| Posted on Thursday, 15 May, 2003 - 03:42 pm: |
A small smooth ring of mass m is threaded on a light inelastic string of length 9l, whose ends are fixed at a point O on a smooth horizontal table and at a point A which is at a height 6l vertically above O. The ring moves in a horizontal circle on the table with the string taut. Given that the ring is moving at a constant angular speed 0.5 sqrt(g/l) calculate the magnitude of the reaction between the ring and the table.
My method:
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Finding the radius
Call the position of the particle; P
OP = radius = r
AP = x
r + x = 9l ....... 1
(6l)2 + r2 = x2 ....... 2
rearrange (1)
x = 9l - r
x2 = (9l - r)2 ...... 3
substitute (3) into (2)
36l2 + r2 = r2 - 18lr + 81l2
45l2 = 18lr
r = 2.5l ...... 4
x = 9l - 2.5l
x = 6.5l ...... 5
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Resolving forces
Call the reaction force R
Call the centripetal force C
Call the angle between the AO and the downward vertical t
Call the tension in the string T
Call the angular speed W
C = T sin(t)
C = mrW2
= m(2.5l)(0.25(g/l))
= 5mg/8
From (4) and (5)
sin(t) = 5/13
cos(t) = 12/13
T sin(t) = 5mg/8
T = 13mg/8
resolving vertically
R + Tcos(t) = mg
R = mg - 1.5mg
which is obviously incorrect, I can't find the error, I would appreciate some help.
Thanks
ANS. = 7mg/12
| Posted on Thursday, 15 May, 2003 - 07:58 pm: |
I think there's something wrong with 'C=Tsin(t)'.
| Posted on Friday, 16 May, 2003 - 08:41 am: |
Good morning,
I was thinking the same way;
The equations I got however were:
Call T the force produced due to the motion of the ring;
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the problem is we have two simultaneous equations and three unknowns , I know I am missing a third link, a hint would be really appreciated
| Posted on Friday, 16 May, 2003 - 07:37 pm: |
You're missing a force in the horizontal equation.
Hint: think about the shape of the string...
| Posted on Saturday, 17 May, 2003 - 08:20 am: |
Matthew,
Thanks a lot mate. I got it! Thanks again.
| Posted on Sunday, 18 May, 2003 - 04:19 pm: |
Good day to everyone,
I can't seem to figure out the following question;
A bowl is made from a smooth spherical shell of radius "r" by cutting away the part which is more than a third "r" above the horizontal plane. A marble is projected from the lowest point of the bowl with speed u. Show that in the subsequent motion the marble will will leave the bowl and not fall back into it provided:
I am just interested in knowing what are the conditions that are required for the ball not to describe full circles and indeed as asked to leave its vertical path (section in bold).
I do beg everyone's pardon if my posts are becoming too many but this is the only site that I find helpful, I will try my best to contribute in the upcoming days once tommorow's exams are over. Once again I thank the nrichment team for being an essential part of my mathematical career.
| Posted on Sunday, 18 May, 2003 - 04:33 pm: |
Masoud,
Suppose the marble has speed v when it reaches the lip of the bowl. Then this is a projectile motion problem. Remember that the marble is necessarily travelling in a plane through the centre of the sphere, so you only have two dimensions to worry about. You know how wide the opening in the bowl is, and the angle of projection (tangential to the sphere at the point of takeoff). So using the equation of trajectory, you can work out whether it will land in the bowl again. Now just calculate v in terms of u using conservation of energy.
(An interesting thing to speculate about is the behaviour of the marble if it does land inside the bowl. If the bowl is smooth, then energy is preserved; so if it has the critical speed Ã?? (17gr/3) then it will circle round indefinitely. If it is slower than this, it will hit the inner surface of the bowl at an angle, and bounce; its behaviour then will be hideously difficult to analyse. But don't let me stop you trying!
)David
| Posted on Sunday, 18 May, 2003 - 04:56 pm: |
Good day,
Thank you for your quick reply, I really appreciate it.
Finding the velocity at the lip;
s = (2/3)Ã?? 8
u = v cos q
v = not needed
a = N/A
t = ?
Find t would require the vertical component
s = N/A
u = v sin q
v = 0
a = -9.8
t = ?
Is this the way you would start or am I missing something?
| Posted on Sunday, 18 May, 2003 - 05:53 pm: |
< font size="2" face= "Verdana,Arial,Helvetica» It's worth being able to quote the formula for the range of a projectile,
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Here we have a = p /2 - b so sin 2a = sin 2b = 2 sin b cos b = 2 (1/3) (2Ã?? 2)/3 = 4 2/9.
So the range will be 4Ã?? 2/(9g) v2 , and we want this to be < 4Ã?? 2/3 r, the width of the opening.
So we need v2 < 3gr.
Now let's work out v2 . Take the bottom of the bowl as the zero of grav.potential energy; then if the marble has mass m, its energy is 1/2 mu2 .
So 1/2 mv2 + 4/3 mgr = 1/2 m u2 , ie v2 = u2 - 8/3 gr.
Thus u2 < (3 + 8/3) gr, or 17/3 gr. QED.
David Here we have a = p/2 - b so
| sin2a = sin2b = 2 sinbcosb = 2 (1/3) (2Ö2)/3 = 4 |
2 9 |
. So the range will be 4Ö 2/(9g) v2 , and we want this to be < 4Ö 2/3 r, the width of the opening. So we need v2 < 3gr. Now let's work out v2 . Take the bottom of the bowl as the zero of grav.potential energy; then if the marble has mass m, its energy is 1/2 mu2 . So 1/2 mv2 + 4/3 mgr = 1/2 m u2 , ie v2 = u2 - 8/3 gr. Thus u2 < (3 + 8/3) gr, or 17/3 gr. QED. David
| Posted on Sunday, 18 May, 2003 - 07:45 pm: |
David,
Thanks a lot, you guys are great!