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Your additional term (Bob) is a restatement of the first term. ie I = r^2 m, your differential term is your angular velocity w.

You are complaining that the equation for a simple pendulum (with a point mass) is not the same as that for a compound pendulum (with a distributed mass). Perfectly true. The art of real life is either to solve the complete equation (which is what engineers have to do when designing oscillating things like cars and bridges) or to design an experiment that eliminates the second-order terms (which is what physicists try to do when measuring things like g).https://en.wikipedia.org/wiki/Kater%27s_pendulum shows how a rigid compond pendulum can be designed to eliminate many of the unknowns.If your freely-rotating bob is on a frictionless axle, there is no torque making it rotate with respect to the vertical, so ω = 0

The question as its written by the author implies that he's modeling the pendulum by taking into account only the translational energy of the bob and is taking the rotational energy of the bob to be insignificant when compared to it and thus ignorable I.e. he's using a model where the bob is being treated as a point particle.In most situations one makes approximations, i.e. the use a model which helps the reader understand the physics at hand and to do that they simplify things. Notice in the photo how long the string is compared to the bob and that the authors mentioned only small oscillations? There's also no reason to assume that the bob rotates with the pendulum. It could have the same orientation in space as it moves. Rotational inertia would see to that. Otherwise it'd be called a compound pendulum.Regarding models in physics see your text. Also use Google to find a good treatment about the use of models in spaceflight. For example; when NASA is plotting a launch of a rocket to Mars they need to know the orbit of Mars. They assume that Mars, the Earth and the Sun are all point objects.

...that the authors mentioned only small oscillations?...

Because inertia and mass are not relative in the case of the pendulum?