Torsional Pendulum

  • Thread starter Bryon
  • Start date
  • #1
99
0

Homework Statement



https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/DuPage/phys2111/fall/homework/Ch-15-SHM/torsion-pendulum/4.gif [Broken]


A uniform meter stick is hung at its center from a thin wire. It is twisted and oscillates with a period of 5 s. The meter stick is then sawed off to a length of 0.76 m, rebalanced at its center, and set into oscillation.

With what period does it now oscillate?

L = 1m
L(final) = 0.76m
T = 5s

Homework Equations



Moment of Inertia: I = (1/12)*M*L^2
period: T = 2pi*sqrt(I/K) Where K is the torsional constant

The Attempt at a Solution



I first found the torsional constant:

K = [(1/12)*M*(1^2)]/[(T/(2pi))^2] = 0.13459M

So no I have K = 0.13459M

T(final) = 2pi*[sqrt((1/12)*M*(0.76^2)/0.13159M)] = 3.8s

The answer looks correct but I'm not sure, any ideas? Thanks!
 
Last edited by a moderator:

Answers and Replies

  • #2
instead of going for so much calculation...
u could have done it like this....

becauz..

T=2*pi*(I/K)^0.5

and u have I=(M*L^2)/12

so jus take the ratio

T1/T2=L1/L2 :smile:

where T2 is the time period when the length is 0.76
directly arrives at your answer ...
 
  • #3
99
0
Oh yeah it does.....cool.
 

Related Threads on Torsional Pendulum

  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
3
Views
2K
Replies
1
Views
3K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
1
Views
7K
Replies
1
Views
4K
  • Last Post
Replies
2
Views
6K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
2
Views
18K
Top