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

Answer:

0.253 m/s

Step-by-step 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 = $\frac{v^{2} }{r}$

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

centripetal force = $\frac{v^{2} }{r}$  = $\frac{v^{2} }{0.013}$

4.905 = $\frac{v^{2} }{0.013}$

$v^{2}$ = 0.063765

v = $\sqrt{0.063765}$ = 0.253 m/s