Torsional Oscillations Question

[akan]

Homework Statement

A thin steel beam 8.0 meters long is suspended from a crane and is undergoing torsional osciallations. Two 75-kg steelworkers leap onto opposite ends of the beam, as shown in the figure (no figure given, they just jump straight towards the center of the beam).

If the frequency of the oscillations diminishes by 20%, what is the mass of the beam?

Homework Equations

Iw_1 = Iw_2 + 2 (m R^2 w_2)
I = (1/12) ML^2
w = 2 (pi) (freq.)

The Attempt at a Solution

I (w_1) = I (w_2) + 2 (m R^2 w_2)
I (w_1) = (w_2) (I + 2 (m R^2))
I (w_1) = ((80%) w_1) (I + 2 (m R^2))
I = (80%) (I + 2 (m R^2))
I = 80% I + 80% 2 (m R^2)
(2/10) I = (8/10) 2 (m R^2)
I = 8 (m R^2)
I = 8 (75 R^2)
I = 600 R^2

But I = (1/12) ML^2, so:

600 R^2 = (1/12) ML^2
M = 7200 R^2 / L^2
M = 7200 (4^2) / (8^2)
M = 1800

Mastering physics does not accept this answer. What is wrong with my solution?

 

[alphysicist]

Hi akan,

I think you are confusing the \omega for angular velocity with the \omega for angular frequency.

This is a torsional pendulum and your book probably has derived formulas for the period, frequency, and angular frequency for them.

 

[akan]

w = 2 pi f
so if f diminishes by 20%
so does w
is that right?

 

[alphysicist]

w = 2 pi f
so if f diminishes by 20%
so does w
is that right?

That's right; now what is the formula for w? For a simple pendulum it is:

<br /> \omega =\sqrt{\frac{g}{\ell}}<br />
for a spring it is

<br /> \omega=\sqrt\frac{k}{m}}<br />

What is it for this torsional pendulum? (Rather than w, you might find a formula for the regular frequency or the period instead.)

 

[akan]

That's right; now what is the formula for w? For a simple pendulum it is:

<br /> \omega =\sqrt{\frac{g}{\ell}}<br />
for a spring it is

<br /> \omega=\sqrt\frac{k}{m}}<br />

What is it for this torsional pendulum? (Rather than w, you might find a formula for the regular frequency or the period instead.)

<br /> \omega =\sqrt{\frac{k}{I}}<br />

However, I do not know what k is, and I also don't know the initial \omega to calculate it. So what do I do?

EDIT:

Nevermind, I see. Haha.

\omega 1 =\sqrt{\frac{k}{I}}

\omega 2 = \frac{80}{100} \sqrt{\frac{k}{I}} = \sqrt{\frac{k}{I + 150 R^2}};<br />

etc., gives M = 800 kg, which is correct.