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 = 289 kg and moves with speed v = 14.08 m/s. The loop-the-loop has a radius of R = 8.2 m.

1)

What is the magnitude of the normal force on the car when it is at the bottom of the circle? (But as the car is accelerating upward.)

N

Your submissions:

Computed value:

92.01

Submitted:

Tuesday, September 16 at 12:17 AM

Feedback:

Computed value:

920.1

Submitted:

Tuesday, September 16 at 12:17 AM

Feedback:

Computed value:

9201

Submitted:

Tuesday, September 16 at 12:17 AM

Feedback:

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

Your submissions:

3)

What is the magnitude of the normal force on the car when it is at the top of the circle?

N

Your submissions:

4)

Compare the magnitude of the cars acceleration at each of the above locations:

abottom = aside = atop

abottom < aside < atop

abottom > aside > atop

Your submissions:

5)

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

For question 1. You are trying to find normal force on the car when it is at the bottom of the loop. You are given the velocity of the car and the radius of the loop. Using the formula from uniform circular motion, you can find the acceleration (a = (v^2)/r.

Your acceleration is 24.18m/s^2

Then you use F=ma to solve for normal force but remember to account for gravity.

F= (289kg)(24.18m/s^2 + 9.8m/s^2)

F=9820.22N

Hope this helps.

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