In short-track speed skating, the track has straight sections and semicircles 16 m in diameter. Assume…
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Question “In short-track speed skating, the track has straight sections and semicircles 16 m in diameter. Assume…”
In short-track speed skating, the track has straight sections
and semicircles 16 m
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
Here, the concept of centripetal force can be used.
To calculate the horizontal force exerted on the skater, first use the expression of centripetal force. Then, convert that force value into a ratio of the calculated force to skater’s body weight.
Here is the expression of the centripetal power:
{F_c} = \frac{{m{v^2}}}{r}m refers to the mass of the skater, v the speed and r the radius of the track.
Here is the ratio of centripetal force and skater’s weight:
\begin{array}{c}\\\frac{{{F_c}}}{W} = \frac{{m{v^2}/r}}{{mg}}\\\\ = \frac{{{v^2}}}{{rg}}\\\end{array}V denotes the velocity. R denotes the radius of the track. G indicates the acceleration caused by gravity.
(A)
The track radius is as follows:
r = \frac{d}{2}In the equation above, substitute 16 m for d.
\begin{array}{c}\\r = \frac{{16{\rm{ m}}}}{2}\\\\ = 8{\rm{ m}}\\\end{array}The horizontal force applied to the skater is:
F = \frac{{m{v^2}}}{r}In equation F = \frac{{m{v^2}}}{r}, substitute 64 kg for m and 11 m/s each for v.
\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 following formula is used to calculate the weight of a skater:
W = mgSubstitute 64 kg to m and 9.8{\rm{ m/}}{{\rm{s}}^2} \begin{array}{c}\\W = \left( {64{\rm{ kg}}} \right)\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 627.2{\rm{ N}}\\\end{array}
This is the ratio:
{\rm{ratio}} = \frac{F}{W}In the equation above, substitute 968 N for F and 622 N for W.
$math_tag_11 Ans: Part A
The horizontal force on skaters is 968 N.
Part B
1.54 is the ratio of the force to skater’s body weight.
Conclusion
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