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(II) On an ice rink two skaters of equal mass grab hands and spin in a mutual circle once every \(2.5 \mathrm{~s}\). If we assume their arms are each \(0.80 \mathrm{~m}\) long and their individual masses are \(60.0 \mathrm{~kg}\), how hard are they pulling on one another?

Short Answer

Expert verified
They pull on each other with approximately 301.26 N.

Step by step solution

01

Understand the Problem

Two skaters with equal mass are spinning in a circle, and we need to find the force with which they are pulling on each other. We have their mass, the rotation period, and the arm length.
02

Calculate Angular Velocity

First, find the angular velocity \( \omega \). The skaters complete one rotation in \( T = 2.5 \) seconds. Angular velocity \( \omega \) is given by \( \omega = \frac{2\pi}{T} \). Calculate \( \omega = \frac{2\pi}{2.5} = 2.51 \) rad/s.
03

Determine the Radius of the Circle

Each skater's rotation circle has a radius of \( 0.80 \) meters, which is equal to the arm length, as they are spinning in a mutual circle by holding hands.
04

Apply Centripetal Force Formula

The centripetal force \( F_c \) required to keep each skater moving in a circle is given by the formula \( F_c = m \omega^2 r \), where \( m \) is the mass of each skater \( (60 \text{ kg}) \), \( \omega \) is the angular velocity \((2.51 \text{ rad/s})\), and \( r \) is the radius \( (0.80 \text{ m})\). Substitute the values: \( F_c = 60 \times (2.51)^2 \times 0.80 \approx 301.26 \text{ N} \).
05

Interpret the Result

The calculated centripetal force is the force with which each skater is pulling on the other to maintain the circular motion. Therefore, each skater is exerting a pull force of approximately 301.26 newtons on the other.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Angular Velocity
Angular velocity is a key concept in understanding rotational motion. It measures how quickly an object rotates around a specified axis. More simply put, it tells us how fast something is spinning.
In mathematical terms, angular velocity is often represented by the symbol \( \omega \). It is calculated using the formula:
  • \( \omega = \frac{2\pi}{T} \)
where \( T \) is the period of rotation, or the time taken for one complete circle or cycle. This formula helps us find angular velocity when we know how long it takes for one rotation.
In the case of the ice skaters, they complete one full spin in 2.5 seconds. Using the formula, \( \omega = \frac{2\pi}{2.5} \), we find that their angular velocity is approximately 2.51 rad/s. This means each skater rotates 2.51 radians every second while spinning. Understanding this allows us to further explore forces involved in their circular motion.
The Concept of Circular Motion
Circular motion occurs when an object moves along the circumference of a circle. This kind of motion can be seen in daily life, such as cars going around a roundabout, or indeed, skaters twirling on ice.
Two main forces play a role here: the centripetal force, which keeps an object moving along a curved path, and the centrifugal force, often perceived by the rotating object, which seems to push it outwards.
Centripetal force acts towards the center of the circle and is essential for circular motion. In the case of the skaters, each skater pulls on the other to maintain their circular path. This is achieved by the centripetal force, calculated using:
  • \( F_c = m \omega^2 r \)
where \( m \) is the mass, \( \omega \) is the angular velocity, and \( r \) is the radius of the circular path. This formula reveals the force required to keep an object moving in a circle at a constant speed.
For our skaters, substituting \( m = 60 \) kg, \( \omega = 2.51 \) rad/s, and \( r = 0.80 \) m gives us \( F_c = 301.26 \) N. This means each skater exerts approximately 301.26 newtons of force to maintain their spinning motion.
Dynamics of Skaters on Ice
Skaters on ice provide a fascinating example of physics in motion. When two skaters hold hands and spin, they exhibit a real-life demonstration of circular motion and the forces involved. The unique surface of ice, with its low friction, allows skaters to spin with ease.
As they rotate, skaters rely heavily on maintaining balance and using the right amount of force to keep their motion steady. Each skater's arm length essentially becomes the radius of their spinning path. This affects how tightly they can spin and the force needed.
By grasping each other's hands, they create a mutual center of rotation, making it crucial that both apply equal force for smooth motion.
  • Ice provides minimal resistance, enhancing their ability to freely rotate while demonstrating principles like conservation of angular momentum.
  • More experienced skaters can control their speed and direction by adjusting their arm positions.
Their spins become a perfect mix of muscular strength, understanding of physics, and graceful execution, providing both a challenging sport and a compelling study of motion physics.

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Most popular questions from this chapter

A roller coaster reaches the top of the steepest hill with a speed of \(6.0 \mathrm{~km} / \mathrm{h}\). It then descends the hill, which is at an average angle of \(45^{\circ}\) and is \(45.0 \mathrm{~m}\) long. What will its speed be when it reaches the bottom? Assume \(\mu_{\mathrm{k}}=0.12\).

(III) The position of a particle moving in the \(x y\) plane is given by \(\overrightarrow{\mathbf{r}}=2.0 \cos (3.0 \mathrm{rad} / \mathrm{s} t) \hat{\mathbf{i}}+2.0 \sin (3.0 \mathrm{rad} / \mathrm{s} t) \hat{\mathbf{j}}\) where \(r\) is in meters and \(t\) is in seconds. ( \(a\) ) Show that this represents circular motion of radius \(2.0 \mathrm{~m}\) centered at the origin. (b) Determine the velocity and acceleration vectors as functions of time. ( \(c\) ) Determine the speed and magnitude of the acceleration. \((d)\) Show that \(a=v^{2} / r .(e)\) Show that the acceleration vector always points toward the center of the circle.

(II) Use dimensional analysis to obtain the form for the centripetal acceleration, \(a_{\mathrm{R}}=v^{2} / r\) .

(III) A 4.0 -kg block is stacked on top of a 12.0 -kg block, which is accelerating along a horizontal table at \(a=5.2 \mathrm{m} / \mathrm{s}^{2}\) (Fig. \(40 ) .\) Let \(\mu_{\mathrm{k}}=\mu_{\mathrm{s}}=\mu .\) (a) What minimum coefficient of friction \(\mu\) between the two blocks will prevent the 4.0 -kg block from sliding off? (b) If \(\mu\) is only half this minimum value, what is the acceleration of the 4.0 -kg block with respect to the table, and \((c)\) with respect to the 12.0 -kg block? (d) What is the force that must be applied to the 12.0 -kg block in \((a)\) and in \((b),\) assuming that the table is frictionless?

Earth is not quite an inertial frame. We often make measurements in a reference frame fixed on the Earth, assuming Earth is an inertial reference frame. But the Earth rotates, so this assumption is not quite valid. Show that this assumption is off by 3 parts in 1000 by calculating the acceleration of an object at Earth's equator due to Earth's daily rotation, and compare to \(g=9.80 \mathrm{~m} / \mathrm{s}^{2},\) the acceleration due to gravity.

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