IIT JEE , AIEEE Forum - Physics , Mechanics

nilesh

hello sir,i am not able to crack the problem given below,i have tried it myself.the answer is close to the actual answer but i am not sure about the solutions approach(to pinpoint it,i am not sure about eqn (3))

please do me a fovour by solving this,also i would be obliged if you tell me where i went wrong

.

hey,and another big problem is that,images are not visible here

i have submitted the figure for this problem,edited in MS word, but it is not visible here plus some of the subscripts are not completely visible.

pleeeeaseee help mr.administrator!

the problem goes like this:-

Q) A small body is released from pt. X, it reaches pt A and enters a track from which a symmetrical portion CB subtending an angle 2µ at the center, is removed as shown in figure. The body then moves to pt C via pt B and thus successfully completes the loop. (The body is in air while traversing form pt B to pt C)

Given H=50cm;R=20cm.find the maximum value of the angle µ.

http://www.goiit.com/templates/default/images/topics/469.gif

my solⁿ:

let vel. at B = v_b

vel. at P= v_p

We have v_b²=3gr-2g(r-x)

Which boils down to, v_b²=g(r+2x)------------------------------(1)

now, cosµ=h/r

cosµ=r-x/r

r.cosµ=r-x

x=r-rcosµ

x=r(1-cosµ)--------------------------------------------------------(2)

v_b²=g[r+2r(1-cosµ)]----------------------------------from(1)&(2)

=10[20+40(1-cosµ)]

=[200+400(1-cosµ)]

now v²=u²+2as

v_p²= v_b² - 2gH---------------------------(3)

(2gr)^1/2=200+400(1-cosµ)-2*10*20

(2*10*20)^1/2=200+400(1-cosµ)-400

20= -200+400(1-cosµ)

cosµ=1-11/20=9/20

µ=63.25degrees

if g=9.8 then µ comes out to be µ=61.93 degrees

but the actual answer given is 60 degrees.

edison

Dear nilesh,

i am unable to imagine shape of the track and figure you want to explain. I further request Administrator to kindly make available the editing of figures to avoid inconvinience, if any, from students and experts as well.

nilesh

hello sir,

check whether this would do-

X

*

* C P. B

* * *

* * + *

* * *

*******

A

where X is the initial position of object

2(alfa)=angle(C+B)

CB is the removed part

Admin

Dear Nilesh,

I know right now we have this big problem with symbols and images. We will be having image upload up and running very soon on the site, and then things will be much simpler. till then pleaseee bear with us.

thanks a mill for the cooperation.

**As a special case, if possible, try mailing the image to contact_at_goiit.com with the link to the post to which you want the image appended. we will try to add it. no promises but for really good problems, we can try to bend some rules.

cheero!

~forum administrator

edison

Dear Nilesh, your approach is perfectly fine upto equation (3), then it should change as explained below.

let vel. at B = v_b

vel. at P= v_p

We have v_b²=3gr-2g(r-x)

Which boils down to, v_b²=g(r+2x)------------------------------(1)

now, cosµ=h/r

cosµ=r-x/r

r.cosµ=r-x

x=r-rcosµ

x=r(1-cosµ)--------------------------------------------------------(2)

v_b²=g[r+2r(1-cosµ)]----------------------------------from(1)&(2)

=10[20+40(1-cosµ)]

=[200+400(1-cosµ)]

now v²=u²+2as

Upto this point its perfect, Now once the body reaches point B, it will not at all traverse circular path but we follow projectile motion from B to C and this path from B to C should therefore be parabolic and hence apply and use the concept of parabolic motion to get the solution. Moreover the body at B goes tangentially to the circle at B and it will go to the highest point where its component of vertical velocity becomes zero.

Use range R = (v²sin2

)/g = BC

Now to get the condition for maxima differentiate R w.r.t.

and equate to zero.

nilesh

first of all,thank you dear administrator & moderator for"bending the rules."!

thank you mr.edison for showing the right direction.

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