,

1/2×(v2²– 0²) = – g(12 – 25)

1/2v2² = 13g

V2² = 2 ×13g = 26×9.80

v2 ² = 254.8

v2 = √254.8 = 15.96m/s ≈ 16m/s.

Energy concepts have been used to solve this problem. In this approach only the start and end points are taking into consideration when doing analysis on any problem of this sort and not what happens in between.

(b) Ghis part is asking us to calculate the force at point B.

At point B the car is moving in a uniform circle of diameter 12m (radius 6m)

So the acceleration of the car at point B

a = v²/r

v = 16m/s, r = 6m

F = ma = m×v²/r = 350×16²/6 = 14933.33N

F = 1.49×10⁴N.

Answer 2

Answer:

a)Velocity of car =v=16 m/s

b)Force against the track at point B=1.15*$10^{4}$N

Explanation:

Given mass of roller coaster=m=350 kg

Position of A=Ha=25 m

Position of B=Hb=12 m

Net potential energy=mg(ha-hb)

Net potential energy=(350)(9.80)(25-12)

Net potential energy=44590 J

Using energy conservation

net kinetic energy=net potential energy

(1/2)mv^2=mg(ha-hb)

m=350

velocity=v=16 m/s

b)There two force acting,centripetal force upward and gravity downward.

Thus net force acting will be

Net force=(mv^2/r)-mg

Net force=14933.33-3430

Net force=1.15*$10^{4}$ N