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 = 207 kg and moves with speed v = 15.72 m/s. The loop-the-loop has a radius of R = 9.6 m. 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 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 What is the magnitude of the normal force on the car when it is at the top of the circle? N Submit Compare the magnitude of the cars acceleration at each of the above locations: a_bottom = a_side = a_top a_bottom a_top Submit What is the minimum speed of the car so that it stays in contact with the track at the top of the loop? m/s Submit

Explanation / Answer

Given

m = 207 kg

v = 15.72 m/s

R = 9.6 m

1) at the bottom of the circle

N = (m*v^2/R) - m*g

N = (207*15.72^2/9.6)-(207 * 9.8)

N = 3299.89 N

2) at side

N = - m*v^2/R

N = - 5328.49 N

3)

At the top

N = (m*v^2/R) + m*g

N = 7357.09 N

4)

abottom > aside > atop

5)

v = sqrt(g*R)

v = sqrt(9.8*9.6)

v = 9.69 m/s

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