can the car do the loopty loop

Question

211.887577 mph

3,164 lb car


circomfrence 90 ft
14ft radius
diameter 28ft

Answer 1

It can do it if:

 

mv²/r ≥ mg

 

v²/r ≥ g   (cancelled the m's)

• v = car's velocity = 211.887577 mph
• r =  14 ft
• g = 32 ft/sec²

 

Convert everything to a consistent system of units then plug the values into the inequality (you have a mix of miles v. feet, and seconds v. hours).  The car's mass and the loop's circumference are irrelevant.  The inequality says the centrifugal acceleration must equal or exceed the acceleration due to gravity.

 

 

UPDATE: Oops, I left out an important step. The car loses velocity as it climbs the loop. It is the car’s velocity at the top of the loop that you should use in the inequality. Car’s Initial Velocity = v₀= 211.887577 mph

 

Convert to ft/sec:

 

(211.887577 mph) (5280 ft/mi)(1/3600 hr/sec) = 310.75 ft/sec

 

The loop has a 28 ft diameter. The change in potential energy is mgh where m is the car’s mass, g is the acceleration due to gravity, and h is the diameter. So the kinetic energy that the car loses as it climbs is:

 

(1/2)mv²= mgh

 

V²= (64)(28)

 

V = 42.33 ft/sec

 

The car’s velocity at the top is:

 

V = 310.75 – 42.33 = 268.42 ft/sec

 

This is the velocity you need to use in the inequality above.