Mike Smith, a 17 year old male from the Internet asks on July 24, 1999,

Can you help me with the physics of an amusement park ride called the Magic Carpet? The Magic Carpet is a ride where you sit on a platform (the carpet) and it starts to swing back and forth like a pendulum. When it gets to the highest point, you don't move for a few moments, then drop back into the swinging. I'm trying to find the apparent weight that you would have at the bottom of the swing. I know the real mass, which is 70.3 kg. The time of fall is 3.18 sec. Acceleration in Gs is 1; deceleration in Gs is 3.25, and the height of the ride is 6.24 meters. I now need the Centripetal Force so I could add it onto the real weight but I am unable to find it. Could you help?viewed 16874 times

Your variables are: mass m = 70.3 kg, pendulum height d = 6.24 m g = local gravitational constant = 9.8 N/kg.

The "trick" to this problem is to realize that the potential energy you must have at the point of release (the top of the pendulum arc) is entirely converted to kinetic energy at the moment the pendulum swings through the bottom of its arc. This is the moment of most rapid centripetal acceleration and therefore the moment the passenger on the carpet ride will feel the greatest force. The potential energy at the point of release = m g d equate this with kinetic energy for a mass of 70.3 kg (0.5 m v^2) i.e., m g d = 0.5 m v^2 to get the maximum velocity. Thus v = square root (2 g d). Now from this velocity you can determine the centripetal acceleration. You must know the radius of the pendulum itself (this works for roller coasters, too, if you can find the radius of the circle which matches the curvature of the bends in the track). Centripetal acceleration equals v^2 / r for acceleration along a circle of radius r. If this is also the pendulum height d (i.e., you are dropped so you are instantaneously in free fall) then the formula is quite simple: centripetal acceleration = (sqrt(2 g d)) ^ 2 / d = 2 g d / d = 2 g As you correctly observed, there's also the extra g contributed by the Earth (it doesn't turn off) so the participant would feel 3 g of force at the bottom of the ride.

Note: All submissions are moderated prior to posting.

If you found this answer useful, please consider making a small donation to science.ca.

- Canadian Nuclear Safety Commission Educational Resources
- National Inventors Hall of Fame
- JUMP Math
- Society for Canadian Women in Science and Technology
- SciQuest e-Solutions for Science
- Natural Sciences and Engineering Research Council of Canada
- Manning Awards for Innovation
- Royal Society of Canada
- Geological Survey of Canada
- Canadian Networks of Centres of Excellence
- Canadian Landscapes at Natural Resources Canada
- Canadian Association of Physicists
- A Century of Innovation
- Understanding Science
- AlphaGalileo
- National Film Board of Canada Youth Science
- PICS Climate Insights 101
- Canadian Association for Girls in Science
- Virtual Library for the History of Science
- The Chemical Institute of Canada
- Canadian Biotechnologist 2.0
- ISI Highly Cited Scientists
- Deep River Science Academy
- Journal of the History of Canadian Science
- Wikipedia
- Innovation Canada
- Mars Society
- Nobel Prize Archive
- science.gc.ca
- Online Science & Engineering Encyclopedia
- CurioCity
- Canadian Nuclear FAQ
- Association of Science Communicators
- Astrofiles
- Wilderness Astronomy