IIT JEE , AIEEE Forum - Physics , Mechanics

waterdemon

Hello to all dear goiitians.
I do not have much time.
Just thought of putting up one question for all of you.Try it.

Q)
A man sits on a swing moving with a small angular amplitude
of "

₀".When the swing passes through equilibrium position
,the man stands quickly and at the instant of swing's max
deviation he sits again.How will the angular amplitude of
the swing be changed during a period if the man's centre
of mass moves up and down through height h.The lenght of
ropes of swing is "L".
Assume L >> H and that mass of swing is negligible to the
mass of Man.

Hope you find the question useful.
Cheers !!!!!!@@@!!!!!!!

Will post the solution tomorrow.
Not a challenge....just for benefit of new question.

ganesha1991

i'll try bro

is it
T²=[ 2

² (A²- (

_o l)² ]/gh

waterdemon

Nopes that is not the answer.
keep an eye on this topic plz.

Best of luck to all who are trying this sum.
Cheers !!!!!!!!!@@@!!!!!!!!

thevyzz

edited:

by law of conservation of energy

let @=theta
gL(1-cos@) = gL(1-cos@' ) + gh

solving

cos @' = cos @ + h/L

anchitsaini

i know this is not the correct solution but still i 'll show my attempt just to say that i attempted it!!!

initially--
mgl(1-cos

₀)=mv²/2

finally--
mg(l - h - lcos

)=mv²/2

thus
l(1-cos

₀) =(l - h - lcos

)
or
- lcos

₀ = - h - lcos

or
cos

= cos

₀ - h/l

as

is small

1 -

² / 2 = 1 -

₀² / 2 - h/l

or

² / 2 =

₀² / 2 + h/l

which gives

² -

₀² =2h/l

taking

² -

₀² =(

-

₀ ) ² (not at all sure about this one

)

the change in angular amplitude is

(2h/l)

if the question is asking for

then
my answer would be

=

(

₀² + 2h/l)

waterdemon

Good Try there by Anchit..But not the correct answer.
Anyways I hoped that there might be some more 10 or
15 replies to this post but I am gonna mail the answer
by today night.

gotta go mates.Bye.Best of luck.
Cheers ! ! ! ! @ @ @ @ ! ! ! !

ashwin123

i think it shud be ^{[2 ]}

theta²_{^{- 2h/L}}

_{^{P.s : May this formula editor burn in hell}}

sameerh522

= cos^-1{ l / l-h }.cos (theta)

suresan

If the man stand upvery time the swing reaches the equillibrium position(Centre), angular velocity is reduced because the change in potential energy from extrem end to the centre is less now(centre of massgoes up). the change potential energy between the extreme and the centre will turn in to the kinetic energy at the centre. As he sits down at the extreme the maximum potential enegy dcreases further. Again this reduces the speed acquired when the swing reaches the centre. This is a sort of forced oscillation in the reverse direction. The ext. energypumped in by the man is used to dampen the oscillations, and the amplitude decreases gradually and eventually the swing stops.
But suppose the man sits everytime the swing reaches the equillibrium and stands up when it reaches the extremes the amplitude will increase gradually. In this case it is a tipical forced oscillation at resonance. The energy is pumped in to the system at double the frequency as to the frequency of the oscillating swing.

waterdemon

The positions of C.G of man are shown by me in the figure
Applying law of conservation of angular momentum,we have:

mvL = mv₁(L-h)

Therefore,
v₁ = vL/(L-h) .......... (1)

Applying law of conservation of energy,we have:

(1/2)mv₁² = mgL'(1-Cos@) ........(2)

&

(1/2)mv² = mgL(1-Cos@₀)..........(3)

Dividing Equation.(2) by equation.(3) , we get :

v₁² = L'(1-Cos@) = (L-h)(1-Cos@) ......(4)
v²     L(1-Cos@₀)       L(1-Cos@)

Substituting value of v₁ from equation.(4) , we get :

L²/(L-h)² = (L-h) (1-Cos@)
                L   (1-Cos@₀)

L³/(L-h)³ = (1-Cos@) = Sin²(@/2)
             (1-Cos@₀) Sin²(@₀/2)

@ = (1 - h/L)^-3/2 = (1 + 3h/2L)
@₀

Therefore,
@ = @₀(1 + 3h/2L)
@ = @₀ + (3h/2L)@₀

@ - @₀ = (3h/2L)@₀

Hope you find it useful.
Cheers !!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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