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

[Gokul43201]

Oops ! Misunderstood the notation. M = M1 + M2 (I guess), so that's fine ! Sorry.

 

[siddharth]

Nu_paradigm, your answers are absolutley correct. The next question is all yours.

 

[nu_paradigm]

Ok... this is not the best question I cud've put up, but here goes...

A charged particle +q of mass 'm' is placed at a distance 'd' from another charged particle -2q of mass '2m' in a uniform magnetic field B as shown(perpendicularly into the plane). The particles are projected towards each other with equal speeds 'v', where
v= Bqd/m. Assuming collision to be perfectly inelastic, find the radius of particle in subsequent motion. (Neglect electric force between the charges.)

 

[nishant]

should we also wproove that first of all that they will actually meet,or is it given that they meet?

 

[nu_paradigm]

It's given that they collide inelastically, i.e, they stick together.

 

[nishant]

Ok,so The Main Thing U Want To Ask Is In Which Way Do Thay Collide?

 

[nu_paradigm]

Ok,so The Main Thing U Want To Ask Is In Which Way Do Thay Collide?

Yes, you have to explain how they move after collision and with what radius. And also, please remember, the electric force between the charges is to be neglected.

 

[nu_paradigm]

Hey is anybody attempting this??

 

[whozum]

I was going to, but I don't really understand the question. Do they suffer some kind of deformation? If they combine, then their radius will be the sum of both their radii, or do they ocmbine into a single spherical body?

 

[Gokul43201]

The question essentially requires the follwing calculations :

0. Ignore direct coulomb interaction between the charges...this will prevent the required inelastic collision

1. Calculate the velocities (vectors) of the two charges just before collision (justify to yourself that they will collide; or find the necessary condition for this)

2. Use momentum conservation to determine the velocity after collision

3. From here, you can determine the cyclotron radius

 

[nu_paradigm]

I was going to, but I don't really understand the question. Do they suffer some kind of deformation? If they combine, then their radius will be the sum of both their radii, or do they ocmbine into a single spherical body?

I think you are getting confused as to which radius is required to be found out. Take the two charges to be point, dimensionless charges. You have to find the radius of the circular motion which they undergo after collision.

Gokul has already outlined all the steps. Only the Lorentz force is needed to be used. With a bit of geometrical considerations the answer is easily found out . And also it is given that they collide and stick together, don't bother finding the necessary condition for that. Try and draw a figure, that will help.

 

[nishant]

the necessary condition should be giving the way they collide.

 

[Gokul43201]

the necessary condition should be giving the way they collide.

Imagine the particles interact only with the B-field, and not with each other. Determine their trajectories. You will find that, at some particular time, they will both be found at the same point. This event defines their collision. From this instant on, they continue as one single particle.

 

[Stefan Udrea]

R=d*sqrt(3)

v= \frac {Bqd} {m}

radius of trajectory of the first particle :
R1=\frac {mv}{Bq}=\frac {m}{Bq} \frac{Bqd}{m}=d
radius for the second particle :
R2=\frac {2m(-v)} {B*(-2q)}=d

The initial component of velocity on the X axis is the same in absolute value and of contrary sign for the two particles ; therefore, they meet at :

x=\frac {d}{2}
y=d \frac {2- \sqrt {3}} {2} [because (d/2)^2+y^2= d^2 ]

for d/2 , the angle made by the radius of the circle of radius d with the horizontal axis is pi/3

the components of velocities on the X and Y axis :
v_x=v*sin(\frac {\pi}{3})=\frac{Bqd}{m}\frac{\sqrt{3}} {2}
v_y=v*cos(\frac {\pi}{3})=\frac {Bqd}{m} \frac {1}{2}



When they collide,the conservation of momentum :

- for the X axis :
mv_x-2mv_x=3m*v_xf (v_x_f =final v_x ;of the new heavier particle) v_x_f= -\frac{v_x}{3}=- \frac{Bqd}{m}\frac {\sqrt{3}} {6}
-for the Y axis :
mv_y+2mv_y=3mv_y_f => v_y_f=v_y=\frac {Bqd}{m}\frac{1}{2}

final v is
\sqrt{{v_x_f}^2+{v_y_f}^2}=\frac {Bqd}{m} \frac{1}{\sqrt{3}}

radius of the trajectory of the "composite" particle is :

R= \frac {3m \frac {Bqd} {m} \frac {1} {\sqrt {3} } } {Bq} =<br /> =\frac {3d} {\sqrt{3}} =d \sqrt {3}

 

[Stefan Udrea]

is this correct ?

 

[nu_paradigm]

Correct !

Stefan, your answer is perfectly correct . Go on... your turn to ask a question.

 

[Stefan Udrea]

Ok,here's the problem :
There are 2 towns on the surface of the Earth, the distance between them is thousands of kilometers.
There is a *straight* tunnnel digged from one town to the other,and a railway in the tunnel.
Show that the train ,which circulates throough the tunnel between the two towns ,doesn't need a ... locomotive.
Neglect the frictions.

(edited to make it bold )

 

[torchbear]

I had to look up some power consumption data. If a human runs at 24 km/h, the power consumption is about 1.7 kW. So it's not a feasible construction (not for me at least).[/QUOTE]

Maybe I am out of the thread. But I wonder how 1.7 kw was reasoned out. If we consider a pure mechanical way( the friction from air and ground,etc), could anybody clear this up for me?

 

[jdavel]

Ok,here's the problem :
There are 2 towns on the surface of the Earth, the distance between them is thousands of kilometers.
There is a *straight* tunnnel digged from one town to the other,and a railway in the tunnel.
Show that the train ,which circulates throough the tunnel between the two towns ,doesn't need a ... locomotive.
Neglect the frictions.

(edited to make it bold )

Assuming both towns are at the same elevation, the potential energy of both locations is the same. So the total work needed to get from one point to another is zero. The train falls to the center point of the track, and then its momentum carries it back up to its destination.

 

[Stefan Udrea]

jdavel, this is too simplistic.Where do you draw the line of zero potential energy ?

 

[extreme_machinations]

Are the towns connected vertically[at a small angle to the axis of rotation ] on the Earth's's surface or otherwise or does it matter .

i think this can be corellated with electrical equipotential surface ,since the work done to move a charge on an equipotential surface is zero ,

if the Earth is a gravitational equipotential surface then the work done to move a body with certain mass, on it from one point to another would be zero .but we do this everyday with considerable loss of energy ,bah ! so there's some loophole in the question which i 'm unable to exploit .

this is just a wild stab in the dark ,i don't think my concepts in the gravitation are clear enough ,but hey i don't want to regret not sharing my brainwave .

i'll go look it up .

 

[jdavel]

jdavel, this is too simplistic.Where do you draw the line of zero potential energy ?

The customary location to define zero potential is at infinity. But I don't see what difference it makes. As long as the PEs at the beginnning and end of the tunnel are the same, no energy is used getting from one to the other. What am I missing?

 

[whozum]

I'm following you jdavel.

 

[Galileo]

The customary location to define zero potential is at infinity. But I don't see what difference it makes. As long as the PEs at the beginnning and end of the tunnel are the same, no energy is used getting from one to the other. What am I missing?

The work done on getting the train from A to B may be zero, but that doesn't mean no locomotive is needed. There might be a potential barrier in between which the train cannot cross.

 

[extreme_machinations]

IF THE TUNNEL IS DUG THROUGH THE CENTRE OF THE EARTH THEN ,THINGS GET A LOT MORE EASIER .

ASSUMING THAT THE TUNNEL IS DUG THROUGH THE CENTRE :
let the train of mass m be at a distance r from the centre of the Earth ,
the portion outside this radius will not exert any net gravitational force on the train .

let the mass inside this radius be Me ,this can be treated as the mass of a particle lying at the centre of the Earth ,
let F be the force b/w the train and Me

F = GmMe/r^2......[eqn 1]

now volume of the sphere of radius r IS[ 4/3 pie r^3]
let P be the density of this sphere of radius r ,

Me=P[4/3pie r^3] ,substituting this in eqn 1

F=4/3{pie}GmPr

4/3{pie}GmP is a constant ,which we take as K

THEREFORE

F=-Kr ,[minus sign indicates that F{vector} and r{vector} have opposite directions]

this is the form of hooke's law ,hence the TRAIN WILL OSCILLATE LIKE A BLOCK ON A SPRING ,

BUT ALL OF THIS HAPPENS UNDER HIGHLY IDEALIZED CONDITIONS ,LIKE DENSITY P BEING A CONSTANT WHICH IS NOT TRUE ,AND THE TUNNEL IS DUG THROUGH THE CENTRE OF EARTH .

 

[Stefan Udrea]

ok, extreme_machinations, could you give it a try with the assumption that the tunnel doesn't go through the center of the Earth ?

 

[jdavel]

The work done on getting the train from A to B may be zero, but that doesn't mean no locomotive is needed. There might be a potential barrier in between which the train cannot cross.

The question specified that the tunnel was "straight". How can there be a barrier if the tunnel is straight? In the first half aren't you always going down hill and in the second half up hill?

 

[DarkEternal]

i remember doing a similar tunnel problem on a problem set which asked for the trip time, which i believe is something like 42 minutes regardless of the location of the cities. use gauss's law to derive a periodic equation of motion; the location-dependent quantities should cancel out. if no one's done it by the time I'm done with finals, i'll write up a solution.

 

[Stefan Udrea]

extreme_machinations,
Because you were the closest to find the solution, you get to ask the next question.

 

[jdavel]

extreme_machinations,
Because you were the closest to find the solution, you get to ask the next question.

What was the solution you were looking for?

 

[extreme_machinations]

thank you sir ,i'll come up with something .

 

[Stefan Udrea]

:smile : don't call me "sir" , I'm just a college student.

jdavel,

In the drawing,
AB is the tunnel , the two cities are in A and B.
T is a random position of the train in the tunnel, at the distance r from the center of the Earth ; r<R
Because the volume of the Earth between r and R has no gravitational influence on the train, we have the gravitational attraction on the train F=kmM/r^2 , where M is the mass of the sphere of radius r.
M_0is the total mass of the Earth
M=M_0 \frac{r^3}{R^3}
The movement of the train is given by the component F_e ,parallel with the tunnel.
Then,
F_e=K \frac {m M_0}{R^3} y =ky
\vec{F_e}=-k \vec{y}

So the movement in the tunnel is a harmonic oscillation given by :

y= \sqrt {R^2-d^2}sin(\omega t+\frac{\pi}{2} )
y=w \sqrt{R^3-d^2}cos(\omega t+\frac{pi}{2})
T_0=2 \pi R \sqrt{\frac{R}{KM_0}}
Earth.bmp

 

[Stefan Udrea]

This wasn't my idea ; I found the problem and the solution in a book.

 

[jdavel]

:smile : don't call me "sir" , I'm just a college student.

jdavel,

In the drawing,
AB is the tunnel , the two cities are in A and B.
T is a random position of the train in the tunnel, at the distance r from the center of the Earth ; r<R
Because the volume of the Earth between r and R has no gravitational influence on the train, we have the gravitational attraction on the train F=kmM/r^2 , where M is the mass of the sphere of radius r.
M_0is the total mass of the Earth
M=M_0 \frac{r^3}{R^3}
The movement of the train is given by the component F_e ,parallel with the tunnel.
Then,
F_e=K \frac {m M_0}{R^3} y =ky
\vec{F_e}=-k \vec{y}

So the movement in the tunnel is a harmonic oscillation given by :

y= \sqrt {R^2-d^2}sin(\omega t+\frac{\pi}{2} )
y=w \sqrt{R^3-d^2}cos(\omega t+\frac{pi}{2})
T_0=2 \pi R \sqrt{\frac{R}{KM_0}}
Earth.bmp

Well...that's a lot more than is necessary to "Show that the train ,which circulates through the tunnel between the two towns ,doesn't need a ... locomotive." The fact that the path never goes above the starting point is sufficient to show what you asked for.

 

[Stefan Udrea]

You gave an intuitive answer.I didn't mean it as a "yes/no" question , I wrote "Show".

 

[extreme_machinations]

now who really gets to ask the next question ,
me or u again stefan ?

 

[whozum]

Were all waitin on you.

 

[Gokul43201]

Extreme : go ahead and post a question.

 

[extreme_machinations]

OK here's the question ,not worth the wait i guess but anyway .


A THIN SUPERCONDUCTING RING OF RADIUS r_0 AND INDUCTANCE{zero resistance } IS HELD ABOVE A VERTICAL CYLINDRICAL MAGNETIC ROD . THE AXIS OF SYMMETRY OF THE ROD AND THE RING ARE SAME .THE CYLINDRICALLY SYMMETRICAL MAGNETIC FEILD AROUND THE RING CAN BE DESCRIBED IN TERMS OF ITS VERTICAL AND RADIAL COMPONENTS AS B_z=B_0(1-aZ) AND B_r=B_0br_0 ,WHERE B_0 ,a ,b are all constants .THE CURRENT IN THE LOOP IS ZERO WHEN IT'S RELEASED FROM ORIGIN z=0 and r=0 .
DETERMINE HOW THE RING MOVES SUBSEQUENTLY.

[ps i don't know how to attach the fig ,so imagine a vertical cylinder and ring kinda levitating on top with magnetic field looking like a fountain coming out of the top part of the cylinder . The vertical and radial components of magnetic field are perpendicular to each other .]duh!

 

[BoTemp]

Should we assume the ring initially accelerates under the influence of Earth gravity?

 

[extreme_machinations]

yes as the ring is placed in the magnetic field , two forces will act on it along the
z- axis one of them being gravity [mg] and the other one is lorentz force ,
you'll have to figure out the directions yourself

 

[extreme_machinations]

is this too tough or too damn stupid to be tried ,
it seems I've killed yer game gokul .

 

[whozum]

I was going to give it a shot just since no one else does, but I have to ask, in your magnetic field equations are the variables that the mag field decreases by Z and R respectively?

 

[BoTemp]

I worked out the problem but I haven't had the time to post it online, and since
my internet at home is out for a little while I probably won't for week or so. Also, is that B_r=B_0br_0 or B_r=B_0br ?

PS I'm not asking anybody to not post a sol'n, btw. Just saying the game is not yet over.

 

[siddharth]

Total flux enclosed is \int B.ds + Li - I

= [Bo(1-az)k + (B_0 b r_0)(r)].(\pi r_0^2)k + Li (k is unit vector in z direction and r is the unit vector in the radial direction)

= B_0(1-az)\pi r_0^2 + Li

Now this value of flux (as written from I) can be taken as constant as
|E|=(d(Flux)/dt)=iR
as R=0, d(Flux)/dt=0.
Therefore Flux enclosed is constant - II

when z=0,i=0.Therefore Flux = (pi)(r0)^2(B0)

From II

(pi)(r0)^2(B0)= = B_0(1-az)\pi r_0^2 + Li


From that we find the current to be
i= \frac {B_0a\pi r_0^2 z}{L}

The force acting on the wire in the z direction is F= B(r)(i)(L)

= - \frac{B_0^2 b r_0 a \pi r_0^2 z}{L}(2\pi r_0)

=-kz (Opposite to direction of displacement of ring)

therefore ma=F-mg

and a=-kz/m - g

this means that the ring undergoes SHM about the mean position zo=-mg/k

So z(t) - zo = Acoswt
and at t=0, z=0, A=-z0

z(t) = -zo(Coswt - 1)
from this it is seen that the z - coordinate is never positve and therefore the magnetic force is always upwards.

P.S: Extreme_machinations, you took this question from the Y.G File right? I knew i did this question sometime ago.

 

[extreme_machinations]

EEEEEEEEEEEEEYYYYUUUPPPP! THAT'S RIGHT .
SIDHARTH ,that ws .."BRILLIANT"........{mabe not} ,HEHEHEHEHEHEHE

SRY BOTEMP early bird takes the worm {B_r=B_0br_0, is correct,
r_0 is constant }

SO I SUPPOSE NEXT ONE'S ALL YOURS SIDHARTH .
I REALLY WISH SOMEONE WOULD POST A THEORETICAL QUESTION INVOLVING MORE PHYSICS AND LESS MATH .

 

[siddharth]

Here is a simple one.
Three balls, each of Mass 'M' ,are connected by means of massless, rigid rods of length 'L' as shown in the diagram so as to form an equilateral triangle. This triangle is placed on a frictionless, horizontal table. A fourth ball of mass 'm', traveling parallel to the base of the Triangle, with velocity v1, collides with the other ball and gets stuck to it.
Find the velocities of each ball after collision and describe the subsequent motion of all four balls.

 

[Doc G]

Here goes my answer based on very little physics background:

Assuming that mass ‘m’ is lesser or equal to ‘M,’ I would imagine that in the collision, it would transfer it’s velocity to the ‘triangle of balls’ which would then move with a velocity of m/3M * V1 along a line parallel to that which the initial ball struck (Each ball in the triangle would be traveling at this velocity of m/3M * V1 ).

Since the surface is frictionless, there would be no forces to cause the triangle to turn.

Having transferred all of its kinetic energy to the ‘triangle’ of balls, the ball of mass ‘m’ would remain in the position where it had struck the triangle of balls.

 

[jdavel]

siddharth,

The physics may be "simple" but unless there's a trick I'm missing, the math looks like a mess!

How about making the top mass of the triangle and the incoming mass both equal to M/2?

 

[siddharth]

How about making the top mass of the triangle and the incoming mass both equal to M/2?

Ok Fine. Take the top mass of the triangle and the incoming mass to be equal to M/2. The bottom two masses are still M.

Let me restate the question

Three balls, two of Mass 'M' , and the top one of mass 'M/2' are connected by means of massless, rigid rods of length 'L' as shown in the diagram so as to form an equilateral triangle. This triangle is placed on a frictionless, horizontal table. A fourth ball of mass 'M/2', traveling parallel to the base of the Triangle, with velocity v1, collides with the other ball and gets stuck to it.
Find the velocities of each ball after collision and describe the subsequent motion of all four balls.


Doc G. The ball after collision with the top of the triangle, gets embedded to it.