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Loop-the-Loop, net force

  • Thread starter KCEWMG
  • Start date
  • #1
11
0

Homework Statement


A small ball of mass m = 0.150 kg is sliding along a frictionless loop-the-loop. The loop-the-loop is standing on a table such that the plane of the loop is vertical. The loop has a radius of r = 0.200 m. What is the magnitude of the net force acting on the ball when it is on the right side and half-way up the loop, and moving upward with a speed of 2.00 m/s?


Homework Equations


F=ma
Centripetal Acceleration= v^2 / r


The Attempt at a Solution


Alright, here's what I've tried:
I drew a free body diagram. Facing down, I put 9.8*.150=1.47 thinking about the gravitational force. I then used the Centripetal Acceleration equation,t (2.00^2)/.2, which resulted in 20. I then multiplied this by the mass and got 3. 3-1.47 is equal to 1.53 N, which is not the correct answer. The correct answer is 3.34 N.
Where am I going wrong?
 

Answers and Replies

  • #2
92
0
Hey KCEWMG.
I think I have the solution. You're trying to add/subtract the forces acting on the ball to get the overall/resulting force acting on it, however the forces aren't collinear, therefore you can't do that. Draw you're free body diagram, with mg in the negative y direction, and the force due to the centripetal acceleration in another direction (think about it, it's always accelerating towards the centre of the of the loop-to-loop).

Hope this helps.
 
  • #3
11
0
Ahh, gotcha.
CENTRIPETAL Acceleration. Thanks!
 

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