Question

In short-track speed skating, the track has straight sections and semicircles 16 m
In short-track speed skating, the track has straight sections and semicircles 16 m m/s .Part A What is the horizontal force on the skater? part B What is the ratio of this force to the skater's weight?kg skater goes around the turn at a constant 11m/s m in diameter. Assume that a 64kg
in diameter. Assume that a 64kg skater goes around the turn at a constant 11m/s .     
Part A

What is the horizontal force on the skater?

part B

What is the ratio of this force to the skater's weight?

Answer 1

Concepts and reason

The concept of centripetal force is used here.

First, calculate the value of horizontal force on the skater using the expression of centripetal force and use that force value to find out the ratio of the calculated force to the skater’s weight.

Fundamentals

The expression of the centripetal force is given as follows:

${F_c} = \frac{{m{v^2}}}{r}$

Here, m is the mass of skater, v is the speed of skater, and r is the radius of track.

The ratio of centripetal force to the skater’s weight is given as follows:

$\begin{array}{c}\\\frac{{{F_c}}}{W} = \frac{{m{v^2}/r}}{{mg}}\\\\ = \frac{{{v^2}}}{{rg}}\\\end{array}$

Here, v is the velocity, r is the radius o track and g is the acceleration due to gravity.

(A)

The radius of the track is given as:

$r = \frac{d}{2}$

Substitute 16 m for d in the above equation.

$\begin{array}{c}\\r = \frac{{16{\rm{ m}}}}{2}\\\\ = 8{\rm{ m}}\\\end{array}$

The horizontal force on the skater is given as:

$F = \frac{{m{v^2}}}{r}$

Substitute 64 kg for m, 11 m/s for v, and 8 m for r in equation $F = \frac{{m{v^2}}}{r}$ .

$\begin{array}{c}\\F = \frac{{\left( {64{\rm{ kg}}} \right){{\left( {11{\rm{ m/s}}} \right)}^2}}}{{\left( {8{\rm{ m}}} \right)}}\\\\ = 968{\rm{ N}}\\\end{array}$

(B)

The skater’s weight is calculated as:

$W = mg$

Substitute 64 kg for m and $9.8{\rm{ m/}}{{\rm{s}}^2}$ for g in the equation $W = mg$ .

$\begin{array}{c}\\W = \left( {64{\rm{ kg}}} \right)\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 627.2{\rm{ N}}\\\end{array}$

The ratio is given as:

${\rm{ratio}} = \frac{F}{W}$

Substitute 968 N for F and 627.2 N for W in the above equation.

$\begin{array}{c}\\{\rm{ratio}} = \frac{{968{\rm{ N}}}}{{627.2{\rm{ N}}}}\\\\ = 1.54\\\end{array}$

Ans: Part A

The horizontal force on skater is 968 N.

Part B

The ratio of the force to the skater’s weight is 1.54.