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

[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
=>\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 < 2/5~ , this value is negative and the ball will roll backwards towards the cue. If v_0/R\omega_0 > 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) <=> \epsilon=\frac {G_t y} {m} <=> \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