E-StoreHome » Ask & Discuss » Physics. » Mechanics « Back to Discussion
Mechanics
20 Nov 2007 19:52:20 IST
0 People liked this
35
2959
IIT 96 sum (Newtons)
None
A Smooth semi-circular wire track of radius R is fixed in a
vertical plane.One end of a massless spring of natural length
(3R/4) is attached to the lower point of O of the wire track
A small ring of mass 'm',which can slide on the track , is
attached to the other end of the spring.The ring is held
stationary at P such that spring akes an angle 600 with the vertical.
The spring constant K = mg/R.
Consider the instant when the ring is released.
a)Determine the tangential acceleration of the ring and the
normal reaction.
The one who answers gets 2 salutes from me.
Challenge for all !!!!!!!!!!
(If answered then salutes but if no answer then I will give
the solution by tomorrow)
Cheers!!!!!!@@@!!!!!!!!!!!!!
Comments (35)
A K
Blazing goIITian
Joined: 13 Mar 2007
Posts: 700
21 Nov 2007 12:34:11 IST
Like
0 people liked this
did u resolve net force along x axis?????
Reply
21 Nov 2007 13:47:06 IST
Like
2 people liked this
Got it!!!
The fbd rquired is attatched.
Original length of the spring = 3R/4
Final length of the string when it is making 60 angle with the vertical = R
.: There is an elongation in the spring ,
so x= R-3R/4 = R/4
Therefore the spring force would act radially inwards.
For tangential acceleration
Ft=mat
Ft=mgcos30+kxcos30.......from the fig
Ft=mg
3/2 + (mg/R)(R/4)3/2=mat
3/2 + (mg/R)(R/4)3/2=mat.: at = g3/2+g3/8 = (53/8)g.........(1)
3/2+g3/8 = (53/8)g.........(1)For, normal reaction.
Now, just at the moment of releasing of the ring centripetal acceleration would be very negligible , so the radial force should be zero at that moment
Fr=0
.: Fr = N+mgsin30-kxcos60=0
N = (3/8)mg {inwards}........(2)
21 Nov 2007 17:47:36 IST
Like
2 people liked this
Okay... I made a silly mistake in the resolving part... Just realised it now after coming back from college :). (Took 60o instead of 30o)
Anyway here is my method, though lots of people seem to have solved it already.
Now the length of the spring in the diagram is R. Elongation is R-3R/4 which is R/4.
Apply newton's law along the tangent at P:
We get ma = mgcos30 + k(R/4)cos30
Which gives a =5
3g/8
Now apply newton's laws along the perpendicular to the tangent (Here I am assuming N acts radially outwards)
So we get N + mgsin30 = kxcos60
or N = mg/8-mg/2
which gives N = -15m/4
Or N = 15m/4 in the inward direction.
This should be the right answer. Please verify waterdemon.
Thanks for pointing that out joy. Btw yeah, you solved it first. I dont deny that. You deserve more credit for that as well. The last time I tried this one was yesterday quite late in the night, and made a mistake in resolving coz of which i got something crappy... you can see my previous post for that. I just saw your method, and yes, I guess they are both the same. As soon as I was back, I went into the topic and clicked the reply button and started typing... didnt realize that so many had already answered.
One more thing, mg/8-mg/2 is -15m/4, I simply put the value of g as 10!! It is the same as 3mg/8. Except the error in the calculation, i dont think there are any other errors.
PS:Just returned from college... so sorry for the late reply :(
Anyway here is my method, though lots of people seem to have solved it already.
Now the length of the spring in the diagram is R. Elongation is R-3R/4 which is R/4.
Apply newton's law along the tangent at P:
We get ma = mgcos30 + k(R/4)cos30
Which gives a =5
3g/8Now apply newton's laws along the perpendicular to the tangent (Here I am assuming N acts radially outwards)
So we get N + mgsin30 = kxcos60
or N = mg/8-mg/2
which gives N = -15m/4
Or N = 15m/4 in the inward direction.
This should be the right answer. Please verify waterdemon.
Thanks for pointing that out joy. Btw yeah, you solved it first. I dont deny that. You deserve more credit for that as well. The last time I tried this one was yesterday quite late in the night, and made a mistake in resolving coz of which i got something crappy... you can see my previous post for that. I just saw your method, and yes, I guess they are both the same. As soon as I was back, I went into the topic and clicked the reply button and started typing... didnt realize that so many had already answered.
One more thing, mg/8-mg/2 is -15m/4, I simply put the value of g as 10!! It is the same as 3mg/8. Except the error in the calculation, i dont think there are any other errors.
PS:Just returned from college... so sorry for the late reply :(
21 Nov 2007 19:46:16 IST
Like
2 people liked this
Well Joy,You have got the answer.
And would also rate Karthik for his effort.
But I am submitting my explanation too , so as to make others
understand.
(Please see the figure prvided by me)
Here OC and CP both are radii and hence.
CPO = 600.OP=R
Therefore,
x = OP - 3R/4
x = R - 3R/4
x = R/4.
Tension:
F = Kx
F = mg/R x R/4 = mg/4.
Considering the equilibrium along the radius of semi-circle,
we have:
N + FCos(60) = mgCos(60)
N + (mg/4)Cos(60) = mgCos(60)
Therefore,
N = 3mgCos(60)/4
N = 3mg*(1/2)*(1/4)
N = 3mg/8
N = 3*m*10/8
N = 15m/4
N = 3.75m.
Further the tangential force will eb given by.
Tangential Force = mgSin(60) + FSin(60)
Tangential Force = mgSin(60) + (mg/4)Sin(60)
Tangential Force = 5mgSin(60)/4
Tangential Force = (5mg/4)*3/2
Tangential Force = (53)mg/8
And now,
m*(Tangential Acceleration) = (5
3)mg/8
Thereofore,
Tangential acceleration = (53)g/8
Tangential acceleration = (253)/4 m/s2.
Hope it helps you all.
Rate if you think it is worth it.
Cheers!!!!!!!!@@@@!!!!!!!!!
And would also rate Karthik for his effort.
But I am submitting my explanation too , so as to make others
understand.
(Please see the figure prvided by me)
Here OC and CP both are radii and hence.
CPO = 600.OP=RTherefore,
x = OP - 3R/4
x = R - 3R/4
x = R/4.
Tension:
F = Kx
F = mg/R x R/4 = mg/4.
Considering the equilibrium along the radius of semi-circle,
we have:
N + FCos(60) = mgCos(60)
N + (mg/4)Cos(60) = mgCos(60)
Therefore,
N = 3mgCos(60)/4
N = 3mg*(1/2)*(1/4)
N = 3mg/8
N = 3*m*10/8
N = 15m/4
N = 3.75m.
Further the tangential force will eb given by.
Tangential Force = mgSin(60) + FSin(60)
Tangential Force = mgSin(60) + (mg/4)Sin(60)
Tangential Force = 5mgSin(60)/4
Tangential Force = (5mg/4)*
3/2Tangential Force = (5
3)mg/8And now,
m*(Tangential Acceleration) = (5
3)mg/8Thereofore,
Tangential acceleration = (5
3)g/8Tangential acceleration = (25
3)/4 m/s2.Hope it helps you all.
Rate if you think it is worth it.
Cheers!!!!!!!!@@@@!!!!!!!!!
Free Sign Up!
Preparing for IIT-JEE ?
Arihant Revision Package for IIT JEE - Books, Practice Tests + Rank Predictor
@ INR 1,995/-
For Quick Info
Find Posts by Topics
Physics.
Topics
9384
20062
3670
2185
7602
2826
2633
Mathematics.
Chemistry.
Biology
Institutes
Parents
Board
Fun Zone
|
|
|
|
|
| 35,229,755 Engineering Questions Attempted |
6,69,530 Visitors Monthly |
79 Engg. Exam Covered |
1251 Students Currently Online |
Connect with Us  |
 |
Vriti Communities
Privacy Policy © Vriti 2011