# Not sure what is it about [*(]

1. Mar 11, 2004

### cristina

Not sure under what subject to name it

An airplane is flying in a horizontal circle at a speed of 480km/hr. banked for this turn; the wings of the plane are tilted to an angle of 40 degree from the horizontal. Assume that a lift force acting perpendicular to the wings holds the aircraft in the sky. What is the radius of the circle in which the plane is flying?

In the entier Newton chapter they never mention anything about radius.
Our professor always like to include one very difficult exercice in each assignment, for he doesn't beleive in full mark! Isn't sweet?
A friend of mine sais its about Motion along a curved path?!
Would you pleaaase help me [b(]

Last edited: Mar 11, 2004
2. Mar 12, 2004

### jamesrc

This problem is a little tricky, but nothing to get too worried about. With most problems, a good first step is to draw out the problem (esp. a free body diagram). Look at the plane at a snapshot in time and try to see the forces that are acting on it. The "Lift," as stated in the problem is pointing perpendicular to the wings, which themselves are at 40 degrees. So you have a component of the lift pointing up (positive y-direction) and a component of the lift pointing toward the origin (in the -x direction as I've drawn mine). What other forces do we have? Well, we've got the weight of the airplane pointing straight down.

Now let's talk about the motion of the plane. The airplane is moving in a "horizontal" circle, so it's not moving up or down. So, we triumphantly write:

$$\Sigma F_y = L\cos\theta - mg = 0$$

where L is the lift force, &theta; is 40 degrees, and m is the mass of the plane. That's one equation to work with.

Now we deal with the horizontal circular motion. Surely you've covered centripetal acceleration in class, right? If so, this equation should make a lot of sense:

$$\Sigma F_x = m\frac{v^2}R = L\sin\theta$$

There's 2 equations and 2 unknowns (L and R; the unknown mass m conveniently drops out of the equations). All that's left for you to do is the algebra it takes to eliminate L and solve for R, the radius of the circle the plane is flying in. (Added reminder: watch your units.)

3. Mar 12, 2004

Centripetal acceleration:

$$a_c = \frac{v^2}{r}$$

This centripetal acceleration is provided by the horizontal component of the lift, which you can calculate based on the given information. v is known, so the only variable left is r. Solve for r.

4. Mar 12, 2004

### cristina

Wich method to use the first or the second?

5. Mar 12, 2004

They're the same. jamesrc just posted before I did.

6. Mar 12, 2004

### cristina

Ok, may you please tell me why it doesn't work when I use your method to solve this?

A 0.20kg stone attached to a 0.8m string is rotated in the horizontal plane. The string makes an angle of 20 degree with the horizontal. Determine the speed of the stone.

7. Mar 12, 2004

I can't tell you unless you show me your work. If you follow jamesrc's method, though, you'll go right through the equation I put (it'll be your last equation). So if you tried his method and it worked and then you tried my method and it didn't, you just made a mistake somewhere the second time around.

As for your second problem, just use the equations we gave you and solve for v this time instead of r.

8. Mar 12, 2004

### cristina

I dont know how to use the characters you write with!
Am I on the write track to solve the new exercice using your equation?
because I feel there is something I'm missing. I will post it using regular character. please help me!

9. Mar 12, 2004

There are three forces acting on the stone. Gravity is acting straight downard. Centripetal acceleration is acting outward (i.e. away from the center of the circle that the stone's going in). The tension force is acting inward and 20 degrees up. All of these forces must balance.

So we have two vertical forces: gravity going downward and the vertical component of tension going upward. Write the equation for the vertical forces and calculate the force of tension from this.

Then you have two horizontal forces: the horizontal component of tension going inward and the centripetal force given by

$$F_c = m\frac{v^2}{r}$$
(notice that this is F = ma, where a is the a I gave you in my first post).

Use this equation to calculate velocity after you've determined tension from the first equation.

Also notice that this procedure is exactly the same as the procedure described by james, with "lift" replaced by "tension", except for the in the very last step you're just solving for something different.

As for how to make the pretty equations, click on one of our equations to see what we wrote to make it appear.