# Mechanics. Circular motion and rotation

1. May 3, 2015

### moenste

1. The problem statement, all variables and given/known data
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?

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

3. 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.

2. May 3, 2015

### billy_joule

You need to include units. Attempt two is close, you ignored gravity. You could have got there faster using your last relevant equation.

3. May 3, 2015

### moenste

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?

4. May 3, 2015

### billy_joule

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.

5. May 3, 2015

### moenste

Not sure where to post a second question, I think I'll better ask it here.

1. The problem statement, all variables and given/known data

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.

2. Relevant 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

3. The attempt at a solution
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?

6. May 3, 2015

### haruspex

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.]

7. May 4, 2015

### moenste

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.

Last edited: May 4, 2015
8. May 4, 2015

### BvU

This give you an idea ?

9. May 4, 2015

### moenste

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.

10. May 4, 2015

### BvU

Try finding how much ?? is ...

11. May 4, 2015

### moenste

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

?

12. May 4, 2015

### BvU

Are you telling me you have difficulty calculating the hypothenuse of a triangle with short sides 6 and 3 ?

13. May 4, 2015

### moenste

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!