Question

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

Answer: The ratio of the speeds of the skaters is 1/2. b) What is the ratio of the magnitudes of the forces acting on each skater? Answer: The ratio of the magnitudes of the forces acting on each skater is 1/2.

Step-by-step solution

Write what we know about skaters A and B

Skater A has a radius \(r_A\), and skater B has radius \(r_B=2r_A\). They have the same period \(T\).

Use the relationship between speed, radius, and period for circular motion

For a circular motion with radius \(r\) and period \(T\), the speed (usually denoted as \(v\)) is related to the radius and period by the formula: \(v = \frac{2 \pi r}{T}\). Let's find the speeds of both skaters A and B.

Calculate speed \(v_A\) for skater A

Speed of skater A: \(v_A = \frac{2 \pi r_A}{T}\)

Calculate speed \(v_B\) for skater B

Speed of skater B: \(v_B = \frac{2 \pi r_B}{T} = \frac{2 \pi (2r_A)}{T}\)

Determine the ratio of speeds \(\frac{v_A}{v_B}\)

Divide equation of \(v_A\) by equation of \(v_B\): $$\frac{v_A}{v_B} = \frac{\frac{2 \pi r_A}{T}}{\frac{2 \pi (2r_A)}{T}} = \boxed{\frac{1}{2}}$$

Calculate centripetal force for each skater

The centripetal force is given by the formula: \(F_c = m \frac{v^2}{r}\), where \(m\) is the mass of the skater. For skater A, the centripetal force is \(F_{cA} = m \frac{v_A^2}{r_A}\). For skater B, the centripetal force is \(F_{cB} = m \frac{v_B^2}{r_B}\).

Determine the ratio of forces \(\frac{F_{cA}}{F_{cB}}\)

Divide equation of \(F_{cA}\) by equation of \(F_{cB}\): $$\frac{F_{cA}}{F_{cB}} = \frac{m \frac{v_A^2}{r_A}}{m \frac{v_B^2}{r_B}} = \frac{v_A^2r_B}{v_B^2r_A}$$ We already know that \(\frac{v_A}{v_B} = \frac{1}{2}\). Therefore, \(v_A = \frac{1}{2}v_B\). Squaring, we get \((v_A)^2 = \frac{1}{4}(v_B)^2\). Also, \(r_B=2r_A\). Now, replace these values in the ratio of forces: $$\frac{F_{cA}}{F_{cB}} = \frac{\frac{1}{4}(v_B)^2(2r_A)}{(v_B)^2r_A} = \boxed{\frac{1}{2}}$$ So, the answers to the questions are: a) The ratio of the speeds of the skaters is \(\frac{1}{2}\). b) The ratio of the magnitudes of the forces acting on each skater is \(\frac{1}{2}\).