# A long, thin, uniform rod, with mass 3.00 kg and length 80.0 cm, is hung vertically and pivoted about axis a distance x from the center of?

A long, thin, uniform rod, with mass 3.00 kg and length 80.0 cm, is hung vertically and

pivoted about axis a distance x from the center of the rod. Assume small-angle oscillations.

(a) Find two different values of x for which the period of small oscillations of the rod is 1.39 s.

(b) There is one particular period for which only one value of x will give that period for small

oscillations. What is this period?

### 1 Answer

- WoodsmanLv 73 years agoFavorite Answer
For a compound pendulum (which this is),

period T = 2π*√(I / mxg)

where I is the moment of inertia about the pivot point.

I = mL²/12 + mx²

Let's let k = x/L. Then

I = mL²/12 + mk²L² = mL²*(1/12 + k²)

and

T = 2π*√[mL²*(1/12 + k²) / mkLg] = 2π*√[L(1/(12k) + k) / g]

T² / 4π² = L(1/(12k) + k) / g

gT² / 4π²L = k + 1/(12k) ← generic solution for uniform rod

Plugging in our values, the LHS is

9.8m/s² * (1.39s)² / (4π²*0.80m) = 0.5995,

and so

0.5995 = k + 1/(12k)

which has roots at

k = 0.219 → which means x = 0.219*80.0cm = 17.5 cm ◄

and k = 0.3805 → which means that x = 0.3805*80.0cm = 30.4 cm ◄

(b) I've shown that T = C*√(k + 1/(12k))

for C some constant.

dT/dk = C * [1 - 1/(12k²)] / [2*√(k + 1/(12k)]

which is zero when

1 - 1/(12k²) = 0

1 = 1/(12k²)

12k² = 1

k = √(1/12) = 0.289

Then

T = 2π*√[0.800m * (1/(12*0.289) + 0.289) / 9.8m/s²]

T = 1.36 s ◄

Email me if you have a question (and have email enabled).

Source(s): (a) http://www.wolframalpha.com/input/?i=0.5995+%3D+k+... (b) http://www.wolframalpha.com/input/?i=d(%E2%88%9A%5...