Q. A roller coaster is designed such that riders experience ‘weightlessness’ as they go round the top of a hill whose radius of curvature is 20 m. The speed of the car at the top the hill is between
A) ${14\ ms}^{- 1}\ and\ {15\ ms}^{- 1}$
B) ${15\ ms}^{- 1}\ and\ {16\ ms}^{- 1}$
C) ${16\ ms}^{- 1}\ and\ {17\ ms}^{- 1}$
D) ${13\ ms}^{- 1}\ and\ {14\ ms}^{- 1}$
Explanation
(a) Balancing the force, we get
$Mg - N = M\frac{v^{2}}{R}$
For weightlessness, $N = 0$
$\therefore\ \frac{{Mv}^{2}}{R} = Mg$
$or\ \ v = \sqrt{Rg}$
Putting the values, $R = 20\ m,\ g = {10.0\ ms}^{- 2}$
$So,\ \ v = \sqrt{20 \times 10.0} = 14.14\ {ms}^{- 1}$
Thus, the speed of the car at the top of the hill is between $14\ {ms}^{- 1}\ and\ {10\ ms}^{- 1}$
[image:4_image84.emf]
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.