Sciencemadness Discussion Board

Brain teaser: rotational problem

blogfast25 - 16-3-2016 at 12:50

Mechanics brain teaser.png - 4kB

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]

wg48 - 16-3-2016 at 14:06

That's an easy one but I will let others have a go.

Metacelsus - 16-3-2016 at 14:10

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]

blogfast25 - 16-3-2016 at 14:20

Quote: Originally posted by Metacelsus  
I get H = 5/2 times the radius of the circle centered at the red point.


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]

woelen - 16-3-2016 at 14:51

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]

woelen - 16-3-2016 at 15:17

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]

wg48 - 16-3-2016 at 16:23

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.

blogfast25 - 16-3-2016 at 16:25

Quote: Originally posted by woelen  

H = 3L/5



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]

wg48 - 16-3-2016 at 17:30

There are two possible solutions. One with the velocity zero and one none zero, my solution and moelen's solutions respectively.

blogfast25 - 16-3-2016 at 18:37

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.


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]

woelen - 16-3-2016 at 23:43

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.

blogfast25 - 17-3-2016 at 06:31

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

woelen.png - 2kB

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.

wg48 - 17-3-2016 at 07:03

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]

___ Physical Principle that the Toy is Based On and How it Works.jpg - 62kB

re

wg48 - 17-3-2016 at 08:05

Related rotational problem

wg48 - 17-3-2016 at 08:28

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]

blogfast25 - 17-3-2016 at 08:31

@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]

Metacelsus - 17-3-2016 at 13:15

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.)

blogfast25 - 17-3-2016 at 14:05

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]

Metacelsus - 17-3-2016 at 14:21

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]

blogfast25 - 17-3-2016 at 14:25

Quote: Originally posted by Metacelsus  
Actually, the moment of inertia of the sphere makes the answer different. I can assure you it is not 5/2 R.


Ah, yes. Totally. My bad. Rolling w/o slipping.

[Edited on 17-3-2016 by blogfast25]

blogfast25 - 17-3-2016 at 14:38

My answer:

$$h=\frac{27}{10}R$$

Metacelsus - 17-3-2016 at 14:57

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

blogfast25 - 17-3-2016 at 15:32

@Metacelsus:

There's some really counter-intuitive stuff that involves rolling-with-slipping. I'll see if I can find some problems like that.