Problem 38 - College Physics

Question

In performing a "figure \(8^{\prime \prime}\) maneuver, a figure skater wants to make the top part of the 8 approximately a circle of radius \(2.20 \mathrm{~m} .\) He needs to glide through this part of the figure at approximately a constant speed, taking \(4.50 \mathrm{~s} .\) His skates digging into the ice are capable of providing a maximum centripetal acceleration of \(3.25 \mathrm{~m} / \mathrm{s}^{2}\). Will he be able to do this as planned? If not, what adjustment can he make if he wants this part of the figure to remain the same size (assume the ice conditions and skates don't change)?

Step-by-Step Solution

Answer

No, the skater can't complete it in 4.50 s. He should take about 5.18 s instead.

Step 1: Calculate the Required Velocity

To find the required velocity, we need to use the formula for the circumference of a circle, which is \(C = 2\pi r\). For a circle with a radius of \(2.20 \, \text{m}\), the circumference would be \[C = 2 \times \pi \times 2.20 = 13.82 \, \text{m}.\] The skater needs to glide this distance in \(4.50 \, \text{s}\), so the required velocity \(v\) is \[v = \frac{13.82}{4.50} \approx 3.07 \, \text{m/s}.\]

Step 2: Calculate the Maximum Possible Velocity

The maximum centripetal acceleration \(a_c\) is given by \(a_c = \frac{v^2}{r}\), where \(v\) is the velocity and \(r\) is the radius (\(2.20 \, \text{m}\)). We are given the maximum centripetal acceleration is \(3.25 \, \text{m/s}^2\). To find the maximum possible velocity, we set up the equation:\[ 3.25 = \frac{v^2}{2.20}. \]Solving for \(v\), we get\[v^2 = 3.25 \times 2.20 \Rightarrow v^2 = 7.15 \Rightarrow v = \sqrt{7.15} \approx 2.67 \, \text{m/s}.\]

Step 3: Compare Velocities and Determine Feasibility

In step 1, we calculated that the required velocity to complete the circle in the given time is approximately \(3.07 \, \text{m/s}\). However, step 2 shows the maximum velocity allowed by the skates' centripetal force is \(2.67 \, \text{m/s}\).Since the required velocity exceeds the maximum possible velocity, the skater will not be able to maintain the circular path as originally planned.

Step 4: Suggest Adjustment for Skater

To complete the maneuver within the limits of the available centripetal acceleration while keeping the radius the same size, the skater could increase the time to finish the circle. By increasing the time, the required velocity (and thus the required centripetal acceleration) decreases.For instance, rearranging the required velocity equation \(v = \frac{13.82}{t}\) and setting it equal to the maximum possible value \(2.67 \, \text{m/s}\), find the time:\[2.67 = \frac{13.82}{t} \Rightarrow t = \frac{13.82}{2.67} \approx 5.18 \, \text{s}.\]The skater could take approximately \(5.18 \, \text{s}\) instead of \(4.50 \, \text{s}\) to circle the top part of the figure 8.

Key Concepts

Figure Skating Mechanics

Figure skating is not just about grace and beauty—there’s a lot of physics at play! When a figure skater performs on the ice, they're using forces and motions that can be described by physics principles. In the case of our figure skater attempting a circle maneuver, understanding these mechanics is crucial:

• Skates on Ice: The skates dig into the ice, applying forces that allow the skater to maintain balance and control.
• Rotational Movements: During a circular move, the skater relies on changing body positions to manage rotational speed and direction.
• Balance and Momentum: A skater's balance is key to making smooth transitions, while momentum allows them to glide gracefully across the ice.

By understanding these principles, figure skaters can perfect their maneuvers, such as the thrilling figure 8.

Circular Motion

Circular motion is a fundamental concept in physics, especially when discussing a skater gliding through a circle path. When moving in a circle:- A constant force called the centripetal force is needed to keep an object on its circular path.- This force keeps pulling the skater towards the center, preventing them from flying off.For our figure skater:

• Centripetal Acceleration: This is the acceleration towards the center of the circle, and it ensures that the skater remains on the curved path. It can be calculated using: \( a_c = \frac{v^2}{r} \)
• Circle Radius: In this exercise, the radius chosen is 2.20 meters.
• Calculation of Path: By using the skater's maximum allowed centripetal acceleration of 3.25 \( \text{m/s}^2 \), we establish whether they can maintain the desired circle path.

Comprehending these aspects of circular motion allows a skater to plan their movements effectively, ensuring their path adheres to physical possibilities.

Velocity Calculation

Calculating velocity in circular motion scenarios is essential for planning the required speed while staying within force limits. Velocity here is how fast the skater is moving along their circular path.The original exercise used the formula for the circumference of a circle \( C = 2 \pi r \) to find how far the skater needed to glide:\[ C = 2 \times \pi \times 2.20 = 13.82 \, \text{m} \]To find the skater's velocity,\[ v = \frac{C}{t} \]where \( t = 4.50 \, \text{s} \) as originally planned, resulting in \( v \approx 3.07 \, \text{m/s} \).

Adjustments for Feasibility

When the calculated velocity exceeds the allowed velocity due to centripetal force limits, adjustments are necessary:

• By increasing the time taken \( t \), the velocity \( v \) becomes less.
• This ensures that the skater stays within the limits of their maximum centripetal force.

For an adjusted time of approximately 5.18 seconds, the feasible velocity would meet the requirements, keeping the figure skating maneuver safe and controlled.