# Homework Help: Circular motion A-Level

Tags:
1. Jan 29, 2017

### QuantumNite

1. The problem statement, all variables and given/known data
An aircraft flies with its wings tilted, in order to fly in a horizontal circle of radius r. The aircraft has mass 4.00 x 104 Kg and has a constant speed of 250ms-1 . with the aircraft flying in this way, two forces are acting on the plane, P [lift] and W [weight which is given as 3.92*105 N] . P is acting at 35 degrees to the vertical. calculate P

2. Relevant equations
centripetal force=mass * velocity2 / radius
centripetal force= mass * angular speed2 * radius
angular speed = theta / time
angular speed = velocity/ radius
angular speed = 2pi * frequency
centripetal acceleration = velocity2 / radius
centripetal acceleration = angular speed2 * radius
Force = mass * acceleration
3. The attempt at a solution
I cant do it. I swear we need more information. can someone enlighten me.

Last edited: Jan 29, 2017
2. Jan 29, 2017

### PeroK

What information do you think you need?

By the way, I think the weight should be $3.92*10^5N$. Personally I would do this question algebraically and just plug the numbers in at the end. That's also a good way to work out what quantities you do need.

3. Jan 29, 2017

### QuantumNite

yes I think that the weight is correct. I couldn't read the sheet correctly.
the info that I think we need : time or radius
I am not sure whether we need to plug it into an equation or just use trig??

4. Jan 29, 2017

### PeroK

You need to think about the forces involved. There's a key equation that will open up the whole problem. And, by equation I mean two quantities that must be equal.

Hint: why doesn't the plan fall out of the sky?

5. Jan 29, 2017

### QuantumNite

Correct me if I am wrong,
but can I say that the vertical component of P is PCos(35), and it doesn't fall out of the sky because Weight is equal to PCos(35)
Therefore, PCos(35) = 3.92 * 105
So, If we rearrange that equation, P = 3.92*105 / Cos(35) == 321107.6 [1dp]
??

6. Jan 29, 2017

### PeroK

Yes, that's it exactly. Although, I prefer:

$P = \frac{mg}{\cos \theta}$

or:

$P = \frac{W}{\cos \theta}$

7. Jan 29, 2017

### QuantumNite

Thanks,
Could you please explain to me how you came to those equations?

8. Jan 29, 2017

### PeroK

I just copied what you had done! I just used letters for the quantities involved instead of plugging in the numbers. There are a lot of advantages to this. Not least, once you plug in the numbers, you lose sight of what is related to what and if you make a mistake, well one number looks very like another, and it is hard to spot an error.

That said, I thought the question was to calculate $r$. I see now that all you need is $P$, which is a lot simpler.

Is part 2 of the question to calculate $r$?

9. Jan 29, 2017

### QuantumNite

Yes it is.

but I am pretty sure how to do it

F=mv2 / r ,, rearrange for r

10. Jan 29, 2017

### PeroK

If you post your answer, I can show you the algebraic approach, which might be interesting and/or enlightening!