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?

Respuesta :

Answer:

211.2 N

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, [tex]F_c = \frac{mv^2}{r}[/tex]

Thus, on putting the values, we get

Magnitude of the centripetal force, [tex]F_c = \frac{33\times4^2}{2.50}[/tex]

or  

Magnitude of the centripetal force, [tex]F_c[/tex] = 211.2 N

Answer:

[tex]F=211.2\ N[/tex]

Explanation:

Given:

  • mass of the child, [tex]m=33\ kg[/tex]
  • radial distance of the child, [tex]r=2.5\ m[/tex]
  • tangential speed of the child, [tex]v_t=4\ m.s^{-1}[/tex]

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

[tex]F=m.\frac{v_t^2}{r}[/tex]

[tex]F=33\times \frac{4^2}{2.5}[/tex]

[tex]F=211.2\ N[/tex]

  • 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.