Moment of Inertia of a Pendulum

[MyNewPony]

Homework Statement

The 20 cm-long wrench in the figure swings on its hook with a period of 0.92s. When the wrench hangs from a spring of spring constant 350 N/m, it stretches the spring 3.5 cm.

What is the wrench's moment of inertia about the hook?

http://session.masteringphysics.com/problemAsset/1070606/9/14.EX25.jpg

Homework Equations

I = m*g*L*T^2/2pi

The Attempt at a Solution

Fsp = Fg
kx = mg
m = kx/g
m = (350)(0.035)/9.8 = 1.25

I = (1.25)(9.8)(0.14)(0.92)^2/2pi = 0.23 kg*m^2

This isn't the correct answer however. Can someone explain what I did wrong?

 

[alphysicist]

Hi MyNewPony,

Homework Statement

The 20 cm-long wrench in the figure swings on its hook with a period of 0.92s. When the wrench hangs from a spring of spring constant 350 N/m, it stretches the spring 3.5 cm.

What is the wrench's moment of inertia about the hook?

http://session.masteringphysics.com/problemAsset/1070606/9/14.EX25.jpg

Homework Equations

I = m*g*L*T^2/2pi

This formula does not look quite right to me. Do you see what it needs to be?

 

[MyNewPony]

Hi MyNewPony,



This formula does not look quite right to me. Do you see what it needs to be?

Ah. I forgot to square the 2pi.

So the equation becomes:

I = mgLT^2/4pi^2

Is that correct?

 

[Doc Al]

I = m*g*L*T^2/2pi

This equation isn't quite right. It's off by a factor of 2pi.

 

[alphysicist]

Ah. I forgot to square the 2pi.

So the equation becomes:

I = mgLT^2/4pi^2

Is that correct?

That looks right to me.

 

[Doc Al]

Ah. I forgot to square the 2pi.

So the equation becomes:

I = mgLT^2/4pi^2

Is that correct?

Yes.

 

[MyNewPony]

That looks right to me.

Yes.

Thanks a bunch!

 

[alphysicist]

Glad to help!

 

[ElTaco]

Sorry for bumping up an old thread, but thought it'd be better than making a new one about the same problem. Can someone tell me where equation for I is derived from? From my knowledge I know that I = cMR^2 (as an estimated value), but what exactly do you plug into get to that point? (I = mgLT^2/4pi^2 )

 

[alphysicist]

Hi ElTaco,

Sorry for bumping up an old thread, but thought it'd be better than making a new one about the same problem. Can someone tell me where equation for I is derived from? From my knowledge I know that I = cMR^2 (as an estimated value), but what exactly do you plug into get to that point? (I = mgLT^2/4pi^2 )

It's a standard form for the period of a physical pendulum. The derivation should be in your book, and you might also look at:

http://hyperphysics.phy-astr.gsu.edu/hbase/pendp.html

 

[ElTaco]

Hi ElTaco,



It's a standard form for the period of a physical pendulum. The derivation should be in your book, and you might also look at:

http://hyperphysics.phy-astr.gsu.edu/hbase/pendp.html

Oh ok, I see, thank you very much!