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

[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.

 

[jdavel]

The center of mass of the triangle structure just after the collision will be 2/3 of the way from the top corner to the base of the triangle. Call this point (x,y) = (0,0). So the top corner of the triangle is at (0, L/sqrt(3))

So, angular momentum about the point (0,0) before the collision is: M/2*v1*L/sqrt(3) = M*L*v1/2sqrt(3)
After the collision the angula momentum is: 3M*v2*L/sqrt(3) = M*L*v2*sqrt(3) where v2 is velocity of the incoming ball in the center of mass frame.
>>Conservation of angular momentum means these must be equal so: v2 = v1/6.

The linear momentum before the collision is M*v1/2.
After the collision, the linear momentum is 3M*vcm, where vcm is the velocity of the center of mass.
>>Conservation of momentum means that these must be equal so: vcm = v1/6

So, immediately after the collision, the two stuck together balls at the top are moving with velocity: v2 + vcm = v1/3 to the right.

The x and y components of the velocity of the bottom left ball are:

vy = v2*sqrt(3)/2 = v1*sqrt(3)/12 and vx = vcm - v2/2 = v1/12

and for the bottom right ball are:

vy = -v1*sqrt(3)/12 and vx = v1/12

Subsequently, all four balls continue to rotate at a speed of v1/6 around the center of mass. This uniform circular motion (in the cm frame) has a constant velocity component v1/6 to the right (+x direction) added to it at all times.

How's that?

 

[siddharth]

Jdavel, that's the right answer. Let me post the full solution anyway for clarity.

First of all, conservation of angular momentum must be done about the center of mass of the system,ie, the center of mass of all four balls and not about the center of mass of the triangle.

This is because after collision the system will rotate only about it's center of mass (in this case, center of mass of all four balls).

As mass of each vertex of the triangle after collision is 'M' the center of mass will be at the centroid, that is, 2/3 distance from the top vertex, which is

$$\frac{2}{3} \frac{\sqrt{3} L}{2}$$

$$= \frac {L}{\sqrt{3}}$$

As the triangle is equilateral, the distance to each vertex is also
$$= \frac {L}{\sqrt{3}}$$

By conservation of angular momentum
$$L_i=L_f$$

$$\frac{Mv_1L}{2\sqrt{3}} = I \omega$$

$$\frac{Mv_1L}{2\sqrt{3}} = 3(\frac{ML^2}{3} \omega)$$

$$\omega = \frac{v_1}{2\sqrt{3} L}$$

By conservation of linear momentum,
$$(\frac{M}{2}) (v_1) = 3M v_ \textrm(com)$$

$$v_ \textrm(com) =\frac{v_1}{6}$$

also the velocity of each vertex is
$$v_ \textrm(com) \vec i + \frac {L}{\sqrt{3}}\omega \vec n$$

$$\vec v = \frac{v_1}{6} \vec i + (\frac {L}{\sqrt{3}})(\frac{v_1}{2\sqrt{3} L}) \vec n$$

$$\vec v = \frac{v_1}{6} \vec i + \frac{v_1}{6} \vec n$$


Taking components, in the x and y directions for each vertex, we get the required answer

 

[Gokul43201]

New question :

According to macroscopic equilibrium thermodynamics, absolute zero can not be reached. However, there is a non-zero probability that the lattice of a finite body will be free from phonons. If a particle can be cooled to a temperature of about a milli-Kelvin (within reach of a good dilution refrigerator - no need for magnetic cooling even !), how small must it be so it can be said to have a lattice temperature of 0K for say, 90% of the time ?

PS : An "order of magnitude" approximation will do.

 

[sreenathb]

I think u don't have to finish off the extinguisher.once u attain a good velocity u don't have to spray more as in space since there is no air,so u won't be slowed down.so u need to spray more only for a change of direction.

 

[Gokul43201]

Hint for present question : Model a phonon (lattice vibration) like it were a vibrating mode of a stretched string. What is the frequency spectrum in such a case ?