Answer to Question #260195 in Physics for grey tyrant

Question

You and your friend ride the merry-go-round at an amusement park. You ride at the outermost horse, with a radius of 3m, while your friend rides on a seat closer to the center of the ride, with a radius of 2m. The ride lasts for 6 revolutions, lasting a total of 3 minutes.

a. What is the period of the merry-go-round ride, in seconds?

b. Do you and your friend experience the same centripetal acceleration? Prove by calculating the centripetal acceleration for both you and your friend.

Answer 1

(a) The period of the merry-go-round ride can be calculated as follows:

$$T=\dfrac{Time}{Number\ of\ revolutions}=\dfrac{3\ min\times\dfrac{60\ s}{1\ min}}{6\ rev}=30\ s.$$

(b) The centripetal acceleration is not the same for the persons that ride on a seats located at different radii from the center of the ride. The person that rides on a seat located farther from the center of the ride experiences larger centripetal acceleration. Let's prove that by calculations:

$$a_{c1}=\dfrac{v^2}{r}=\dfrac{4\pi^2r_1}{T^2}=\dfrac{4\pi^2\times2\ m}{(30\ s)^2}=0.088\ \dfrac{m}{s^2},$$ $$a_{c2}=\dfrac{v^2}{r_2}=\dfrac{4\pi^2r_2}{T^2}=\dfrac{4\pi^2\times3\ m}{(30\ s)^2}=0.132\ \dfrac{m}{s^2},$$ $$a_{c2}>a_{c1}.$$

As we can see from the calculations, the person that rides on a seat located farther from the center of the ride experiences larger centripetal acceleration.