Question

A roller coaster car is going over the top of a 13-mm-radius circular rise. At the top of the hill, the passengers "feel light," with an apparent weight only 50 %% of their true weight. How fast is the coaster moving?

Answer 1

0.253 m/s

Explanation:

radius r of the circular rise = 13 mm = 0.013 m

apparent weight loss = 50%

acceleration of the new weight = 0.5 x 9.81 = 4.905 m/s^2

centripetal acceleration = 9.81 - 4.905 = 4.905 m/s^2

centripetal acceleration =
`(v^(2) )/(r)`

where v is the acceleration at the rise and r is the radius of the rise

centripetal force =
`(v^(2) )/(r)` =
`(v^(2) )/(0.013)`

4.905 =
`(v^(2) )/(0.013)`

`v^(2)` = 0.063765

v =
`√(0.063765)` = 0.253 m/s