Solving for Period of a Physical Pendulum - Homework Help

[lackos]

Homework Statement

A very light rigid rod with a length of 0.516 m extends straight out from one end of a meter stick. The combination is suspended from a pivot at the upper end of the rod as shown in the following figure. The combination is then pulled out by a small angle and released.
a) determine the period of oscillation?
b)By what percentage does the period differ from a simple pendulum of length 1.016m?

sorry i don't know how to put a picture in. its fairly simple pictue just a rod and metre stick connected to a wall.

Homework Equations

omega = sqrt(I/mgd)
omega = 2*Pi/T where I is moment of inertia, T is period and d is distance from point of rotation to center of mass

However I am not sure if these are correct

The Attempt at a Solution

So far just fumbling with the equation. I have got T as a function of g and d

 

[tiny-tim]

hi lackos!

(have a square-root: √ and a pi: π and an omega: ω )

A very light rigid rod with a length of 0.516 m extends straight out from one end of a meter stick …

(do you mean 0.016 ? and is the light rod at the top, next to the pivot?)

when you're not sure, just go back to the basic equation, τ = Iα

 

[lackos]

sorry about the late reply. yes the light rigid rod is connected to the roof(pivot point).

also sorry about not using the proper syntax. I didnt know you had them so i just wrote in maple syntax.

also although I am familiar with
[tex]\tau=\alpha*r[/tex] in terms of rotation I am not sure how it applies in this context

 

[tiny-tim]

hi lackos!

(just got up :zzz: …)

are you thinking of a = αr ?

τ = Iα is the equivalent of F = ma …

τ is the torque of the weight (in this case) acting at the centre of mass, about the pivot point

 

[lackos]

thanks for the quick reply

ahh yes my bad, i got the formulas mixed up.

if i were to use the torque would it be that
T=2*[tex]\pi*sqrt{}I/mgd[/tex]

where mg is the torque and therefore
T=2*[tex]\pi*sqrt{}1/d*\alpha[/tex]

where d is the distance to the centre of mass (in this case being the length of the rod plus 0.5 metres as the rod has no maths)

it feels lkie I've done something wrong

 

[tiny-tim]

hi lackos!

your first equation looks correct, but did you just guess it?

though mg is the weight, not the torque of the weight

(i don't understand your second equation at all )

write out τ = Iθ'' carefully …

what do you get? ​

 

[lackos]

my first equation come from the formula for a physical pendulum.
[tex]\omega=\sqrt{}I/mgd[/tex]

then subbing in for [tex]\omega=2*\pi/T[/tex]

second equation was just subbing in my (wrong) assumption that [tex]\tau=I*\alpha[/tex]
and then the I's canceled out, but alas it was wrong

okay I am sorry but i don't know what you mean by writing out the equation carefully, the only common link seems to be the moment of inertia, so would i rearrange and sub in for that.

also since the angle is small I am not sure how the torque would play any major part, due to torque being a cross product of force and angle

sorry about my ineptitude lol, this is taking longer than it seems it shouldawesome I am starting to understand how the syntax works

 

[Doc Al]

my first equation come from the formula for a physical pendulum.
[tex]\omega=\sqrt{}I/mgd[/tex]

This equation isn't quite right. (Check the units!)

Once you fix this equation, that is all you'll need. (You should express it in terms of T, of course.)

What's the rotational inertia (I) of the object about the pivot point? What's d?

 

[lackos]

okay my bad i think i got the fraction the wrong way around, i was thinking of the period function.

for pivitol point at end, I=(1/3)*ML^2

okay so i have

T=2*[tex]\pi*/\sqrt{}L²/3*gd[/tex]

okay so my thought is that L=1m, d=0.758m(halfway along the pendulum) or just 1.016(which is half the distance of the ruler plus the length of the rod)

would someone please check my ideas, I am unsure about the value of d due to the effect of the LIGHT rod, and therefore wondering if it factors into the length of d

 

[Doc Al]

for pivitol point at end, I=(1/3)*ML^2

But the pivot point is not at the end of the meter stick. Hint: Use the parallel axis theorem.

would someone please check my ideas, I am unsure about the value of d due to the effect of the LIGHT rod, and therefore wondering if it factors into the length of d

d is the distance from the pivot point to the center of mass of the object.

 

[lackos]

im sorry what's the parallel axis theroem, i don't think we did that in our course. I've heard of it but i don't know how to apply it

 

[Doc Al]

im sorry what's the parallel axis theroem, i don't think we did that in our course. I've heard of it but i don't know how to apply it

Look it up! It explains how to calculate the moment of inertia of an object about an axis given its moment of inertia about the center of mass. In order to apply your physical pendulum formula, you need the moment of inertia of the object about the pivot point, which is a distance d from the center of mass.

 

[lackos]

okay so just to clarify, you are saying that the pivotal point isn't actually the end of the rigid rod, but the end of the ruler?

 

[Doc Al]

okay so just to clarify, you are saying that the pivotal point isn't actually the end of the rigid rod, but the end of the ruler?

No, just the opposite. The pivot point is the end of the light rigid rod, not the end of the ruler.

 

[lackos]

okay applying the axis theorem

I=(1/3)*ML²+ Md²

where L is the length of the ruler, and d is the length of the rod.

 

[lackos]

okay but then when i use the formula is the d in mgd the distance to the centre of mass, not just the length of the rod.

 

[Doc Al]

okay applying the axis theorem

I=(1/3)*ML²+ Md²

To use the parallel axis theorem, you need the rotational inertia about the center of mass, not the end.

where L is the length of the ruler, and d is the length of the rod.

d should be the distance from the center of mass to the pivot.

(Draw yourself a picture.)

 

[lackos]

okay i think i got it now.
so using parallel axis theorem it is actually,
I=(1/12)*ML²+Md²

where d is 0.5+0.516 and l is 1.

does this seem right.

then subbing into final equation

for all values

T=2*[tex]\pi[/tex]*[tex]\sqrt{}(1+1.016²)/g*1.016[/tex]

what do you think

 

[Doc Al]

okay i think i got it now.
so using parallel axis theorem it is actually,
I=(1/12)*ML

Yes. (But L², not L.)

 

[lackos]

Yes. (But L², not L.)

yes sorry bout that i hit submit instead of go advanced so i could put the power on

 

[lackos]

okay so the answer i get is 2.0298 which is wrong, any suggestions for errors.also i would just like to thank you doc Al and also tiny-tim for your help

 

[Doc Al]

okay i think i got it now.
so using parallel axis theorem it is actually,
I=(1/12)*ML²+Md²

where d is 0.5+0.516 and l is 1.

does this seem right.

Yes.

then subbing into final equation

for all values

T=2*[tex]\pi[/tex]*[tex]\sqrt{}(1+1.016²)/g*1.016[/tex]

what do you think

Double check your value for I. (And careful how you mix Latex and non-Latex. Use Latex for the entire equation and you'll get better results.)

 

[lackos]

Yes.Double check your value for I. (And careful how you mix Latex and non-Latex. Use Latex for the entire equation and you'll get better results.)

im kind of struggling with how latex works atm, i will go read a tutorial after this.

and yes i realized afterwards that i didnt factor in the 1/12. however when i do this i get an answer that the system says is wrong.

did you get the value 2.0298

okay my issue was i squared the 1/12 which was wrong,

so now my new result is 2.103, does this seem right, i just want to make sure as i only have 1 submission left, as long as this doesn't break the ethics of the site

 

[Doc Al]

so now my new result is 2.103, does this seem right, i just want to make sure as i only have 1 submission left, as long as this doesn't break the ethics of the site

That looks OK. (Depends on how fussy your online system is.)

 

[lackos]

you sir are a god.

but seriously thank you for guiding me through it i appreciate it greatly, and i will forever retain THE PARALLEL AXIS THEOREM