Problem 3 - College Physics

Question

A \(52 \mathrm{~kg}\) ice skater spins about a vertical axis through her body with her arms horizontally outstretched, making 2.0 turns each second. The distance from one hand to the other is \(1.50 \mathrm{~m}\). Biometric measurements indicate that each hand typically makes up about \(1.25 \%\) of body weight. (a) Draw a free-body diagram of one of her 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

(a) Draw an FBD showing centripetal force at the wrist. (b) The wrist exerts a force of 76.48 N on the hand. (c) This force is about 12 times the hand's weight.

Step 1: Understand the situation

A figure skater is spinning, with her arms outstretched and the hands moving in a circular motion. We need to analyze the forces involved in this circular motion, starting with drawing a diagram of the forces acting on one of her hands.

Step 2: Draw a Free-Body Diagram (FBD)

In the free-body diagram of her hand, consider the forces acting. The main force involved is the centripetal force that acts horizontally towards the center of the circle. There would also be a force from the wrist providing this centripetal force.

Step 3: Calculate the weight of one hand

The hands make up 1.25% of the body weight each. Therefore, the weight of one hand, given the total body weight of 52 kg, is: Weight of hand = 0.0125 x 52 kg x 9.81 m/s² This results in approximately 6.375 N.

Step 4: Determine centripetal force needed

Centripetal force is needed to keep the hand moving in a circle. The formula for centripetal force is \[ F_c = \frac{mv^2}{r} \]where \( m \) is the mass of the hand, \( v \) is the tangential speed, and \( r \) is the radius of the circular path.

Step 5: Calculate the mass of one hand

The mass of one hand can be found from its weight:Mass of hand = \( \frac{6.375 \text{ N}}{9.81 \text{ m/s}^2} \approx 0.65 \text{ kg} \).

Step 6: Find the tangential speed of her hands

The skater makes 2 turns per second, so the rotational speed \( \omega \) is 4\( \pi \) \( \text{rad/s} \). The radius \( r \) is half the distance between her hands, i.e., 0.75 m.The linear speed \( v \) is given by \( v = r\omega \). \[ v = 0.75 \text{ m} \times 4\pi \text{ rad/s} \approx 9.42 \text{ m/s} \].

Step 7: Calculate the centripetal force

Substitute the values into the centripetal force formula:\[ F_c = \frac{mv^2}{r} = \frac{0.65 \text{ kg} \times (9.42 \text{ m/s})^2}{0.75 \text{ m}} \approx 76.48 \text{ N} \].This is the force the wrist must exert on the hand horizontally.

Step 8: Express the force as a multiple of the hand's weight

The weight of the hand is 6.375 N. The ratio \( \frac{F_c}{ ext{weight of hand}} \) is:\[ \frac{76.48 \text{ N}}{6.375 \text{ N}} \approx 12 \]Therefore, the force the wrist exerts is approximately 12 times the weight of the hand.

Key Concepts

Centripetal Force

Centripetal force is essential in understanding circular motions, such as a skater's hand while spinning. It is a force that keeps an object moving in a circular path by constantly pulling it toward the center of the circle. This force ensures that the object does not travel in a straight line due to inertia.

To calculate the centripetal force, we use the formula:

• \( F_c = \frac{mv^2}{r} \)

where \( m \) represents the mass of the rotating object, \( v \) is the tangential or linear speed, and \( r \) is the radius of the circle. In the case of the ice skater, her wrist supplies the necessary centripetal force to keep her hand moving in a circle.

In the example of the ice skater, her wrist exerts a force that is much larger than the weight of her hand, demonstrating how significant centripetal forces can be even for normal movements.

Free-Body Diagram

A free-body diagram (FBD) is a key tool in physics to visually represent the forces acting on an object. For the ice skater's hand, drawing an FBD helps us understand how different forces interact to maintain her circular motion.

In the FBD of the skater's hand, we identify:

• Centripetal force directed towards the body (center of the circular path).
• A reactive force from the wrist which supplies the centripetal force.

These diagrams eliminate overwhelming details and focus on forces like gravity, tension, and contact that influence the object's motion. By doing so, we can simplify what might otherwise be a complex situation into manageable pieces.

Rotational Motion

Rotational motion is when an object moves around a fixed point or axis, like the ice skater spinning about her vertical axis. This kind of motion can be described through various quantities, such as rotational speed (ω) and tangential speed (v).

The tangential speed of the skater's hand helps us determine how fast the hand is moving along its circular path. It relates to rotational speed through the formula:

• \( v = r\omega \)

In this context, \( r \) is the radius of the path, and \( \omega \) is the angular speed, which is measured in radians per second. This indicates how quickly the skater is completing rotations. Understanding these calculations gives insight into how variables such as speed and radius affect the energy required to maintain rotational motion.

Physics in Sports

Physics plays a significant role in sports, helping athletes enhance performance and understand the mechanics of their movements. In figure skating, principles like centripetal force and rotational motion are crucial.

Consider how a skater uses physics. Increased velocity helps in gaining more control and maintaining balance during complex spins. By understanding forces involved, athletes can optimize their techniques for better energy efficiency and precision.

Sports such as skating provide excellent real-world examples of physics at work, from simple balance to intricate maneuvers involving rotational dynamics. Athletes who master these principles can often achieve a higher level of skill and efficiency, showing how intertwined the physical laws are with sports performance.