# Homework Help: Help quite hard pendulum problem

1. May 19, 2012

### charmedbeauty

1. The problem statement, all variables and given/known data

A very light rigid rod with a length of 0.062 m extends straight out from one end of a metre stick. The combination is suspended from a pivot at the upper end of the rod as shown in the following figure. The combination is then pulled out by a small angle and released.

2. Relevant equations

3. The attempt at a solution

well using parallel axis Theorem

I= 1/12mL2+md2

ω=√mgd/I

T=2∏/ω

T=2∏√(I/mgd)

= 2∏√((1/12)(m)(L)2+md2)/mgd

m's cancel

=2∏√((1/12)(L)2+d2)/gd

but what should d be??

is it a half L, I'm confused

is this the right method as well?

2. May 19, 2012

### tiny-tim

hi charmedbeauty!
looks ok

you're using the parallel axis theorem, and your axis through the c.o.m. is perpendicular to the rod, so d is the perpendicular distance from that axis to the pivot, ie the distance PC from the pivot P to the c.o.m. C

(if you're wondering how the moment of inertia can be the same whatever the angle θ of the stick, just do a bit of geometry using the cosine formula for two points A and B on the stick equidistant from C … you'll find that PA2 + PB2 is independent of θ )

3. May 19, 2012

### charmedbeauty

so say I have the pivot point which is connected to a rod which in turn is connected to a ruler then d would be the length of the the rod+half that of the ruler?? i.e. 0.5+0.062=d

http://www.webassign.net/serpse8/15-p-034-alt.gif

thats the pic there with all details given in question.

4. May 19, 2012

### tiny-tim

(oh it's straight!)

yes

5. May 19, 2012

### charmedbeauty

Thanks a bunch tiny-tim

6. May 19, 2012

### charmedbeauty

and just to double check L is the length of the rod (0.062) and not the length of the ruler?

7. May 19, 2012

### tiny-tim

no, L is in 1/12 mL2

L is the length of the thing that has mass m (in this case, the metre stick)

8. May 19, 2012

### charmedbeauty

oh right so I can just leave it out since it is 1 metre. cool.