Circular motion-what is the radius of the loop de loop in meters

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Snoopy's warplane is performing a loop de loop at a speed of 180 km/hr, which converts to 50 m/s. While at the bottom of the loop, he experiences a weight four times his normal weight, indicating a net upward force contributing to centripetal acceleration. The net force acting on him is calculated as 3mg, leading to an acceleration of 29.4 m/s². Using the centripetal acceleration formula, the radius of the loop is determined to be approximately 85 meters. The discussion emphasizes the importance of identifying forces and applying Newton's laws to solve for the radius in circular motion problems.
  • #31
vela said:
There are only two forces on Snoopy: the gravitational force and normal force. The net force on Snoopy results in his centripetal acceleration.

Yes that is correct.
But I am looking at the force that had put the body in circular motion.
What can supply this force?
As OP has stated in one of his equations, Fc.
What are the sources that supply Fc

Surely one of the candidates is the spring of the scale.
 
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  • #32
if I got an answer R=255m... does this seem plausible?
 
  • #33
Show us your calculations.
 
  • #34
v=180km/h=50m/s

ac=v2/R
(9.8m/s2)=(50m/s)2/R

solved for R=255.10m... Probably wrong
 
  • #35
Close, but the value you used for ac isn't correct. You need to find the acceleration using F=ma. Find the net force F on Snoopy and divide by his mass to find a. That will be the value of a you need in your calculation.
 
  • #36
but I don't have his mass... or is his mass 4*g=4*(9.8m/s^2)?
 
  • #37
[Snoopy] is sitting on a set of bathroom scales that him he weighs four times what he normally.
Say he has mass m. It'll turn out to cancel out in the end, as is often the case, so we don't actually need to know its value.

All these F=ma problems start with identifying what forces act on the body. You have to get this part right, otherwise the rest won't work out. That's why it's almost always a good idea to draw a free-body diagram with the forces. In this problem, there are two forces acting on Snoopy, his weight and the normal force. So start by answering these questions.

In terms of m and g, what is his weight w? In what direction does the force of gravity pull on him?

In terms of m and g, what is the magnitude of the force N exerted on him by the scale? In what direction does this force act on him?

Now add the two forces together, remembering they are vectors so that they're directions matter. What is the net force on Snoopy?
 
  • #38
You keep saying in terms on m and g, but I thought m canceled out? Sorry its hard for me to picture it without seeing it in front of me
 
  • #39
It cancels out in the end when you solve for the acceleration, but it's present in the expressions for weight and the normal force.

What I want you to do is write down the equation that will let you solve for the acceleration, and you will see that when you solve for it, the mass drops out in the end. It's often the case you won't know what a particular quantity, like Snoopy's mass, is equal to, but you just give it a name and use it in the equations and see where it leads.
 
  • #40
is the a=g/4?
 
  • #41
No. Answer these questions and post your answers here:
In terms of m and g, what is his weight w? In what direction does the force of gravity pull on him?

In terms of m and g, what is the magnitude of the force N exerted on him by the scale? In what direction does this force act on him?

Now add the two forces together, remembering they are vectors so that they're directions matter. What is the net force on Snoopy?
 
  • #42
There's Fg pulling downwards and 4FN pushing up?
 
  • #43
his weight would be Fg=m*g=9.8m/s2 pulling downwards?

FN=4*m*a? pulling upwards?
 
  • #44
the m is messing me up and i don't know why
 
  • #45
dani123 said:
There's Fg pulling downwards and 4FN pushing up?
Those are the correct directions. The 4FN isn't quite right. See below.
dani123 said:
his weight would be Fg=m*g=9.8m/s2 pulling downwards?
Good. Fg=mg. You can't get rid of the m just yet.
FN=4*m*a? pulling upwards?
No, this isn't right. The problem says the scale reads four times what he normally weighs. He normally weighs Fg, which we know is equal to mg, so FN is four times that, that is, FN=4mg.

Now sum the two forces. What's the net force on Snoopy?
 
  • #46
net force on Snoopy would be 5mg?
 
  • #47
No, you have to remember forces are vectors. FN points upward while Fg points downward, so they partially cancel.
 
  • #48
Oh ops! I knew that... so its 3mg right?
 
  • #49
Right! So you have Fnet=3mg, and Newton's 2nd law tells you that equals ma.
 
  • #50
So now I solve for a=3g=29.4m/s2

and then plug that into my acceleration equation from earlier and solve for the radius which should give me R=85m ... Does this seem reasonable?
 
  • #51
Yup! That's right!
 
  • #52
thank you so much for your help and patience! along with everyone else who helped, it is greatly appreciated!
 

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