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(a) A 19.0 kg child is riding a playground merry-go-round that is rotating at 45.0 rev/min. What centripetal force (in N) must she exert to stay on if she is 3.00 m from its center? N
(b) What centripetal force (in N) does she need to stay on an amusement park merry-go-round that rotates at 3.00 rev/min if she is 5.00 m from its center? N
(c) Compare each force with her weight. force from (a) weight = ________? force from (b) weight = _________?

Respuesta :

Answer:

(a) 1264.49 N

(B) 9.36 N

(C) For part a [tex]\frac{F}{mg}=\frac{1264.49}{19\times 9.8}=6.791[/tex]

For part b [tex]\frac{F}{mg}=\frac{9.36}{19\times 9.8}=0.05[/tex]

Explanation:

We have given mass of the child m = 19 kg

Angular speed [tex]\omega =45rpm=\frac{2\times \pi \times 45}{60}=4.71rad/sec[/tex]

Distance from the center r = 3 m

So centripetal force [tex]F=m\omega ^2r=19\times 4.71^2\times 3=1264.49N[/tex]

(b) Angular speed [tex]\omega =3rpm=\frac{2\times \pi \times 3

{60}=0.314rad/sec[/tex]

Distance from the center r = 5 m

So centripetal force [tex]F=m\omega ^2r=19\times 0.314^2\times 5=9.36N[/tex]

(C) Now comparison with weight

For part a [tex]\frac{F}{mg}=\frac{1264.49}{19\times 9.8}=6.791[/tex]

For part b [tex]\frac{F}{mg}=\frac{9.36}{19\times 9.8}=0.05[/tex]

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