Question

A 33.0 kg child is riding a playground merry-go-round. If the child is 2.50 m from the center of the merry-go-round and has a constant tangential speed of 4.00 m/s. What is the magnitude of the centripetal force that is necessary to keep her on the merry-go-round at this radius?

Answer 1

`F=211.2\ N`

Step-by-step explanation:

Given:

• mass of the child,
  `m=33\ kg`

• radial distance of the child,
  `r=2.5\ m`

• tangential speed of the child,
  `v_t=4\ m.s^(-1)`

Now from the given data we find the centrifugal force acting on the child mass:

`F=m.(v_t^2)/(r)`

`F=33* (4^2)/(2.5)`

`F=211.2\ N`

• So for the child to be in state of equilibrium with respect to the merry-go-round an equal amount of centripetal force must be acting on the child to keep it stationary with respect to the merry-go-round.

• Do note that a centrifugal force always acts away from the center of rotation and a centrifugal force always acts towards the center of rotation.

Answer 2

211.2 N

Step-by-step explanation:

Data provided in the question:

Mass of child, m = 33.0 kg

Distance from the center, r = 2.50 m

Tangential speed, v = 4.00 m/s

Now,

Magnitude of the centripetal force,
`F_c = (mv^2)/(r)`

Thus, on putting the values, we get

Magnitude of the centripetal force,
`F_c = (33*4^2)/(2.50)`

or

Magnitude of the centripetal force,
`F_c` = 211.2 N