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

Answer:

(a)11.24 m/s

(b)7.44 m/s

(c)409 N

(d)$539.55\mu$

(e) 0

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:

$a = \frac{v^2}{R} = \frac{11.24^2}{17} = 7.44 m/s^2$

(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

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