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Urgent - Loop de Loop problem, Normal Forces

  • #1

Homework Statement



In Fig. 8-28, a small block of mass m = 3.0 kg can slide along the frictionless loop-the-loop. The loop has radius R = 0.80 m. The block is released from rest at point P, at height h = 2.9R above the bottom of the loop.

I attatched a picture

(a) Find the speed of the block when it reaches point Q.

(b) Find the normal force on the block at point Q.

(c) Find the normal force on the block when it is at the very bottom.

(d) Find the normal force on the block when it is at the top of the little loop (i.e., at a height 2R above the ground)




Homework Equations



KEi + PEi = KEf +PEf

KE=1/2mv^2

v = w (r)

The Attempt at a Solution



So for the velocity isnt it just converting the PE at the top for the KE at the bottom?

For all the normal forces i am completely stumped
 

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Answers and Replies

  • #2
So for a) v = 3(.98) 1.9(.8) = 1/2(3)v^2
 
  • #3
tiny-tim
Science Advisor
Homework Helper
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hi gcharles_42! :smile:
So for the velocity isnt it just converting the PE at the top for the KE at the bottom?
yes

but i don't understand the LHS of your equation :redface:
So for a) v = 3(.98) 1.9(.8) = 1/2(3)v^2
for the normal forces, you need to use Ftotal = ma in the normal direction

(you'll need the centripetal acceleration formula)
 

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