Simple Energy Equation for Pendulum

[joe_23]

Homework Statement

Solve for the equation of motion of a simple pendulum using energy methods assuming it is a massless rod with a point mass at the end

Homework Equations

Translational Kinetic Energy T = 1/2mv^2

m is mass and v is linear velocity

Rotational Kinetic Energy T = 1/2Iw^2

I = mr^2

I is inertia, and w is angular velocity, r is mass radius

The Attempt at a Solution

I've read many times that the total kinetic energy of a mechanical system is a combination of the translational energy and the rotational energy. If my understanding is correct, then taking your mouse cursor and moving it in a circle would be only translational motion since the cursor's orientation is always the same relative to an inertial RF (it is upright at all times). For a pendulum, the mass at the end would rotate with the rod (its not upright at all times- instead it tilts from one side to the other), and therefore has rotational motion in addition to the linear motion of the path of its CM . Is there something wrong in my logic? Using Newtons or eulers method gives you the same result as omitting the rotational kinetic energy term (or linear since theyre equal) so I would like to know why my logic is not correct.

 

[joe_23]

This question comes from pretty much all physics textbooks; is the answer that they model the mass at the end as being very small so that I is zero and can be ignored?

 

[Redbelly98]

Welcome to Physics Forums!

This question comes from pretty much all physics textbooks; is the answer that they model the mass at the end as being very small so that I is zero and can be ignored?

Yes. Point masses have zero moment of inertia, so the kinetic energy is completely due to translational motion.

Don't forget about the potential energy.