Moments & Pendulum: Different Lengths, Same Oscillation?

[sgstudent]

if i have two forces acting at two different angles at a same spot will the addition of the moments of the two forces be the same as the moment whereby i take the net force of the two forces and get the moment of the net force of the same two forces?

is there a way to explain why for the same mass of the bob, two different lengths of the pendulum will have different times to have a complete oscillation using moments to explain it. Thanks for the help!

 

[tiny-tim]

hi sgstudent!

if i have two forces acting at two different angles at a same spot will the addition of the moments of the two forces be the same as the moment whereby i take the net force of the two forces and get the moment of the net force of the same two forces?

yes … r x a + r x b = r x (a + b)

is there a way to explain why for the same mass of the bob, two different lengths of the pendulum will have different times to have a complete oscillation using moments to explain it.

yes, the moment tells you the angular acceleration

 

[sgstudent]

hi sgstudent!


yes … r x a + r x b = r x (a + b)


yes, the moment tells you the angular acceleration

Thanks tiny Tim. However, I don't really know the explanation using moments on why a longer pendulum will have longer periods than shorter ones. Since they are longer so won't their moment be greater? Thanks!

 

[HallsofIvy]

The longer the pendulum, the longer the distance it has to cover to complete one period. Why would you think the time to cover this longer distance would be shorter?

If you want a detailed answer:
Suppose the pedulum has mass m centered at distance L from the pivot. There is a downward force of strength -mg. But the pendulum mass can only move around the circumference of the circle or radius L and the component of force parallel to the circumference is -mg sin(\theta) where \theta is the angle the pendulum makes with the vertical.

If we measure \theta in radians, the angle \theta corresponds to a distance around the arc of s= L\theta and so the linear velocity is v= ds/dt= L d\theta/dt and the acceleration is a= dv/dt= L d^2/theta/dt^2. Since "mass times acceleration= force", the motion is given by
ma= mL d^\theta/dt^2= -mg cos(\theta)

That is a badly "non-linear" equation so there is no simple exact solution but there are a number of ways to approximate it. One is to note that for small angles, cos(\theta) is approximately \theta itself so we can approximate the equation by
mLd^2\theta/dt^2= -mg\theta

That is a second order linear equation with constant coefficients. It has "characteristic equation" Lr^2=-g which has "characteristic roots" \pm\sqrt{g/L}. That, turn, tells us that two independent solutions are sin(\sqrt{gt/L} and cos(\sqrt{gt/L}. That will have period given by gt/L= 2\pi so that t= 2L/g which is directly proportional to L- the larger L, the longer the period.