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

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, $F_c = \frac{mv^2}{r}$

Thus, on putting the values, we get

Magnitude of the centripetal force, $F_c = \frac{33\times4^2}{2.50}$

or  

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

Answer 2

Answer:

$F=211.2\ N$

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:

$F=m.\frac{v_t^2}{r}$

$F=33\times \frac{4^2}{2.5}$

$F=211.2\ N$

• 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.