Q 44E - University Physics with Modern Physics

Question

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

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,

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

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

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

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

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

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

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

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

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.