# Classical mechanics, simple pendulum

1. Jul 17, 2012

### xzibition8612

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

See attachment "question"

2. Relevant equations

3. The attempt at a solution

See attachment "work"

I did the work for (1) and (2). I end up with two equations: the first is the tension T, the second is the angular acceleration. I'm not so sure if I made any mistakes in solving the equations of motion, but I'm not really comfortable with these two equations and feel like something went wrong. I just can't see it. Remember the pendulum is a point mass, hence for the point mass the moment of inertia is ml^2. Now for question (3), how do I find the angular velocity? None of my two equations contain this quantity, hence I feel like something went wrong. For (4), I do indeed have T= 9.83/sin(theta). Now how do I plot this in relation to time? Because from this equation I can only plot the tension T in relation to the change in angle, not time. I'm not going to worry about (5) right now, gotta get (1)-(4) right first.

Thanks for the help.

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2. Jul 17, 2012

### TSny

Why did you set the x and y components of acceleration equal to zero? When you set up equations of motion, they should be valid for any instant of time of the motion. In fact an "equation of motion" is meant to be an expression for the components of acceleration in terms of velocity and position components.

Have you worked with polar coordinates? If so, I would recommend using them instead of cartesian coordinates.

Are you meant to use a computer to assist with the solution of the equations of motion? (There is no way to write the solution for $\theta$ as a function of time in terms of elementary functions.)

Last edited: Jul 17, 2012
3. Jul 17, 2012

### xzibition8612

I set x'' and y'' to zero because the particle itself does not inherently have any acceleration, its acceleration is due to gravity which is an external force. Might be wrong about this, but that's my way of thinking. I'm meant to use a computer to graph, not to solve the equations...I'm pretty sure. You mean there's no way to get angular velocity by calculation?

4. Jul 17, 2012

### TSny

Even at the instant of release, $\ddot{y}$ $\neq$0 (Think freefall). So, your expression for the tension is not correct.

Your equation for $\ddot{θ}$ looks good except for the sign.

5. Jul 17, 2012

### TSny

Right. You can get a simple expression for $\dot{θ}$ in terms of θ, but there are no simple expressions for $\dot{θ}$ and θ as functions of time. They can be expressed in terms of certain "functions of higher mathematics".