- #1
Nikitin
- 735
- 27
Hey! Where is the error in my reasoning?:
The acceleration of the centre of mass in a swinging physical pendulum in simple harmonic motion is given by:
##M \ddot{\vec{r_{cm}}} = \sum m_j \ddot{\vec{r_j}} = \vec{g} M##
If ##x## is the coordinate distance measured along the swinging-arc of the CM, ##\theta## is the angular displacement from the vertical line thru ##x=0## and ##R## is the distance from the pivot and to the CM, and if we assume the angular amplitude is small:
##\ddot{x_{cm}} \approx -g \theta_{cm} = -\frac{x_{cm}}{R} g##
Thus the period the CM is swinging with is identical to that of a simple pendulum.
Why is this wrong? The acceleration of the centre of mass in a physical pendulum has the same form as that of the ball in the simple pendulum, so why can't I just ignore moment of inertia, torque and all that and just set up a normal DE?
The acceleration of the centre of mass in a swinging physical pendulum in simple harmonic motion is given by:
##M \ddot{\vec{r_{cm}}} = \sum m_j \ddot{\vec{r_j}} = \vec{g} M##
If ##x## is the coordinate distance measured along the swinging-arc of the CM, ##\theta## is the angular displacement from the vertical line thru ##x=0## and ##R## is the distance from the pivot and to the CM, and if we assume the angular amplitude is small:
##\ddot{x_{cm}} \approx -g \theta_{cm} = -\frac{x_{cm}}{R} g##
Thus the period the CM is swinging with is identical to that of a simple pendulum.
Why is this wrong? The acceleration of the centre of mass in a physical pendulum has the same form as that of the ball in the simple pendulum, so why can't I just ignore moment of inertia, torque and all that and just set up a normal DE?
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