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Chapter 9: Problem 48

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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