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

[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

 

[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.