Question

A student has a weight of 655 N. While riding a roller coaster they seem
to weigh 1.96x103 N at the bottom of a dip that has a radius of 18.0 m. What
is the speed of the roller coaster at this point?

Answer 1

Explanation

• For the question to be meaningful, the value mentioned ($\color{red}\footnotesize\bold{1.96\times103\,N}$ ) should be corrected to $\color{red}\footnotesize\bold{1.96\times10^3\,N}$.
• 655N weight happens to be when he is at rest.
• The phenomenon of apparent weight occurs when there is a net force acting on the body (any acceleration, deceleration, upthrust etc).
• What happens here is, when he take the path at the bottom of the dip he enters into a rotational motion in which a generated centrifugal force pushes him down towards the bottom with an intention to getting supplied the needed centripetal force to complete the motion.
• There he feels as if his weight is increased.
• We calculate the weight of an object by measuring the resultant force( as in the picture 1)
• So you can see that the resultant force is increased while in the motion(in picture 2), hence the weight is increased (=apparent weight).

Notations

• Please refer to the sketch attached

Calculations

• Applying F=ma on the person down at the bottom towards the center of the circular path.

$\qquad \begin{aligned} \footnotesize R_1-mg&=\footnotesize \frac{mv^2}{r} \cdots\cdots\cdots R_1=\footnotesize mg+ \frac{mv^2}{r}\cdots(weight \,is\, increased\,=\,1.96\times10^3N)\\ \footnotesize v&=\footnotesize \sqrt{\frac{r(R_1-mg)}{m}}\\ & = \footnotesize \sqrt{ \frac{18m(1960N-655N)}{65.5kg}}\\ & =\footnotesize \bold {18.94 ms^{-1}} \end{aligned}$

• Velocity of the roller coaster equals that of the person riding it.