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

[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

 

[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 ?