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Mechanics

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11 Nov 2008 10:52:34 IST

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A bead is free to slide down a smooth wire tightly streched between points A & B on vertical circle . If the bead starts from rest at A , the highest point on the circle .

(a) its velocity v on arriving at B is proportional to sin( theta )

(b) its velocity v on arriving at B is proportional to tan( theta )

(d) time to arrive at B is independant of theta

Hence , Find Velocity & Time with proper Explantaion

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Comments (16)

tejas

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11 Nov 2008 10:55:38 IST

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wherzz the Q??

VARUN RAJ

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11 Nov 2008 11:54:02 IST

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SEE IT IS A GOOD SUM

LET US DRAW THE FBD OF THE FIGURE THEN THE FORCES ACTING ALONG THE DIRECTION OF MOTION R

MG/COS@-T =MA (BUT SINCE T=MG/COS@) THE RESULATANT FORCE ACTING IS 0

THUS WE HV TO APPLY THE PRINCIPLE OF CONSERVATION OF ENERGY

MGABCOS@=1/2MV^2 (BUT AB =R /COS@)

MGR=1/2MV^2

2GR=V^2

THUS V IS PROP TO NOTHING

LENGTH =RCOS@ and v 2GR BUT DISTANCE IS PROP TO COS @ THUS TIME IS INVE PROP TO COS@

pls rate me if find me sueful

!!!!!!cheers!!!!!!!!

VARUN RAJ

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11 Nov 2008 12:35:28 IST

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it is a highly confusing sum
i cant ge tthe geometry part right
dammn

Shivam Mishra

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11 Nov 2008 14:56:24 IST

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Sorry bro , gud try , but u are wrong
Pls dont mind it yaar
..................!!
: )

Bhargav Reddy

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11 Nov 2008 16:41:44 IST

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acceleration of the bead is

which implies time to arrive at B is proportional to

Shivam Mishra

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11 Nov 2008 16:46:34 IST

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Bharghav , sorry u are incorrect

harry

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11 Nov 2008 17:28:26 IST

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va = g cos o
s =2 r cos o

from v^2-u^2 =2as
v=4gr cos^2 o

now use t =v/a

we get time =2 sq root r/g
o = theta

harry

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11 Nov 2008 17:29:04 IST

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am i correct

harry

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11 Nov 2008 17:37:29 IST

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see s will be 2r cos theta not r cos theta

Celestine Preetham

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11 Nov 2008 18:05:36 IST

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time is independent of teta
i got v proportional to cos teta

Shivam Mishra

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12 Nov 2008 00:54:14 IST

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Celestine , cud u prove it for velocity........

rohan ghosh

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12 Nov 2008 16:35:10 IST

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its a very easy ques

velocity = root(2*a*s)

a=acceleration along the bead =gcostheta

s=2*r*cos(theta) (as it is a right angled triangle)

so v is directly proportional to theta

further s=(1/2)*a*t^2

s=2rcos(theta) and a=gcos(theta) => time is independent of theta

nikhil

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12 Nov 2008 20:17:23 IST

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i also got velocity prapotional to cos theta

if u need i will post u the solution

but say me how time is independent to theta

Shivam Mishra

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12 Nov 2008 22:56:51 IST

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Most of u answered my question correctly ,

Nikhil for u ,

a = gcos(theta) ; From right angled triangle dist = 2rcos(theta)

Now apply s = 0(t) + (0.5)at^2

Cos(theta) gets cancelled on both sides when 'a' & dist are substitued

Hope , u Get it ................!!!

Karthik M

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13 Nov 2008 00:56:00 IST

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As for the first one, v is proportional to cos@.

Conserve energy to get it. But your options seem to be different.

rumki chakraborty

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13 Nov 2008 07:06:46 IST

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yes, true , just apply conservation of energy ;

at below point B, loss of PE = gain in KE, and u ll get v = answer

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