Physics Q&A Game: Calculate Minimum Power for Man-Powered Helicopter

[Gokul43201]

... i hope it's clear why that should be the reason for different functioning ;)

But jdavel makes it clear the you should be able to determine whther the watch speeds up or slows down.

 

[hemmul]

But jdavel makes it clear the you should be able to determine whther the watch speeds up or slows down.

excuse me, note taken!
IMHO it should run slower - u know friction, additional inertia due to the dust particles etc... to say the truth i was unable to find the charts/plots/diagrams of the mechanisms of hand watches from the beginning of XX'th century, to make further assumptions :) hope anyone else will succeed

 

[jdavel]

hemmul,

I bet you're kicking yourself for letting me go next!

The answer isn't messy (like particles getting in or not). It's real physicsy (if that's a word), and (at least I think) pretty cool!

In fact (and this is a hint) if you knew the mass of the astronaut and the mass and dimensions of all the watch components, you could calculate quantitatively how much the watch worn by the astronaut will run slow or fast. That's a BIG hint!

 

[Davorak]

Ok I had to do some research,
http://elginwatches.org/technical/adjustments.html
http://rustyrobin.com/ElginHistory/ElginHistory.htm

The watch will get heated up through conduction from the astronaut. The balance wheel in the watch will expand causing the moment of inertia to go up and the watch to slow down.

<br /> \omega = \sqrt{\frac{m g L}{ I}}=\sqrt{\frac{g}{L}}<br />

Right?

After ~1920 "invar" and "elinvar" metals were introduced witch expand less when heated. Before that the balance wheel had adjustable screw to adjust it for different temperatures.

edit:
On rereading the article, there is another reason why the warm clock would run slower:
"The colder the watch gets, the stiffer the hairspring gets and the quicker the watch runs."

In fact the effect mentioned first does not make a noticable change.

 

[hemmul]

hemmul,
I bet you're kicking yourself for letting me go next!

hehe :) why so? its pleasure to think on non-standart question, and I'm happy its your turn ;)

The answer isn't messy (like particles getting in or not). It's real physicsy (if that's a word), and (at least I think) pretty cool!

In fact (and this is a hint) if you knew the mass of the astronaut and the mass and dimensions of all the watch components, you could calculate quantitatively how much the watch worn by the astronaut will run slow or fast. That's a BIG hint!

hmmm... i shall think on it...

 

[Davorak]

Am I right jdavel? I think I added in the edit after you saw my post. I would PM you but you do not seem to have that feature enabled.

 

[jdavel]

Davorak,

Well, it' not the solution I had in mind!

In my post #53 (written at almost the same time you were writing you solution) I gave a hint by saying that if you knew the dimensions and mass of the watch components and the astronaut you could get a quantitative solution to how slow the watch worn by the astronaut will run. That wouldn't be the case with your termperature change solution, since you have no info about relevant temperatures.

Hint: the solution is based purely on Newtonian mechanics.

 

[Davorak]

Well he did find them floating in space and still in working condition that places some restraint on the temperature. I guess I was still going off the original "no calculations necessary."

I gave a hint by saying that if you knew the dimensions and mass of the watch components and the astronaut you could get a quantitative solution

means of interaction:
electromagnetic "touch effects" or physical contact
electromagnetic radiation.
weak
strong
gravity

With your last post you have ruled out temperature, and therefore most the electromagnetic radiation interaction through inferred radiation.

The weak and strong forces can be ignored.

This only leaves gravity in my mind andelectromagnetic "touch effects". Your hint implies this by including mass as an important factor.

But Crossman’s answer has already been ruled out so no GR, SR. This leaves Newtonian gravity. Now Newtonian gravity with 90 kilo astronaut and 1 kilo watch at 0.5 meters would give 2.40228 × 10-08 Newton. Too small to have a large effect. If the gravitational force was much larger I could see it slowing the watch down through friction, but no friction was mentioned and you have implied that we could do a quantitative analysis just know the mass and the construction.

So effectively gravity is eliminated, unless I am missing something.

Now the Astronaut moved the watch as well witch could cause the watch to run faster or slower if it is an oddly made watch. Ok I know I am stretching here. It is probably something simple which has slipped every one as of yet.

None of the above answers are satisfying. So I will go out on a limb and say that these early clock have no anti recoil action on the hair spring. In essence as the watch ticks and the hair spring unwinds the watch will spin do to conservation of angular momentum. When the watch is attached to the astronaut the astronaut becomes part of the initial system.
Iastronaut>>Iwatch
The watch will not rotate because of the hairspring unwinding or watch hands moving. Thus the hair spring will act with more effective force on the rest of the watch making it run faster then the watch in space.
I guess this would fall under electromagnetic touch effects.

So I am stumped and no one else seems to have come up with anything I think it is time for you to set a timetable to reveal the answer even if no one gets it by then.

edit:
In addition many old clock can not run when weightless. So when put on by the asronaut the watch starts to work while before it was not working at all. This is a more extrem example of the last example above.

 

[jdavel]

Davorak,

Your analysis based on conservation of angular momentum is on the right track (or at least the track I had intended). But if I understand your conclusion, it's that the watch worn by the astronaut runs faster than the one left floating. I'm quite sure it's the other way around.

But since you've cracked through all my misdirection and obfuscation (only some of which was intentional!) maybe I should summarize the physics by simplifying my problem statement.

In the absense of any outside force, a mechanical watch will run slower when it is attached to another object (something with non-zero mass and volume) than when it is not attached to another object.

Hint: Mechanical watches have (at least) two springs. The drive spring provides the energy to move the hands while the balance spring is part of the system that regulates the average angular speed of the hands. The function of the drive spring is not affected by the watch being attached to another object.

 

[Davorak]

Mind posting some links to some pictures, it would help with understanding of the watches workings which seems necessary to solve the problem. I have found:
http://www.timezone.com/library/horologium/horologium631673198118416858
http://www.timezone.com/library/archives/archives631703030492773204

Here is how I came to the conclusion that the watch would be worn by the astronaut would run faster then the free floating watch.

To my limited understanding of mechanical watches, the hair spring, some times called the balance spring is attached to the frame of the watch. This is an assumption on my part since I could not find the specific information any where.

When the hair spring unwinds it provides a force on both ends of the spring and through fore a torque on the balance wheel and the frame of the watch.

This provides equal angular momentum to both the astronaut and the balance wheel, however a greater amount of energy is transferred to the balance wheel when compared to the astronaut since it has a greater freedom to move.

When a hair spring gets cold it will cause the clock to run faster, because it gets stiffer.
Since the frame of the watch is to be held still the hairspring can provide more force to the balance wheel, thus causing the watch to run faster.

That is how I thought about it. Ready to share why you think it would move slower?
I don’t know how to include the watches isochronic Adjustments into the picture. Before 1850 these adjustments apparently did not help much and were greatly improved after 1850. Any isochronic adjustment should simply reduce the time dilation between the watches.

 

[jdavel]

Davorak,

Nice pictures you found; ain't the www grand!?

Since there doesn't seem to be a whole lot of interest left in this question anymore, I think I'll declare you the winner, and give my answer.

My thinking goes like this. The balance wheel system is a simple harmonic oscilator, consisting of the hair spring, the balance wheel and either 1) the rest of the watch or 2) the rest of the watch plus the astronaut.

In short, the greater the mass of a SHO the lower the frequency. So with the astronaut included, the watch runs slow.

In more depth, for a SHO with two masses (connected by a spring) if m1 >> m2, then the end of the spring attached to m1 will hardly move, so the frequency will be almost exactly (k/m2)^1/2. But if m1 = m2, it's the center of the spring that doesn't move. So each mass oscillates as if it were attached to a spring that's half the length of the original spring. But when you cut a spring, k, in half, each piece has a spring constant of 2k. So when m1=m2 the frequency is (2k/m2)^1/2. So as m1 increases from m1 = m2 to m1>>m2, the frequency decreases. Attaching the astronaut doesn't make much difference in the frequency because even without him, m1>>m2. But with him m1>>>m2. So the watch slows down a little.

Whatdya think?

Ok, Davorak, your turn.

 

[Gokul43201]

Nice one Davorak. It's your turn now. If you don't post a new question here by tomorrow, I'll jump in again with a question.

 

[Gokul43201]

NEW QUESTION : A billiard ball of radius R is hit with a cue such that it starts out with a velocity v_0 and a backspin rate of \omega _0. Calculate the subsequent motion of the ball. What is required for the ball to return towards the cue ?

 

[siddharth]

Subsequent motion is sliding of the ball. That is, the ball slides on the ground with friction slowing down both the velcoity and angular velocity.

The condition fot the ball to come back must be that when v=rw, v=0.
There fore the relation between v0 and w0 is v0=(2/5)rwo.

 

[siddharth]

If v=rw when v=0, then the body won't come back at all. For any other value of w greater than this, then the sphere will undergo pure rolling as it comes back. Therefore it will then reach the inital point

Intially, a=f/m
there fore
v(t)=v0-(f/m)t
and
w(t)=w0-(5f/2mr)t

when v=0, t=(mv0/f)
therefore, t=(mv0)/f
at this time, w>0
wo-(5f/2mr)(mv0/f)>0
ie,
wo>(5v0/2r)

 

[Gokul43201]

That was quick ! Siddharth's reply is correct in all essential details.

For the sake of completeness, I shall paraphrase Siddharth's post with a few additional details thrown in.

Choose the positive direction of \omega, so that the velocity at the point of contact, v_c, is given by v + R \omega. Friction, F = \mu mg acting backwards (only because of the backspin - this is important) at this point, is a constant. So, the motion of the center is given by

v = v_0 \mu gt and the rotation about the center, by

\omega = \omega _0 -\mu Rmgt/I~~;~~~I = (2/5)mR^2
=&gt;\omega = \omega _0 - (5 \mu g t/2R)

The ball is sliding forwards with a constant deceleration, and spinning backwards at a constant angular deceleration. Plugging into the first equation, the velocity of the point of contact evolves as :

v_c = v_0 + R\omega _0 -(7/2) \mu gt

When a ball is rolling without slipping, there is no relative motion between the point of contact and the floor (hence no kinetic friction on a rolling ball). So here, slipping stops and rolling begins at the instant when v_c = 0. Using this condition in the previous equation tells us when this happens.

t = 2(v_0 + R \omega_0)/7 \mu g = \tau~,~~say

The value of v at t = \tau is given by R\omega _0(5v_0 /R \omega_0 - 2)/7. This gives the speed of the rolling ball after slipping has stopped. If v_0/R\omega_0 &lt; 2/5~ , this value is negative and the ball will roll backwards towards the cue. If v_0/R\omega_0 &gt; 2/5~ , the ball will roll forwards.

This makes sense. To get the ball to roll back it seems intuitive that you want large backspin, \omega _0 and small velocity v_0. The subsequesnt motion is that of a ball rolling at a uniform velocity (which in reality, decays due to rolling friction).

***********

Siddharth, your turn now to ask a question...

 

[siddharth]

The subsequesnt motion is that of a ball rolling at a uniform velocity (which in reality, decays due to rolling friction).

I thought the ball slows down because of the fact that when the ball is in contact with the ground, the surface of contact is deformed. As a result, there is an area of contact rather than a point of contact where the front part pushes the table more than the back part. As a result the normal does not pass through the centre and is shifted to the front. This Normal force produces a torque which slows the ball down.

Anyway, here's another question involving mechanics.

A ring of radius R is made from two semi circular rings of mass M1 and M2. The ring is then released from an inclined plane with angle of inclination of 30 degrees. If the ring rolls without slipping ,find
i) Angular Acceleration
ii) Normal Reaction
iii) Frictional force

Take M1=2kg, M2=4kg, and 2R=1m

 

[Gokul43201]

I thought the ball slows down because of the fact that when the ball is in contact with the ground, the surface of contact is deformed. As a result, there is an area of contact rather than a point of contact where the front part pushes the table more than the back part. As a result the normal does not pass through the centre and is shifted to the front. This Normal force produces a torque which slows the ball down.

This is exactly what constitutes rolling friction ! You will not have rolling friction without deformation.

I'll repeat your question (for clarity). The standing question is the following :

A ring of radius R is made from two semi circular rings of mass M1 and M2. The ring is then released from an inclined plane with angle of inclination of 30 degrees. If the ring rolls without slipping ,find
i) Angular Acceleration
ii) Normal Reaction
iii) Frictional force

Take M1=2kg, M2=4kg, and 2R=1m

 

[Stefan Udrea]

I don't know how to solve the problem, but I have a little observation :
Suppose A and B are the points where the 2 semi-circles join.
If,in the initial position,the ring is placed so that the tangential component of the gravity is ortogonal to AB,then the ring either slips or it doesn't move at all (if friction is high enough),in any case,it cannot rotate.

 

[Stefan Udrea]

With the notation above, if AB is on the horizontal axis of coordinates and the heavier semi-circle above the axis, the origin in the middle of the AB ,the center of mass of the ring has the coordinates :

X=0
Y= \pi \frac {R} {12} = \frac {\pi } {24} meters

 

[marlon]

NEW QUESTION : A billiard ball of radius R is hit with a cue such that it starts out with a velocity v_0 and a backspin rate of \omega _0. Calculate the subsequent motion of the ball. What is required for the ball to return towards the cue ?

this is a great question. I'll remember it for the students that i tutor at college

marlon

 

[Stefan Udrea]

the Y of the center of mass of only 1 of the semi-circles :
y=\frac {y_1 + y_2 + ... + y_n} {n} = lim \frac {1} {n} \sum_{1}^{n}y_k = \frac {1} {2R} \int_{-R}^{R} y(x) dx = \frac {1} {2R} \int_{-R}^{R} \sqrt{R^2-X^2} dx =\pi \frac {R} {4}

the Y of the center of the total mass is :

Y=\frac {2 kg \frac {\pi R} {4} + 4 kg * (- \frac {\pi R} {4} ) } {6 kg } = - \frac {\pi R } {12} = - \frac { \pi} {24}

 

[Stefan Udrea]

the angular acceleration \epsilon can be obtained from :

tangential component of gravity G_t * (Y of the mass center )=(total mass) * ( \epsilon=angular acceleration) &lt;=&gt; \epsilon=\frac {G_t y} {m} &lt;=&gt; \epsilon=\frac {Y G sin 30 } {m}
\epsilon=\frac {10 \frac {m}{s^2} 6 kg \frac {\pi} {24} } {6 kg} = \frac {\pi} {2.4} radian / second = 1.308 rad/sec /

 

[siddharth]

Ok, for clarity's sake, initially the common axis is perpendicular to the inclined plane. That is, one of the points where the two semi-circular rings are joined is in contact with the inclined plane.

 

[Stefan Udrea]

I forgot sin 30 = 0.5 , so my "answer" is 0.65 rad/s
So ,the next question : what is wrong here :D

 

[Stefan Udrea]

Siddarth, this is exactly the situation that lookst to me like it makes it impossible for the ring to rotate, because the center of the mass of the ring is in the line of the vector of the tangential component of the gravity .

 

[Stefan Udrea]

I'll drop it here, this is obviously above my head

 

[siddharth]

I forgot sin 30 = 0.5 , so my "answer" is 0.65 rad/s
So ,the next question : what is wrong here :D

Stefan, that is not the correct answer. Isn't the equation you used dimensionally incorrect?

 

[Stefan Udrea]

I used the total mass instead of the inertial moment.
It should be
\epsilon=\frac {G_t y}{I}
I=6 kg * Y^2
\epsilon = \frac {G_t Y} {6 kg * y^2 }=\frac {G sin 30 } { 6 kg Y}
\epsilon = \frac {g sin 30 m/s*s} {Y m} = \frac {5 } {pi/24 s^2} = 38.21656 rad /s^2

(this is the final edit )

 

[Stefan Udrea]

and now I'm off for 3 or 4 hours

 

[Stefan Udrea]

Actually it's the normal gravity which rotates the ball,so it's cos 30,not sin 30 .
So it should be

\epsilon = 64.7885 \frac {rad }{s^2}

 

[Stefan Udrea]

ii) the normal reaction is
-G_n Y = - 6 * 9.8 * \frac {\sqrt {3}} {2} \frac {\pi}{24}=-6.65 Newtons
iii) the frictional force :
I don't know.I know that it's miu * N , but I don't know how to calculate miu

 

[siddharth]

Stefan, neither answer is correct.

I advise you to do more problems involving rolling (like a sphere rolling down an inclined plane) before you attempt this, as it would help in clearing any conceptual doubts you have.

This is a tricky question, and there are a couple of important ideas involved.

 

[Gokul43201]

siddharth, are you suggesting that the angular acceleration is constant ?

The condition I get for the initial orientation where the ring would not roll is

cos \alpha = 3 \pi \mu / 2 \sqrt{2}

where \alpha is the angle made by the line joining the CoMs of the two halves (or the diametric line perpendicular to the one connecting the joints) and the horizontal.

 

[siddharth]

Gokul,
I am sorry if the question is not clear enough. The initial orientation is as in the diagram. The heavier part is on the lower side while the lighter part is on the higher side.
It is given that the ring rolls without slipping on the surface.
When I first posted the question, I thought that the angular acceleration will be constant, but now I am not so sure.

So, to be on the safer side, Find the INITIAL values of angular acceleration, Normal force and frictional force acting on the ring when it is released from the configuration as shown in the diagram. That will not affect the numerical answer in any way. Once those values are found, we can try to find if the angular acceleration is constant or not.

So here is the modified question. Find the INITIAL values of angular acceleration, Normal force and frictional force acting on the ring when it is released from the configuration as shown in the diagram.

 

[nishant]

1.I think the angular accn. has to be a cons here bacause the antitorqe components cannot increase aftera certain value.
siddharth,this is a long question mate and I 'll give the important steps in words and u tell me that those r correct or not.,first of all calculte the position of the centre of mass of this arrangement,it will obviously lie closer to the 4m mass,normal reaction can simply be foud by the component of weight perpendicular to the plane,after that the MI can be found about the centroidal axis of rotation,about this axis the weight and the frictional forces can be found,both will amplify each other and the body starts rolling anti clockwise,now at any time the translational velocity has to be equal to the rotational velocity so that the body does not slip,for this the angular accn is the thing to be found which can be found by dividing the normal torqe and the torqe due to the weight by MI about the centroidal axis,EQUATING THE 2 EQUATIONS WILL give us the frictional force

 

[siddharth]

first of all calculte the position of the centre of mass of this arrangement,it will obviously lie closer to the 4m mass

Yes, that idea is right.
The location of the center of mass of a semi-circular ring of radius 'r' first needs to be found. Then the location of the Center of mass of the system can be found.

normal reaction can simply be foud by the component of weight perpendicular to the plane

Not quite. There is one more step here. What is the direction of acceleration of the center of mass of the ring in this case? How can it be found? Where is the instantaneous center of rotation (ie, the point about which the ring will rotate)?

now at any time the translational velocity has to be equal to the rotational velocity so that the body does not slip

Could you elabotate on that? What is the relation between v(com) and w?

 

[nu_paradigm]

The centre of mass of each semi circular ring is :
+2R/pi and -2R/pi on the line perpendicularly bisecting the line where they are joined.
So that gives us the centre of mass of the system, i.e,

[M1(2R/pi) + M2(-2R/pi)]/(M1+M2)

= (2R/pi)[(M1-M2)/(M1+M2)]
Now taking the larger mass to be M1, we get
Com = (2 x 0.5/pi)(1/3) = 1/3pi on the side of the greater mass.

Now the two components of the weight acting will produce torque with opposite sense, taking the point of contact as the instantaneous point of rotation.

 

[Gokul43201]

Looks good so far ...

 

[nishant]

why is here the normal reaction not equal to the weight of the ring,which r the only 2 forces perpendicular to the plane of the wedge

 

[Gokul43201]

why is here the normal reaction not equal to the weight of the ring,which r the only 2 forces perpendicular to the plane of the wedge

What about the acceleration (of the CoM) of the object along this (normal) direction ? Is it zero ?

 

[Stefan Udrea]

nu_paradigm,
Could you please explain how did you get to 2R/pi ?

 

[nu_paradigm]

Sorry, didnt complete it earlier, so continuing...

The frictional force and normal reaction will not produce any torque as they pass thru the instantaneous point of rotation.

Hence, total torque acting

= r x Mgcos(30) - r x Mgsin(30) (Take -ve to be anticlockwise, as per Siddharth's diagram)

Now r = [R^2 + (1/3pi)^2]^1/2
= [0.25 + (0.0112)]^1/2
= 0.511 m

and angle between r and mgcos30 :

(theta) = 90 + sin^-1 (0.5/0.511)
= 90 + 78.1 (approx)

and angle between r and mgsin30 :

phi = 90 + 11.9 (approx)

Therefore, torque acting :
= 0.511 (6) (9.8) (1.732/2) [sin (theta)] - 0.511 (6) (9.8) (1/2) [sin (phi)]
= 5.365 - 14.7
= - 9.335 Nm (-ve implies anti clockwise direction)

Now, I (alpha) = Torque

and I = 2MR^2 (about axis perpendicular to plane thru point of contact)

so, alpha = 9.335/2(6)(0.25)
= 3.111 rad/s^2

Now the forces acting are constant, so alpha is constant. This means angular velocity is constantly increasing, so to equal it out velocity has to constantly increase (using v=wR).

Differentiating v = wR w.r.t time

we get, a = (alpha) R
gsin(30) + friction/M = (3.111) (0.5)
Mgsin30 + friction = (3.111) (0.5) M
friction = 6[1.555 - 4.9]
= - 20.07 N (-ve sign implies it is up the plane)

Now for finding normal reaction ... I was thinking divide frictional force by mu, but mu is not given.

 

[nu_paradigm]

For a semi circular ring, if M is mass, R is radius.
Take the diameter as X axis and line perpedicularly bisecting it as Y axis. Now take a radius making an angle theta with X axis. Rotate this radius by a small angle d(theta). This small angle has length of R [d(theta)] on the wire. Now "co-ordinates of this element" are (Rcos theta, R sin theta).

The wire is uniform, mass per unit length = M/pi R
so mass of small element dm = (M/ pi R) [R d(thetha)] = (M/pi) [d(thetha)]

Since Com of a uniform body is given by (1/M) integral of x dm between limits.

Com X = 1/M integral of x dm = 1/M integral of R cos thetha dm between 0 and pi
= 0

Com Y = 1/M integral of R sin thetha dm between 0 and pi
= 2 pi / R

The Com is at (0, 2 pi/R)



So

 

[nu_paradigm]

Sorry that was (0, 2R/ pi) !

 

[nu_paradigm]

Hey Siddharth is my reasoning correct? Let me know.

 

[nu_paradigm]

Hey one mistake i just noticed I've taken the heavier side to be on top. The only difference that would make is that the torques wud add up and both will be in anti clockwise direction. Also the angles will have to be recalculated.

Stupid blunder on my part. But still the reasoning is the same.

 

[nu_paradigm]

Let me edit my answers:


Hence, total torque acting

= - r x Mgcos(30) - r x Mgsin(30) (Take -ve to be anticlockwise, as per Siddharth's diagram)

Now r = [R^2 + (1/3pi)^2]^1/2
= [0.25 + (0.0112)]^1/2
= 0.511 m

and angle between r and mgcos30 :

(theta) = 90 + sin^-1 (0.5/0.511)
= 90 + 78.1 (approx)

and angle between r and mgsin30 :

phi = 78.1 (approx)

Therefore, torque acting :
= - 0.511 (6) (9.8) (1.732/2) [sin (theta)] - 0.511 (6) (9.8) (1/2) [sin (phi)]
= -5.365 - 14.7
= - 20.065 Nm (-ve implies anti clockwise direction)

Now, I (alpha) = Torque

and I = 2MR^2 (about axis perpendicular to plane thru point of contact)

so, alpha = 20.065/2(6)(0.25)
= 6.688 rad/s^2

Now the forces acting are constant, so alpha is constant. This means angular velocity is constant increasing, so to equal it out velocity has to constantly increase (using v=wR).

Differentiating v = wR w.r.t time

we get, a = (alpha) R
gsin(30) + friction/M = (6.688) (0.5)
Mgsin30 + friction = (6.688) (0.5) M
friction = 6[3.344 - 4.9]
= - 9.336 N (-ve sign implies it is up the plane)

 

[Gokul43201]

Hey Siddharth is my reasoning correct? Let me know.

Are you sure about the moment of inertia : I = 2MR^2 ?

You can edit your posts simply by using the green "EDIT" button at the bottom right.

 

[nu_paradigm]

I think so... I'm taking the axis at the point of contact, right, so the moment of inertia will have to be about an axis perpendicualr to the ring and tangential to it, so i used the parallel axis theorem to get the M.I about that axis.

i.e. M.I = MR^2 + MR^2