Dynamics of uniform circular motion

Click For Summary
SUMMARY

The discussion focuses on calculating the speed of a rider at the top of a vertical loop-the-loop ride using centripetal acceleration principles. Given a radius of 21 meters and normal forces of 770 N at rest and 350 N at the top, the rider's mass is calculated as 78.4 kg. The final speed at the top of the loop is determined to be 17.3 m/s using the formula for centripetal force, confirming the setup and calculations are correct.

PREREQUISITES
  • Centripetal acceleration in vertical circular motion
  • Understanding of normal force and gravitational force
  • Basic physics equations for force and motion
  • Knowledge of unit conversions (N to kg)
NEXT STEPS
  • Study the principles of centripetal force in circular motion
  • Learn how to derive equations for vertical circular motion
  • Explore the effects of varying radius on speed in circular motion
  • Investigate real-world applications of centripetal acceleration in amusement park rides
USEFUL FOR

Physics students, mechanical engineers, and amusement park ride designers will benefit from this discussion, particularly those interested in the dynamics of circular motion and force analysis.

bulbasaur88
Messages
57
Reaction score
0
A special electronic sensor is embedded in the seat of a car that takes riders around a circular loop-the-loop ride at an amusement park. The sensor measures the magnitude of the normal force that the seat exerts on a rider. The loop-the-loop ride is in the vertical plane and its radius is 21 m. Sitting on the seat before the ride starts, a rider is level and stationary, and the electronic sensor reads 770 N. At the top of the loop the rider is upside down and moving, and the sensor reads 350 N. What is the speed of the rider at the top of the loop?

I think this problem is a great problem for centripetal acceleration in vertical circular motion; however, I cannot check to see if I understand the concept behind the problem because there is no answer provided. Could someone please let me know if at least my set-up is correct?

-----------------------------------------------------------------------------------------
Given
R = 21 meters
W = 770 N
m = 770/9.8 kg
FN = 350 N at the top

At the top:
Fc = mg + FN = mv2 / R
R[FN + mg] = mv2

Plug-n-Chug to solve for V
V = 17.3 m/s
 
Physics news on Phys.org
I haven't checked the numbers, but your set up looks fine to me :approve:
 
FANTASTIC THANK YOU :biggrin:
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 21 ·
Replies
21
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
12K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K