- #1

lc99

- 161

- 3

## Homework Statement

Assume the balance wheel of a torisonal pendulumn is doubled in size but the torison constant of the spring remains the same. The density of the disk remains the same, but bot the radius and the thickness double. What happens to the period of the pendulumn?

a) Period doubles

b) period goes up by a factor of sqrt(2)

c) period goes up by factor of four

d) period goes up by factor of 4sqrt(2)

e) none of the other answers

## Homework Equations

omega = sqrt(k/I)

k= torsional constant

I = 1/2MR^2

R is doubled and M is doubled (since thickness/size doubled)

## The Attempt at a Solution

My answer is e) none of the answers because i calculated that the period actually goes down...

w = sqrt(k/I)

I = 1/2MR^2 = 1/2 (2M)(2R)^2 = 8 (1/2MR^2)

--> w = sqrt(k/(8*I)) = 1/ sqrt(8) * sqrt(k/I)

W = 2pi/T --> T = 2pi / (sqrtk/I)

--> 1/sqrt(8) T

the period went down by sqrt(8).