What is the acceleration of a plane based on a pendulum's angle?

[tjackson3]

Homework Statement

Granted, I may be thinking too much into this.

You are sitting in a jet airplane as it accelerates at a constant rate down the
runway. Being a good physics student you hold the string of a small pendulum of length
l = 0.75m and mass m = 1 kg. You then measure the angle between the string and a
vertical line and $\theta = 37^{\circ}$. Assume $\sin\ 37 = 3/5; \cos\37 = 4/5; \tan\ 37 = 3/4$, and take $g = 10 m/s^2$

a.) What is the magnitude of the acceleration of the plane?
b.) What is the tension on the string?

Homework Equations

Nonrelativistic coordinate transformation: $x' = x - vt$

The Attempt at a Solution

There are a couple of ways I can see doing this, but I think both show that I don't completely understand what's going on. My first thought was that the net acceleration is zero, meaning that $a = g\cos\theta = 8 m/s^2$. This would make the tension $\sqrt{m^2g^2 + mg^2\cos^2\theta} = \sqrt{100 + 100.5625} = \sqrt{200.5625}$ but I don't think that's correct.

Alternatively, it could involve circular motion, although I have no idea what to plug in for the velocity in the centripetal force equation.

Thanks for the help!

 

[ehild]

I suggest you to start every problem-solving with a figure. Here you have to show the forces acting on the moving object, the bob of the pendulum. When the pendulum is stationary with respect to you, it moves together with the aeroplane, with a constant horizontal acceleration. As the bob is connected to the string only, the horizontal component of the tension T provides the force needed to this horizontal acceleration. The other force is gravity, it cancels with the vertical component of T.

ehild

 

[tjackson3]

So is it as simple as $a = g\tan 37 = 7.5$? Since the diagram looks like

http://img29.imageshack.us/img29/3688/penddia.png

 

[ehild]

Yes, it is so simple...

ehild

 

[paranormal]

It is important to note that in this scenario, the pendulum is not in uniform circular motion. Therefore, we cannot use the centripetal force equation to find the tension on the string.

Instead, we can use the nonrelativistic coordinate transformation to find the acceleration of the plane in the frame of reference of the pendulum. Since the plane is accelerating at a constant rate, the acceleration can be represented as a = \frac{\Delta v}{\Delta t} = \frac{v}{t} where v is the final velocity and t is the time it takes for the plane to reach that velocity.

Using the given values, we can calculate the final velocity of the plane as v = at = (8 m/s^2)(3 s) = 24 m/s. Then, using the coordinate transformation x' = x - vt, we can find the acceleration of the pendulum in the frame of reference of the plane as a' = -vt = -(24 m/s)(3 s) = -72 m/s^2.

However, since we are only concerned with the magnitude of the acceleration, we can take the absolute value of this value to get a' = 72 m/s^2. This means that the pendulum is experiencing an acceleration of 72 m/s^2 in the opposite direction of the plane's acceleration.

To find the tension on the string, we can use Newton's second law, F = ma, where F is the net force on the pendulum, m is the mass, and a is the acceleration. Since the only force acting on the pendulum in the horizontal direction is the tension, we can set F = T and solve for T. Plugging in the values, we get T = ma = (1 kg)(72 m/s^2) = 72 N.

Therefore, the magnitude of the acceleration of the plane is 8 m/s^2 and the tension on the string is 72 N. It is important to note that these values are only valid in the frame of reference of the pendulum, which is accelerating with the plane. In the frame of reference of the ground, the pendulum would appear to be at rest and the tension on the string would be equal to the weight of the pendulum, mg = 10 N.