Question

You are standing a distance of 17.0 meters from the center of a merry-go-round. The merry-go-round takes 9.50 seconds to go completely around once and you have a mass of 55.0 kg.

a. What will be your speed as you move around the center of the merry-go-round?
b. What will be your centripetal acceleration as you move around the center of the merry-go-round?
c. What will be the magnitude of the centripetal force necessary to keep your body moving around the center of this merry-go-round at the calculated speed?
d. How much frictional force will be applied to you by the surface of the merry-go-round?
e. what is the minimum coefficient of friction between your shoes and the surface of the merry-go-round?

Answer 1

(a)11.24 m/s

(b)7.44 m/s

(c)409 N

(d)
`539.55\mu`

(e) 0

Step-by-step explanation:

The period for 1 circle
`2\pi` of the merry go around is 9.5s. It means the angular speed is:

`\omega = \theta / t = 2\pi / 9.5 \approx 0.661 rad/s`

(a)The speed is

`v = \omega * R = 0.661 * 17 = 11.24 m/s`

(b) Centripetal acceleration:

`a = (v^2)/(R) = (11.24^2)/(17) = 7.44 m/s^2`

(c) Magnitude of the force that keeps you go around at this acceleration

`F = ma = 55 * 7.44 = 409 N`

(d) let the coefficient of friction by
`\mu`. The frictional force shall be this coefficient multiplied by normal force reverting gravity of the man

`F_f = mg\mu = 55*9.81\mu = 539.55\mu`