Help quite hard pendulum problem

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Homework Statement



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.


Homework Equations





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:confused:

is this the right method as well?
 
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hi charmedbeauty! :smile:
charmedbeauty said:
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.

but what should d be??

is it a half L, I'm confused:confused:

is this the right method as well?

looks ok :smile:

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 :wink:)
 
tiny-tim said:
hi charmedbeauty! :smile:


looks ok :smile:

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 :wink:)

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.
 
charmedbeauty said:
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

(oh it's straight!)

yes :smile:
 
tiny-tim said:
(oh it's straight!)

yes :smile:
Thanks a bunch tiny-tim:approve:
 
charmedbeauty said:
Thanks a bunch tiny-tim:approve:

and just to double check L is the length of the rod (0.062) and not the length of the ruler?
 
no, L is in 1/12 mL2

L is the length of the thing that has mass m (in this case, the metre stick)
 
tiny-tim said:
no, L is in 1/12 mL2

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

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