Friction and centripetal force

Ok, in summary the child slides down a slide with a 29 degree incline and her speed is at two-thirds of what it would be if the slide was frictionless. If there was no friction, her velocity would be 3/4 of what it is. Her height change is related to the length of the slide. When there is friction, her velocity is multiplied by W_f, the coefficient of kinetic friction. Finally, she has to solve for F_f, the normal force, and figure out what cancels it out.
  • #1
I feel like there's not enough information to solve this:

A child slides down a slide with a 29 degree incline, and at the bottom her speed is precisely two-thirds what it would have been if the slide had been frictionless. Calculate the coeffieicnt of kinetic friction between the slide and the child.

Also, I can't explain why, in those "rotor-rides" at the carnival where you're spun around fast enough to overcome static friction and not slide down, why you feel like you're being pressed against the wall but you're really not. Any help would be greatly appreciated.
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  • #2
Welcome to the forums! :smile:

However in order to get help you need to SHOW US what progress you have made since we don't provide solutions we simply guide you along the way

ok first of all let's just assume that the child started to slide from rest

so then if there was no friction involved
[tex] \Delta K + \Delta U = 0 [/tex]

calculate the velocity (use variables only)
the height change you used here is somehow related to the length of the slide itself
Now that you have a velocity 2/3 of this velocity is the velocity you will use in the next part

assume ther was friction

[tex] \Delta K + \Delta U = W_{f} = F_{f} d = \mu F_{N} l [/tex]

where L is the length of the slide
sub in what you have so far for the velocity (remember to square it!) you know the change in potential energy.

Draw a fre body body diagram of the child nad figure out the normal force and see waht cancels out. Your answer should have no need for the mass or the length of the slide.
  • #3
For the second problem, I'll give you a hint: You get pressed against the wall right? What kind a reference frame are you in? What kind of a reference frame is a person on the ground in? What is the difference? (Sorry for making it look all so obvious...but you still got to reason it yourself mate).

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