Solve Pendulum Problem: Find Inertia and Period

[Daniiel]

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For part a

We know Icm = m/12 (L^2 + W^2) where W is width and L is length

using parallel axis theorem to find the I with the pivot L/4 down the ruler

I = Icm + m(L^2/16)

so then with the I we find the torque

T = IO O = theta double dot (d^2 theta / dtheta ^2)

then the torque would be = L/4 (-mg sin(theta))

so IO = L/4 (-mg sin(theta)) for small theta sin theta > theta

then the sin term is omega squared
so

T = 2pi/omega and yea that's the period

is that correct?
im not sure if the parallel axis theorem is correct.

and for B, how do you find the new inertia with the mass on the top of the ruler? is it the same kind of thing?

 

[AdrianMay]

I don't see why you can't just find the centre of gravity then treat it like a simple pendulum.

 

[Rayquesto]

period=2pie times the square root of Length divided by the gravitational force in angles less than 15 degrees. So, period would be 2pie times the square root of .015meters (length measured from axis of rotation which i believe is the pivot point) divided by 1.962Newtons which is .5494seconds for every revolution(period). then the next one is simply just adding .02kg to the mass and calculating the period, which is .5236seconds.

 

[Daniiel]

Thanks Ray, but it says don't neglect width and in the example i have the period is not dependent on m, but i think when you add a mass to the top of the ruler it will change the center of mass, I am just not sure how to incorporate that into I.

 

[RTW69]

This looks like a physical pendulum where the period=2*Pi*Sqrt(I/(m*g*d) where d is the distance from CM to pivot, in this case L/4

I=I_cm+m*d^2 which you have

for part b I think you add the m*r^2 for the clay weight to I where r is the distance of the weight to the pivot. Can someone confirm this?

 

[Daniiel]

Oh cool thanks, Yea that's what we were thinking for the mass but we weren't sure what to use as the m in the m r^2, because when we do m d^2 is the mass of the system
but for the clay term m r^2 would we use the clays mass or the total mass including the clay. And also when we do m d^2 after the clay has been placed, would the m also be the total mass including the clay?

 

[Rayquesto]

Length is measure from the pivot point to the end of the object tha the gravity force exerts. So L is .015meters. I think when they say don't neglect the width means don't neglact the width's mass, because the mass of the width above the pivot point does effect the period, not it's length, since L is measured from the pivot point to the end point the gravity force exerts force upon.

 

[Daniiel]

Solved it using the tecnique I did, and for b RTWs way worked, you just find the new CM and then the new torque and yea it all makes sense

 

[Rayquesto]

IDK lol it all makes sense the way I did it though. I can't think of another way to do it.