blogfast25
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Brain teaser: rotational problem
A bob is connected by means of flexible string of length L to static point O. On release the string hits a thin horizontal bar
perpendicular to the page (red point), at distance H from O. This causes the bob to starts rotating about the bar.
Determine the distance H so that when the bob reaches the highest point O', the tension T in the string is zero.
Ignore all friction or drag. The string is massless.
[Edited on 16-3-2016 by blogfast25]
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wg48
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That's an easy one but I will let others have a go.
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Metacelsus
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I get H = 5/2 times the radius of the circle centered at the red point.
How I solved it:
v^2 / r = g, where v is the velocity at the top of the loop (this equation comes from the fact that tension will be 0).
gh = 1/2*v^2, where h is the height above the top of the loop (this equation comes from conservation of energy)
rg= 2gh -> r = 2h -> h = r/2
Since this height is the height above the top of the circular path, the total height H is r/2 + 2r = 5/2*r
(Another assumption I'm making is that the string is massless).
Edit:
I see I mixed up the definitions of the variables in my derivation, as I was just looking at the diagram instead of reading the text carefully. This
means that 5/2*R = L, not H, and so H = 3/5 L (as Blogfast and Woelen showed below).
[Edited on 3-17-2016 by Metacelsus]
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blogfast25
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And what is the radius of the circle centred at the red point?
I.o.w., express H as a function of L?
[Edited on 16-3-2016 by blogfast25]
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woelen
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R: Radius of circle around bar
H + R = L
conservation of energy: at top of small circle: ½mv² = mg*(H-R)
Balance of centripetal force and gravity at top: mv²/R = mg
mv² = R*mg
This yields two equations:
H + R = L
½R = H - R ==> H = 3R/2
R = 2L/5
H = 3L/5
[Edited on 16-3-16 by woelen]
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woelen
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Quote: Originally posted by Metacelsus  | [...]Since this height is the height above the top of the circular path, the total height H is r/2 + 2r = 5/2*r
[...] |
Your solution is almost correct, but I think this quoted equation is a mistake.
In your terms you do not have H = r/2 + 2r, but:
L = r/2 + 2r
H = r/2 + r
From this follows:
r = 2L/5
H = r/2 + r = L/5 + 2L/5 = 3L/5
[Edited on 16-3-16 by woelen]
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wg48
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Assuming no losses the mass can return to its original height when its velocity will be the same as the stating velocity, zero hence no tension in the
string.
So the start height must be equal to the final height so H equals r.
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blogfast25
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Yes, woelen's got it.
Conversion of potential to kinetic energy:
$$\Delta U=\Delta K$$
$$mgL-2mg(L-H)=\frac12 mv^2$$
$$(2H-L)g=\frac12 v^2$$
$$v^2=2g(2H-L)$$
For:
$$T=0$$
Then:
$$g=\frac{v^2}{R}$$
$$g=\frac{2g(2H-L)}{L-H}$$
$$H=\frac35 H$$
[Edited on 17-3-2016 by blogfast25]
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wg48
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There are two possible solutions. One with the velocity zero and one none zero, my solution and moelen's solutions respectively.
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blogfast25
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| Quote: Originally posted by wg48 | | There are two possible solutions. One with the velocity zero and one none zero, my solution and moelen's solutions respectively.
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The solution with zero velocity is only physically possible if there's no bar at distance H as a constraint (analytically: H = 0).
That solution occurred to me as well but it's hardly in the spirit of the problem though. 
In that case the bob would continue swinging until its height was L again: then (obviously) v = 0. It would behave like a (large
angle) pendulum. That's not the case here.
[Edited on 17-3-2016 by blogfast25]
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woelen
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| Quote: Originally posted by wg48 | | There are two possible solutions. One with the velocity zero and one none zero, my solution and moelen's solutions respectively.
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Initially I also thought that there was a second solution with R equal to L/2 and H equal to 0, but this is
not the case. In that situation, the velocity would be equal to 0, but the gravitational pull would be m*g. There would be negative tension in the
wire. In reality this does not occur. Before the bob reaches the top of the circle, it will fall down already, while it still has a horizontal
velocity component. The bob will not move along a circle in the higher parts of its movement, the wire will not be stretched all the time. The
trajectory in fact will be quite interesting. The bottom part will be part of a circle, the top part will be part of an inverted parabola and these
two will be smoothly connected.
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blogfast25
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| Quote: Originally posted by woelen | | Initially I also thought that there was a second solution with R equal to L/2 and H equal to 0, but this is not the case. In that situation, the
velocity would be equal to 0, but the gravitational pull would be m*g. |
I don't see it.
Higher up you correctly wrote:
H + R = L
so R = L - H, with H = 0, R = L
As the bob arrives in point 2, its velocity is zero, the centripetal force is zero and mg acts perpendicularly to the string. So T =
0.
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wg48
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| Quote: Originally posted by woelen | | Quote: Originally posted by wg48 | | There are two possible solutions. One with the velocity zero and one none zero, my solution and moelen's solutions respectively.
|
Initially I also thought that there was a second solution with R equal to L/2 and H equal to 0, but this is
not the case. In that situation, the velocity would be equal to 0, but the gravitational pull would be m*g. There would be negative tension in the
wire. In reality this does not occur. Before the bob reaches the top of the circle, it will fall down already, while it still has a horizontal
velocity component. The bob will not move along a circle in the higher parts of its movement, the wire will not be stretched all the time. The
trajectory in fact will be quite interesting. The bottom part will be part of a circle, the top part will be part of an inverted parabola and these
two will be smoothly connected. |
Yes I think your correct, my solution was wrong. When playing with toy cars (not recently lol) I had assumed the effect you described was due to
friction. So the minimum starting height for a car or rollercoaster to make a loop is 1.5 times the loop height/diameter (see pic) luckely I have
never designed a roller coaster lol.
[Edited on 17-3-2016 by wg48]
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wg48
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re
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wg48
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Related rotational problem
Design a daredevil track similar to the track, shown in my pic above, with the addition of a horizontal track inside of the loop such that when the
car falls of the loop it lands on the track having done precisely 180deg flip and lands with the minimum possible vertical velocity.
What is the height of the starting point and the height of the additional track relative the bottom of the loop as function of the loop diameter.
Assume the car is unpowered, frictionless and negligible wheel base length.
I don't know the answer or if there is only one. I just thought it was an I intriguing problem following on from woelen's comments. I guess a 360deg
flip is not possible. Feel free to manipulated the problem to get an answer or a sensible answer (whatever that means).
[Edited on 17-3-2016 by wg48]
[Edited on 17-3-2016 by wg48]
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blogfast25
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@wg48:
Yes, if the toy car's tangential velocity became zero at the highest point of the loop, the only net force acting on it would be mg, so it
would have to fall down by Newton's Law. The minimum speed allowable is that where:
$$mg=\frac{mv^2}{R}$$
$$v^2=gR$$
If we launch a toy car from h above the bottom of the loop (and v0 = 0), then
$$mg(h-2R)=\frac12 mgR$$
So:
$$h=\frac52 R$$
[Edited on 17-3-2016 by blogfast25]
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Metacelsus
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New brain teaser (I hope it's OK if I post this in this thread):
A sphere of uniform density rolls (without slipping) down a track like the one in wg48's picture. What is the minimum release height h, as a function
of R, for it to make it all the way along the loop? You can assume that the radius of the sphere is small, and that there is no energy lost to
friction.
(This was actually an exam question in my high school physics class.)
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blogfast25
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| Quote: Originally posted by Metacelsus | New brain teaser (I hope it's OK if I post this in this thread):
A sphere of uniform density rolls (without slipping) down a track like the one in wg48's picture. What is the minimum release height h, as a function
of R, for it to make it all the way along the loop? You can assume that the radius of the sphere is small, and that there is no energy lost to
friction.
(This was actually an exam question in my high school physics class.) |
Errrmm. That has already been solved, Metacelsus. In the post just above yours. Toy cars or uniform spheres, as long as there's no friction involved
the answer is always:
$$h=\frac52 R$$
http://www.sciencemadness.org/talk/viewthread.php?tid=65628#...
[Edited on 17-3-2016 by blogfast25]
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Metacelsus
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Actually, the moment of inertia of the sphere makes the answer different. I can assure you it is not 5/2 R.
Hint: KE = 1/2 mv2 + 1/2 Iω2
[Edited on 3-17-2016 by Metacelsus]
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blogfast25
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Ah, yes. Totally. My bad. Rolling w/o slipping.
[Edited on 17-3-2016 by blogfast25]
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blogfast25
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My answer:
$$h=\frac{27}{10}R$$
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Metacelsus
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Yes, that's correct.
Derivation:
As with the non-rolling particle, v2 = gR
Here, KE = 1/2 mv2 + 1/2 Iω2
= 1/2 mv2 + 1/2 (2/5*mr2)ω2 (from the moment of inertia of a sphere)
= 7/10 mv2 (from the condition that the sphere is rolling without slipping)
The initial potential energy is mgh, the final potential energy is mg(2R), and the final kinetic energy is 7/10 mv2
Therefore, mgh = mg(2R) + 7/10(mv^2)
Substituting v^2 = gR,
mgh = mg(2R) + mg(7/10 R)
so h = 27/10 R
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blogfast25
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@Metacelsus:
There's some really counter-intuitive stuff that involves rolling-with-slipping. I'll see if I can find some problems like that.
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