Mechanics. Circular motion and rotation

AI Thread Summary
An airplane loops a vertical circle with a radius of 200 m and a speed of 40 m/s, leading to a tension of 160 N in the pilot's strap at the top of the loop. The calculations involved using equations for circular motion, where gravity was a crucial factor. A second problem discussed involved a particle in a hemispherical bowl, where the total reaction force was determined to be 6.7 N, incorporating both horizontal and vertical components. The confusion arose from the interpretation of forces and the correct application of trigonometric relationships. Ultimately, basic principles of geometry clarified the calculations needed to arrive at the correct answer.
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Homework Statement


An aeroplane loops the loop in a vertical circle of radius 200 m, with a speed of 40 m s^-1 at the top of the loop. The pilot has a mass of 80 kg. What is the tention in the strap holding him into his seat when he is at the top of the loop?

Answer: 160 N.

Homework Equations


A-Level Physics, chapter 6 formulas:
w = Angle / t
T = 2Pi / w
v = wr
a = w^2 * r
a = v^2 / r

The Attempt at a Solution


r = 200 m
v / u ? = 40 m s^-1
m = 80 kg

Attempt 1. 1/2 mv^2 = mgr -> 64 000 = 160 000. Wrong.
Attempt 2. F = ma -> a = w^2 * r -> v = wr -> 40 = w*200 -> w = 0.2 rad s -> a = 0.2^2 * 200 = 8 -> F = 80*8 = 640 N. Wrong.

Thank you in advance for any suggestions.
 
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You need to include units. Attempt two is close, you ignored gravity. You could have got there faster using your last relevant equation.
 
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billy_joule said:
You need to include units. Attempt two is close, you ignored gravity. You could have got there faster using your last relevant equation.
You are right indeed about a = v^2 / r, thank you.

So F = ma - mg = 640 - 80*10 (for simplicity reasons) = -160 N ?
or F = mg - ma = 800 - 640 = 160 N. Does fit the book answer but why mg first? Maybe a silly question, but still.

And by units you mean F = ma = 80 kg * 8 m s^1 instead of 80*8?
 
Units: yes, makes it easier to understand your work, and many markers will subtract marks for missing units.

The question asks for tension, which is always positive.
 
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Not sure where to post a second question, I think I'll better ask it here.

1. Homework Statement

A particle of mass 0.30 kg moves with an angular velocity of 10 rad s^1 in a horizontal circle 20 cm inside a smooth hemispherical bowl. FInd the reaction of the bowl on the particle and the radius of the bowl.

Answer: 6.7 N, 22 cm

Homework Equations


R sin angle = mv^2 / r
R cos angle = mg
tan angle = v^2 / gr
F sin angle = mv^2 / r
F cos angle = mg

The Attempt at a Solution


https://www.physicsforums.com/threads/circular-motion-particle-in-a-bowl.252962/
I am a bit lost because in this link with a wrong (according to the book the answer is 6.7 N) 6 N answer Thanuka got 22.3 cm, a relatively correct answer... And how to get the 6.7 N answer as in the book?
 
moenste said:
this link with a wrong (according to the book the answer is 6.7 N) 6 N answer
The 6N at that link is for the centripetal force, which is only the horizontal component of the reaction. The 6.7 N is the entire reaction.

[It is generally preferred that you start a new thread for each problem.]
 
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haruspex said:
The 6N at that link is for the centripetal force, which is only the horizontal component of the reaction. The 6.7 N is the entire reaction.

[It is generally preferred that you start a new thread for each problem.]
Thank you. But in that case horizontal F = m(w^2)*r = 0.3 kg * 10^2 rads^-1 * 0.2 m = 6 N. And vertical F = mg = 0.3 kg * 10 (simplified g) = 3 N. In that case 6 N + 3 N = 9 N and not 6.7 N. Could you elaborate a bit on how to get 6.7 N? The bowl is smooth so I consider the only vertical force is gravity (F=mg).

UPD: and for the second part I don't quite understand why it is
Rcos theta= 6N as centripetal force is 6N
Rsin theta = 3N as mass of the particle is 0.3kg

and not R sin = 6 N for hor and R cos = 3 N for ver. Everywhere in the book the horizontal is sin and vertical is cos. Plus the angle which is created by the normal reaction also shows the same.
 
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This give you an idea ?

Add_vecs.jpg

Did you make your own drawing already (to see what's this sine and cosine business) ?
 
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BvU said:
This give you an idea ?

Did you make your own drawing already (to see what's this sine and cosine business) ?
I actually made a same drawing though instead of 3 N and 6 N I had R cos angle and R sin angle. Now I get the second part: cos angle = 6 N / R -> R cos angle = 6 N and sin angle = 3 N / R -> R sin angle = 3 N.

But I still don't quite understand how to get the 6.7 N answer from the book for the first part.
 
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Try finding how much ?? is ...
 
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BvU said:
Try finding how much ?? is ...
R = 3 / sin angle
R = 6 / cos angle

3/sin angle = 6/cos angle
3 cos angle = 6 sin angle
cos angle = 2 sin angle

?
 
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Are you telling me you have difficulty calculating the hypothenuse of a triangle with short sides 6 and 3 ?
 
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  • #13
BvU said:
Are you telling me you have difficulty calculating the hypothenuse of a triangle with short sides 6 and 3 ?
3^2+6^2 = 45, root 45 = 6.7 N... I really was looking for something complicated and totally forgot the basics. Thanks a lot!
 
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