In a loop-the-loop ride a car goes around a vertical, circular loop at a constan

Question

In a loop-the-loop ride a car goes around a vertical, circular loop at a constant speed. The car has a mass m -235 kg and moves with speed v = 17.4 m/s. The loop-the-loop has a radius of R = 11.9 m. 1) What is the magnitude of the normal force on the care when it is at the bottom of the circle? (But as the car is accelerating upward.) N Submit 2) What is the magnitude of the normal force on the car when it is at the side of the circle (moving vertically upward)? N Submit 3) What is the magnitude of the normal force on the car when it is at the top of the circle N Submit 4) Compare the magnitude of the cars acceleration at each of the above locations: O abottom- aside atop O abottom

Explanation / Answer

here,

mass , m = 235 kg

v = 17.4 m/s

R = 11.9 m

a)

the magnitude of normal force on the car at the bottom , Nb = m * ( v^2 /R + g)

Nb = 235 * ( 17.4^2 /11.9 + 9.81) N

Nb = 8284.2 N

b)

the magnitude of normal force on the car at the side of loop , Ns = m * ( v^2 /R )

Ns = 235 * ( 17.4^2 /11.9) N

Ns = 5978.9 N

c)

the magnitude of normal force on the car at the top , Nt = m * ( v^2 /R + g)

Nt = 235 * ( 17.4^2 /11.9 - 9.81 ) N

Nt = 3673.5 N

d)

accelration , a = net force /effective mass

so,

ab > as > at

e)

let the minimum speed at the top be v

the magnitude of normal force on the car at the top , Nt = m * ( v^2 /R + g)

0 = 235 * ( v^2 /11.9 - 9.81 ) N

v = 10.8 m/s

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