Work Energy Theorem Loop-The-Loop problem

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http://img530.imageshack.us/img530/5664/looptheloophg9.gif [Broken]

The two problems below are related to a cart of mass M = 500 kg going around a circular loop-the-loop of radius R = 8 m, as shown in the figures. All surfaces are frictionless. In order for the cart to negotiate the loop safely, the normal force exerted by the track on the cart at the top of the loop must be at least equal to 0.4 times the weight of the cart. You may neglect the size of the cart. (Note: This is different from the conditions needed to "just negotiate" the loop.)

a) For this part, the cart slides down a frictionless track before encountering the loop. What is the minimum height h above the top of the loop that the cart can be released from rest in order that it safely negotiate the loop?

b) For this part, we launch the cart horizontally along a surface at the same height as the bottom of the loop by releasing it from rest from a compressed spring with spring constant k = 10000 N/m. What is the minimum amount X that the spring must be compressed in order that the cart "safely" (as defined above) negotiate the loop?

c) When the car is descending vertically (ie at a height R above the ground) in the loop, what is its speed |v|?

d) At the bottom of the loop, on the flat part of the track, the cart must be stopped in a distance of d = 20 m. What retarding acceleration |a| is required?




Centripedal Acceleration: V^2/R
Work Energy Theorem



I just don't know where to start? This problem confuses the heck out of me.
 
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Answers and Replies

  • #2
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Ok, you can work out the weight of the cart using: [tex] F = ma [/tex].

Then you can work out the centripetal force required to keep it on the track: [tex] F = \frac{mv^2}{r} [/tex]

After that you will have the velocity it must travel at. Because the surfaces are frictionless, the magnitude of this velocity will be the same just before it enters the loop.

use conservation of energy to calculate h: [tex] \frac{1}{2}mv^2 = mg\Delta h [/tex]
 
  • #3
Doc Al
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I just don't know where to start? This problem confuses the heck out of me.
Start by identifying the forces acting on the cart when its at the top of the loop. Then apply Newton's 2nd law and your knowledge of centripetal acceleration to solve for the speed at the top.
 
  • #4
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The two forces are Fgravity, Fnormal.
Fgravity = 500kg * 9.81m/s^2 = 4905N
Fnormal = 0.4 * 500kg = 200N

4905N + 200N = (500kg * V^2)/8m
Speed at the top: V= 9.04m/s

Then placing that in the conservation of energy equation:
(1/2)(500kg)(9.04m/s)^2 = (500kg)(9.81m/s^2)(Change in height)
Change in height = 4.17m, which is not right.
 
  • #5
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You wrote: Fnormal = 0.4 * 500kg = 200N
Can you see what is wrong with this?
HINT: You will see if you check units.
 
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  • #6
Doc Al
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The two forces are Fgravity, Fnormal.
Good.
Fgravity = 500kg * 9.81m/s^2 = 4905N
Good.
Fnormal = 0.4 * 500kg = 200N
Oops! 0.4 times the weight, not just the mass.
 
  • #7
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It says in the problem that the normal force of the cart on the track has to be at least 0.4 times the weight of the cart. That is where I'm getting the normal force from. I'm guessing this is not correct.
Is the centripetal force: (m*v^2)/r??? I thought that was the centripetal acceleration?
 
  • #8
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[tex] \frac{mv^2}{r} [/tex] is centripetal force

[tex] \frac{v^2}{r} [/tex] is centripetal acceleration
 
  • #9
Doc Al
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It says in the problem that the normal force of the cart on the track has to be at least 0.4 times the weight of the cart. That is where I'm getting the normal force from. I'm guessing this is not correct.
You just made a mistake. You calculated the weight of the cart. What's 0.4 times that?
Is the centripetal force: (m*v^2)/r???
Sure.
I thought that was the centripetal acceleration?
The centripetal acceleration is just v^2/r. You multiply it by m to get the centripetal force. (That's just Newton's 2nd law.)

You're doing fine. Just correct your error.
 
  • #10
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Alright yeah that was my problem, dumb mistake :)

On part b now i used all the information that i have to find this distance:

KE(spring)+PE(spring) = KE(top) + PE(top)

0+(1/2)(10000N/m)x^2 = (1/2)(500kg)(5.598m/s)^2 + (500kg)(9.81m/s^2)(16m)

x = 4.16m, which isn't correct
I feel i set it up right, but i must be missing something.
 
  • #11
Doc Al
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KE(spring)+PE(spring) = KE(top) + PE(top)

0+(1/2)(10000N/m)x^2 = (1/2)(500kg)(5.598m/s)^2 + (500kg)(9.81m/s^2)(16m)
The set up is perfect, but where did you get that speed?
 
  • #12
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I think i'm having problems today haha, i used the height from part a as the speed. The actual speed is 10.48m/s. I continued on and figured out the rest of the parts. I was getting messed up with all the different equations. I think that's what this problem was trying to do. Thanks for all your help Doc Al and Rake-MC!
 
  • #13
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I think i'm having problems today haha, i used the height from part a as the speed. The actual speed is 10.48m/s. I continued on and figured out the rest of the parts. I was getting messed up with all the different equations. I think that's what this problem was trying to do. Thanks for all your help Doc Al and Rake-MC!

Hi I have a very similar question. In pt. b it says you just you the speed for part a but Awwnutz says a different speed is needed. How do I calculate this speed?
 
  • #14
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The forces have to sum up to equal the centripetal force. So the force of gravity + the normal force have to equal the centripetal force. You know everything else in the equation, but the velocity at the top of the loop. So solve for that and you'll get your speed.
 
  • #15
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The forces have to sum up to equal the centripetal force. So the force of gravity + the normal force have to equal the centripetal force. You know everything else in the equation, but the velocity at the top of the loop. So solve for that and you'll get your speed.

So I'm assuming that the Normal force at that point is zero right?
 
  • #16
Doc Al
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So I'm assuming that the Normal force at that point is zero right?
Not in this problem. The normal force is given.
 

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