Q44E - University Physics with Modern Physics

Question

A 52-kg ice skater spins about a vertical axis through her body with her arms horizontally out stretched; she makes 2.0 turns each second. The distance from one hand to the other is 1.50 m. Biometric measurements indicate that each hand typically makes up about 1.25% of body weight. (a) Draw a free-body diagram of one of the skater’s hands. (b) What horizontal force must her wrist exert on her hand? (c) Express the force in part (b) as a multiple of the weight of her hand.

Step-by-Step Solution

Answer

(b) The horizontal force her wrist exert on her hand is, 77 N.

(c) the force as a multiple of the weight of her hand is, 10.9.

Step 1: identification of given data

The given data can be listed below as,

• The ice skater’s weight is, $w=52 kg$ .
• The distance from one hand to other is, d = 1.50 m .

Step 2: Significance of centripetal force

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.

Step 3: Determination of free body diagram of one of the skater’s hand.

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

Step 4: determination of horizontal force exert on her hand.

The expression for the velocity can be expressed as,

 

 $$\begin{gathered} v=\frac{dis\tan ce}{time} \\ V=\frac{d}{t} \\ v=\frac{no. of turns\times 2\times \mathrm{\pi }\times \mathrm{r}}{t} \end{gathered}$$

 

For $r=0.75 m$ and $t=1 s$ , the above equation becomes-

 

 $$\begin{gathered} v=\frac{2.0\times 2\times 3.14\times {\displaystyle \frac{1.50 m}{2}}}{1 sec} \\ =9.42 m/s \end{gathered}$$

Hence, the velocity is, .

 

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

 

$$\begin{gathered} F=\frac{0.0125\times 52 kg\times {\left(9.42m/s\right)}^2}{{\displaystyle \frac{1.50 m}{2}}} \\ =77 N \end{gathered}$$ 

                                                                  

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

Step 5: Determination of force as multiple of weight of her hand.

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

 

$W=m g$ 

 

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

 

For $m=0.0125\times 52 kg$ , and $9.8 m/s^2$ , the above equation becomes,

$$\begin{gathered} W=0.0125\times 52 kg\times 9.8 m/s^2 \\ =10.9 N \end{gathered}$$ 

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,

$$\begin{gathered} k=\frac{wrist force}{weight of hand} \\ =\frac{F}{W} \end{gathered}$$   

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

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

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

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