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Two skaters, \(A\) and \(B,\) of equal mass are moving in clockwise uniform circular motion on the ice. Their motions have equal periods, but the radius of skater A's circle is half that of skater B's circle a) What is the ratio of the speeds of the skaters? b) What is the ratio of the magnitudes of the forces acting on each skater?

Short Answer

Expert verified
Answer: The ratio of the speeds of skater A and skater B is 1/2 and the ratio of the magnitudes of the forces acting on them is 1/4.

Step by step solution

01

Formula for speed in uniform circular motion

The formula for the speed of an object in uniform circular motion is given by: \(v = \frac{2\pi r}{T}\) where \(v\) is the speed, \(r\) is the radius of the circular path, \(T\) is the period of motion, and \(2\pi\) is a constant.
02

Compare the speeds of the skaters

We know that the period for both skaters is the same, so we can represent it as \(T\). The radius of skater A's circle is half that of skater B's circle, so we can represent them as \(r_A = \frac{1}{2}r_B\). Now, using the formula for speed, we can find \(v_A\) and \(v_B\): \(v_A = \frac{2\pi r_A}{T} = \frac{2\pi (\frac{1}{2}r_B)}{T}\) \(v_B = \frac{2\pi r_B}{T}\) Now, we can find the ratio of their speeds: \(\frac{v_A}{v_B} = \frac{\frac{2\pi (\frac{1}{2}r_B)}{T}}{\frac{2\pi r_B}{T}}\)
03

Simplify the speed ratio

To simplify the speed ratio, we can cancel out common factors: \(\frac{v_A}{v_B} = \frac{2\pi (\frac{1}{2}r_B)}{T} \times \frac{T}{2\pi r_B} = \frac{1}{2}\) So the ratio of the speeds of the skaters is \(\frac{1}{2}\).
04

Formula for the centripetal force in uniform circular motion

The centripetal force acting on an object in uniform circular motion is given by: \(F_c = m\frac{v^2}{r}\) where \(F_c\) is the centripetal force, \(m\) is the mass of the object, \(v\) is the speed, and \(r\) is the radius of the circular path.
05

Compare the magnitudes of the forces acting on the skaters

Since the skaters have equal mass, their masses can be represented as \(m\). From step 3, we know the ratio of their speeds is \(\frac{1}{2}\), so we can represent their speeds as \(v_A = \frac{1}{2}v_B\). Using the formula for centripetal force, we can find \(F_{cA}\) and \(F_{cB}\): \(F_{cA} = m\frac{v_A^2}{r_A} = m\frac{(\frac{1}{2}v_B)^2}{\frac{1}{2}r_B}\) \(F_{cB} = m\frac{v_B^2}{r_B}\) Now, we can find the ratio of their forces: \(\frac{F_{cA}}{F_{cB}} = \frac{m\frac{(\frac{1}{2}v_B)^2}{\frac{1}{2}r_B}}{m\frac{v_B^2}{r_B}}\)
06

Simplify the force ratio

To simplify the force ratio, we can cancel out common factors: \(\frac{F_{cA}}{F_{cB}} = \frac{m\frac{(\frac{1}{2}v_B)^2}{\frac{1}{2}r_B}}{m\frac{v_B^2}{r_B}} \times \frac{r_B}{\frac{1}{2}r_B} = \frac{\frac{1}{4}v_B^2}{v_B^2} = \frac{1}{4}\) So the ratio of the magnitudes of the forces acting on the skaters is \(\frac{1}{4}\).

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Key Concepts

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

Centripetal Force
Centripetal force is the force that keeps an object moving in a circular path, directed towards the center of the circle. This force does not act in the direction of motion but rather perpendicular to it.
This is why, even though the object is moving at a constant speed, continuous force is needed to maintain the circular path. For an object of mass \(m\) moving at a speed \(v\) in a circle of radius \(r\), the centripetal force \(F_c\) is given by the formula:
  • \( F_c = m\frac{v^2}{r}\))
The centripetal force increases with both the square of the speed and as the radius of the circular path decreases.
This means if you double the speed of an object, the centripetal force quadruples, and if you halve the radius, the centripetal force doubles.
Uniform Circular Motion
Uniform circular motion describes an object moving in a circle with constant speed.
Despite the constant speed, the object is accelerating because its direction is constantly changing. This type of motion is key to understanding centripetal force. In uniform circular motion, even though the speed remains constant, the velocity does not, because velocity is a vector that includes both speed and direction.
It's important to note that although the speed is constant, the object accelerates towards the center of the circle due to the centripetal force.
  • The formula for speed in uniform circular motion is: \( v = \frac{2\pi r}{T} \), where \( v \) is the speed, \( r \) is the radius, and \( T \) is the period of rotation.
This formula shows how the speed depends on the radius and the period, a relationship critical in problems involving circular motion.
Speed Ratio
When comparing the speed of objects in circular motion, the radius of their paths and the period of motion are crucial factors.
For skaters moving in circles with the same period, their speeds can be different if their circular paths have different radii. Given the formula for speed \( v = \frac{2\pi r}{T} \), if two skaters have periods \( T \) that are equal but different radii \( r_A = \frac{1}{2}r_B \), skater A will have half the radius of skater B.
Thus, the speed ratio \( \frac{v_A}{v_B} \) simplifies to \( \frac{1}{2} \), indicating skater A moves at half the speed of skater B despite both completing one circle in the same amount of time.
Radius and Period Relationship
In circular motion, the radius and the period of rotation have a direct relationship to the speed of an object.
This relationship helps determine various parameters such as speed and the forces involved. The formula \( v = \frac{2\pi r}{T} \) signifies the speed of an object depends directly on the radius of its circular path and inversely on its period.
  • As the radius increases, assuming the period stays constant, the speed increases.
  • Conversely, if the radius decreases, the speed decreases when the period remains unchanged.
For skaters with equal periods but different radii, these relationships indicate how their paths alter their speeds and the resulting centripetal forces required to keep them in motion.

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Most popular questions from this chapter

Calculate the centripetal force exerted on a vehicle of mass \(m=1500 .\) kg that is moving at a speed of \(15.0 \mathrm{~m} / \mathrm{s}\) around a curve of radius \(R=400 . \mathrm{m} .\) Which force plays the role of the centripetal force in this case?

A 20.0 -g metal cylinder is placed on a turntable, with its center \(80.0 \mathrm{~cm}\) from the turntable's center. The coefficient of static friction between the cylinder and the turntable's surface is \(\mu_{s}=0.800\). A thin, massless string of length \(80.0 \mathrm{~cm}\) connects the center of the turntable to the cylinder, and initially, the string has zero tension in it. Starting from rest, the turntable very slowly attains higher and higher angular velocities, but the turntable and the cylinder can be considered to have uniform circular motion at any instant. Calculate the tension in the string when the angular velocity of the turntable is 60.0 rpm (rotations per minute).

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A particular Ferris wheel takes riders in a vertical circle of radius \(9.0 \mathrm{~m}\) once every \(12.0 \mathrm{~s} .\) a) Calculate the speed of the riders, assuming it to be constant. b) Draw a free-body diagram for a rider at a time when she is at the bottom of the circle. Calculate the normal force exerted by the seat on the rider at that point in the ride. c) Perform the same analysis as in part (b) for a point at the top of the ride.

A carousel at a carnival has a diameter of \(6.00 \mathrm{~m}\). The ride starts from rest and accelerates at a constant angular acceleration to an angular speed of 0.600 rev/s in \(8.00 \mathrm{~s}\). a) What is the value of the angular acceleration? b) What are the centripetal and angular accelerations of a seat on the carousel that is \(2.75 \mathrm{~m}\) from the rotation axis? c) What is the total acceleration, magnitude and direction, \(8.00 \mathrm{~s}\) after the angular acceleration starts?

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