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 = 288 kg and moves with speed v = 14 m/s. The loop-the-loop has a radius of R = 8.1 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.) What is the magnitude of the normal force on the car when it is at the side of the circle (moving vertically upward) What is the magnitude of the normal force on the car when it is at the top of the circle? Compare the magnitude of the cars acceleration at each of the above locations: a_bottom = a_side = a_top a_bottom atop What is the minimum speed of the car so that it stays in contact with the track at the top of the loop?

Explanation / Answer

1)

Fn = nrmal force in upward direction

mg = weight of car in downward direction

Force equation is given as

Fn - mg = mv2/r

Fn = mg + mv2/r

Fn = 288 (9.8) + 288 (14)2 /8.1

Fn = 9791.3 N

2)

Fn = normal force acting towards center

force equation is given as

Fn = mv2/r

Fn = 288 (14)2 / 8.1

Fn = 6968.89 N

3)

Fn = normal force in downward direction

mg = weight of car in downward direction

Force equation is given as

Fn + mg = mv2/r

Fn = mv2/r - mg

Fn = 288 (14)2 /8.1 - 288 (9.8)

Fn = 4146.5 N

4)

the acceleration of car is centripetal acceleration which remains constant since speed is constant.

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