Centripetal Force of merry-go-round

  • Thread starter nautica
  • Start date
  • #1
nautica
a) A 22 kg child is riding a merry-go-round that is rotating at 40 rpm. What centripetal Force must she exert to stay on if she is 1.25 m from its center???

40 rpm = 4.19 rad/s
r = 1.25 m

Ac = (4.19 rad/s)^2 * (1.25 m) = 21.95 m/s^2

F = (22 kg) * ( 21.95 m/s^2) = 482.9 N


b) What cent Force does she need to stay on one that rotates at 3 rpm if she is 8 m from its center

3 rpm = .314 rad/s

Ac = (.314 rad/s)^2 (8 m) = .789 m/s^2

Force = (22 kg) (.789 m/s^2) = 17.36 N

c) compare each force with her weight.

Force 1 is 21.95 x's her weight
Force 2 is .789 x's her weight

how does this look
thank
nautica
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
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Looks good to me.
 
  • #3
nautica
Cool, thanks

Nautica
 
  • #4
This is the wrong forum. This should be posted in K12 section not College.
 

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