Simple pendulum (1 Viewer)

[MongMan]

For a simple (ie back and forth, not horizontally circular) pendulum which direction is the centripical force?

Tcos(theta) or just T?

T being the tension force along the length of the string.

Diagram attached.

 

[lychnobity]

The direction of the centripetal force would be towards point F

 

[MongMan]

The direction of the centripetal force would be towards point F

Centripical force being from F to P. That's what I thought.

I'm stuck on (ii) of this question.

Tcosø = mg
T = mlω² (l = r string)

T = mg/cosø

mg/cosø = mlω²
g/lcosø = ω²

so,
s'' = l(d/dø)(1/2 * ω²)
s'' = l(d/dø)(1/2 * g/lcosø)
question wants s'' = gsinø

what am I doing wrong?


(Edit: image attached)

 

[untouchablecuz]

i drew this up for someone today

ALMOST the same situation

the Q these images refer to is in the 2000 paper

 

[MongMan]

i drew this up for someone today

ALMOST the same situation

the Q these images refer to is in the 2000 paper

So how would you get to s'' = -gsin∂

 

[untouchablecuz]

given that s''=tangential acceleration

derived above that the tangential acceleration=-gsin(theta)

.'. s''=-gsin(theta)

 

[MongMan]

given that s''=tangential acceleration

derived above that the tangential acceleration=-gsin(theta)

.'. s''=-gsin(theta)

I see. Geometric. Very cute.

Could one extrapolate therefore, that:

s'' = l(d/dø)(1/2 * -g/lcosø) = -gsin(theta)

or are my equations above broken?

 

[untouchablecuz]

I see. Geometric. Very cute.

Could one extrapolate therefore, that:

s'' = l(d/dø)(1/2 * -g/lcosø) = -gsin(theta)

or are my equations above broken?

no

because w²=g/Lcos(ø) was derived in a different reference frame to -gsin(ø)

also, no point in extrapolating further; just equate L(d/dø)(0.5*w²)=-gsin(ø) and integrate with respect to ø

 

[MongMan]

no

because w²=g/Lcos(ø) was derived in a different reference frame to -gsin(ø)

also, no point in extrapolating further; just equate L(d/dø)(0.5*w²)=-gsin(ø) and integrate with respect to ø

how is
s''=l(d/dø)(1/2 * -g/lcosø) a different reference frame?
same angle, same time, same set of forces (ie gravity, tangential, and centrifugal)

I'm geniunely curious

 

[untouchablecuz]

how is
s''=l(d/dø)(1/2 * -g/lcosø) a different reference frame?
same angle, same time, same set of forces (ie gravity, tangential, and centrifugal)

I'm geniunely curious

s''=l(d/dø)(1/2 * -g/lcosø) - this is not a different reference frame; this is wrong because it is the combination of results from 2 different reference frames

w²=g/Lcos(ø) was derived with respect to directly horizontal and vertical axis

a_tangential=-gsin(ø) was derived with respect to a tangential and radial axis (this is perfectly valid, because by circle geometry, they are perpendicular)

hence L(d/dø)(0.5*w²) =/= -gsin(ø)

HOWEVER

if you resolved the tangential force into components with respect to the horizontal and vertical axis, then you may equate L(d/dø)(0.5*g/Lcosø)=a_{some component}

ask if you would like me to do it both ways

 

[MongMan]

hence L(d/dø)(0.5*w²) =/= -gsin(ø)

Ah, but I'm just following the logic from the question, as follows:
(i) Show that tangential accleration of P is given by:
s'' = l(d/dø)(1/2 * w²)
where s is arc length

(ii) Show that the equation of motion of the pendulum is

l(d/dø)(1/2 * w²) = -gsinø

Sure, I would love to see your proof on this.:balloon:

 

[untouchablecuz]

i will do a proof resolving in the vert and horizon directions l8r