Question

The passengers in a roller coaster car feel 50% heavier than their true weight as the car goes through a dip with a 30.0 m radius of curvature.

What is the car's speed at the bottom of the dip?
Need Step by step. Sorry its late. Thanks.

Answer 1

Concepts and reason

The concept required to solve this problem is the Newton’s second law of motion.

Initially, write an equation for the apparent weight of the passengers. Later, apply the Newton’s law of motion. Finally, solve for the speed of the car at the bottom of the dip.

Fundamentals

Newton’s second law of motion states that the net force on an object is directly proportional to its acceleration and is given by,

$F = ma$

Here, m is the mass and a is the acceleration.

The expression for the apparent weight of the passenger is as follows:

$R = W + \frac{W}{2}$ …… (1)

Here, W is the net weight acting downwards.

The net weight acting downward on the car is as follows:

$W = mg$

Here, m is the mass of the car and g is the acceleration due to gravity.

Substitute mg for W in the equation (1).

$R = mg + \frac{{mg}}{2}$ …… (2)

The centripetal force acting on the car is as follows:

${F_{\rm{c}}} = \frac{{m{v^2}}}{r}$

Apply the Newton’s law of motion on the car as follows:

$R = mg + \frac{{m{v^2}}}{r}$ …… (3)

Substitute $mg + \frac{{mg}}{2}$ for R in the equation (3).

$\begin{array}{c}\\mg + \frac{{mg}}{2} = mg + \frac{{m{v^2}}}{r}\\\\\frac{{mg}}{2} = \frac{{m{v^2}}}{r}\\\\\frac{g}{2} = \frac{{{v^2}}}{r}\\\end{array}$

Solve for v.

$v = \sqrt {\frac{{gr}}{2}}$ ……. (4)

Substitute $9.8{\rm{ m/}}{{\rm{s}}^2}$for g, 30.0 m for r in the equation (4) and solve for v.

$\begin{array}{c}\\v = \sqrt {\frac{{\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {30.0{\rm{ m}}} \right)}}{2}} \\\\ = 12.12{\rm{ m/s}}\\\end{array}$

Ans:

The magnitude of the car’s speed at the bottom of the dip is equal to 12.12 m/s.