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Moments and pendulum 
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#1
Apr112, 07:45 AM

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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! 


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Apr112, 03:23 PM

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hi sgstudent!



#3
Apr112, 05:44 PM

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#4
Apr112, 06:10 PM

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Moments and pendulum
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 [itex]mg sin(\theta)[/itex] where [itex]\theta[/itex] is the angle the pendulum makes with the vertical. If we measure [itex]\theta[/itex] in radians, the angle [itex]\theta[/itex] corresponds to a distance around the arc of [itex]s= L\theta[/itex] and so the linear velocity is [itex]v= ds/dt= L d\theta/dt[/itex] and the acceleration is [itex]a= dv/dt= L d^2/theta/dt^2[/itex]. Since "mass times acceleration= force", the motion is given by [tex]ma= mL d^\theta/dt^2= mg cos(\theta)[/tex] That is a badly "nonlinear" 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, [itex]cos(\theta)[/itex] is approximately [itex]\theta[/itex] itself so we can approximate the equation by [tex]mLd^2\theta/dt^2= mg\theta[/tex] That is a second order linear equation with constant coefficients. It has "characteristic equation" [itex]Lr^2=g[/itex] which has "characteristic roots" [itex]\pm\sqrt{g/L}[/itex]. That, turn, tells us that two independent solutions are [itex]sin(\sqrt{gt/L}[/itex] and [itex]cos(\sqrt{gt/L}[/itex]. That will have period given by [itex]gt/L= 2\pi[/itex] so that [itex]t= 2L/g[/itex] which is directly proportional to L the larger L, the longer the period. 


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