Weightless roller coaster

Classical Mechanics Level 2

A roller coaster goes over a hill of radius $30\text{ m}$ . If the roller coaster moves at a great speed, then it will lose contact with the ground and fall freely in a parabolic trajectory. Under the free fall, people sitting in the roller coaster will feel weightless.

What is the minimum speed (in $\text{m/s}$ ) required to achieve the weightlessness?

Give your answer to 2 decimal places.

Take $g = 9.8\text{ m/s}^{2}$ .

The answer is 17.15.

1 solution

Chew-Seong Cheong
Jan 1, 2017

Weightlessness is attained when the centripetal force of the roller coaster balances its weight. That is:

$\begin{aligned} \frac {mv^2}r&=mg \\ \frac {v^2}r&=g \\ v^2 &=rg \\ \implies v &=\sqrt {rg} \\ &=\sqrt {30 \times 9.8} \\ & \approx \boxed {17.15} \end{aligned}$

There is no 'centrifugal force' associated here. When the gravitational force equals the 'centripetal' force, you achieve weightlessness. Centrifugal force (mainly a pseudo force) can exist only when you're observing the event from a non-inertial frame of reference.

Atomsky Jahid - 4 years, 5 months ago

Thanks. I will change it.

Chew-Seong Cheong - 4 years, 5 months ago

I think you have to prove it is minimum speed, otherwise the minimum speed can be at angle with vertical.

Prince Loomba - 4 years, 5 months ago

Weightless roller coaster

The answer is 17.15.

1 solution

0 pending reports