91影视

The given data can be listed below as,

• The ice skater鈥檚 weight is,$w=52\text{kg}$.
• The distance from one hand to other is, $d=1.50\text{m}$.

When an object moves in a circular motion than an outward force acting on it in order to balance the movement that outward force is known as centripetal force.

The free body diagram of one of the skater鈥檚 hand can be expressed as,

The expression for the velocity can be expressed as,

$$\begin{gathered} v=\frac{\text{distance}}{\text{time}} \\ v=\frac{d}{t}\backslash \mathrm{hfill} \\ v=\frac{\text{no.}\text{of}\text{turns}脳2脳蟺脳r}{t} \end{gathered}$$

For$r=0.75\text{m}$and $t=鈥�1\text{s}$, the above equation becomes-

$\begin{array}{rcl}v& =& \frac{2.0脳2脳3.14脳\frac{1.50\text{m}}{2}}{1\sec }\\ & =& 9.42\text{m/s}\end{array}$

Hence, the velocity is, $9.42\text{m/s}$.

The expression for the centripetal force can be expressed as,

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

Here m is the mass, v is the velocity of object, r is the radius of circular path.

For $m=0.0125脳52\text{kg}$, $v=9.42\text{m/s}$, and $r=\frac{1.50\text{m}}{2}$, the above equation becomes,

$\begin{array}{rcl}F& =& \frac{0.0125脳52\text{kg}脳{(9.42\text{m/s})}^2}{\frac{1.50\text{m}}{2}}\\ & =& 77\text{N}\end{array}$

Hence, required horizontal force exerted on her hand is, 77 N.

The expression for the weight of hand can be expressed as,

$W=mg$

Here m is the mass, and g is the gravitational acceleration of earth.

For $m=0.0125脳52\text{kg}$, and$g=9.8{\text{m/s}}^{\text{2}}$, the above equation becomes,

role="math" localid="1668420038477" $\begin{array}{rcl}W& =& 0.0125脳52\text{kg}脳\text{9.8}{\text{m/s}}^2\\ & =& 10.9\text{N}\end{array}$

Hence, the weight of hand is, 10.9 N.

The expression for the force as the multiple of the weight of the hand can be expressed as,

$\begin{array}{rcl}k& =& \frac{\text{wrist}\text{force}}{\text{weight}\text{of}\text{hand}}\\ & =& \frac{F}{W}\end{array}$

Here F is the wrist force, and W is the weight of the hand.

Substitute, 77 N for F, and 10.9 N for W in the above equation.

$$\begin{gathered} =\frac{77\text{N}}{10.9\text{N}} \\ =7.06 \end{gathered}$$

Hence, required force as a multiple of weight of hand is, 7.06.