Question

A roller-coaster track has six semicircular "dips" with different radii of curvature. The same roller-coaster cart rides through each dip at a different speed.

For the different values given for the radius of curvatureR and speed v, rank the magnitude of the force of the roller-coaster track on the cart at the bottom of each dip.

A) R=60m v=16 m/s B) R=15m v= 8m/s C) R=30m v= 4m/s D) R=45m v= 4m/s E) 30m v= 16 m/s F) R= 15m v=12 m/s

Answer 1

Concepts and reason

The main concept used to solve the problem is centripetal force.

Initially, use the expression to calculate the centripetal force to calculate the force. Later, calculate the magnitude of centripetal force for all the six items. Finally, rank them according magnitude.

Fundamentals

The expression to calculate the centripetal force is,

${F_C} = \frac{{m{v^2}}}{R}$

Here, ${F_C}$ is the centripetal force, m is the mass, v is the velocity, and R is the radius of the circle.

The expression to calculate the centripetal force is,

${F_C} = \frac{{m{v^2}}}{R}$

Here, ${F_C}$ is the centripetal force, m is the mass, v is the velocity, and R is the radius of the circle.

For item A,

Substitute 60 m for R and $16\,{\rm{m/s}}$ for v in expression ${F_C} = \frac{{m{v^2}}}{R}$.

$\begin{array}{c}\\{F_C} = \frac{{m{{\left( {16\,{\rm{m/s}}} \right)}^2}}}{{60\,{\rm{m}}}}\\\\ = 4.27\,{\rm{m}}\\\end{array}$

For item B,

Substitute 15 m for R and $8\,{\rm{m/s}}$ for v in expression ${F_C} = \frac{{m{v^2}}}{R}$.

$\begin{array}{c}\\{F_C} = \frac{{m{{\left( {8\,{\rm{m/s}}} \right)}^2}}}{{15\,{\rm{m}}}}\\\\ = 4.27\,{\rm{m}}\\\end{array}$

For item C,

Substitute 30 m for R and $4\,{\rm{m/s}}$ for v in expression ${F_C} = \frac{{m{v^2}}}{R}$.

$\begin{array}{c}\\{F_C} = \frac{{m{{\left( {4\,{\rm{m/s}}} \right)}^2}}}{{30\,{\rm{m}}}}\\\\ = 0.534\,{\rm{m}}\\\end{array}$

For item D,

Substitute 45 m for R and $4\,{\rm{m/s}}$ for v in expression ${F_C} = \frac{{m{v^2}}}{R}$.

$\begin{array}{c}\\{F_C} = \frac{{m{{\left( {4\,{\rm{m/s}}} \right)}^2}}}{{45\,{\rm{m}}}}\\\\ = 0.356\,{\rm{m}}\\\end{array}$

For item E,

Substitute 30 m for R and $16\,{\rm{m/s}}$ for v in expression ${F_C} = \frac{{m{v^2}}}{R}$.

$\begin{array}{c}\\{F_C} = \frac{{m{{\left( {16\,{\rm{m/s}}} \right)}^2}}}{{30\,{\rm{m}}}}\\\\ = 8.534\,{\rm{m}}\\\end{array}$

For item F,

Substitute 15 m for R and $12\,{\rm{m/s}}$ for v in expression ${F_C} = \frac{{m{v^2}}}{R}$.

$\begin{array}{c}\\{F_C} = \frac{{m{{\left( {12\,{\rm{m/s}}} \right)}^2}}}{{15\,{\rm{m}}}}\\\\ = 9.6\,{\rm{m}}\\\end{array}$

Ans:

The rank of magnitude of the force of the roller-coaster track at the bottom of each dip is.