Help Needed: Determining Speed of Dart After Being Fired

[Pyrrhus]

Actually, dextercioby is right, because when the bullet hits the block and it swings up to a theta max then the block will accelerate back to the lowest point because of gravity where it will only have max velocity, therefore the acceleration will be 0, and this is definately a simple pendulum or mathematical pendulum where the dimensions of the block are not taken into account.

 

[apchemstudent]

Actually, dextercioby is right, because when the bullet hits the block and it swings up to a theta max then the block will accelerate back to the lowest point because of gravity where it will only have max velocity, therefore the acceleration will be 0, and this is definately a simple pendulum or mathematical pendulum where the dimensions of the block are not taken into account.

What? So what if it has max velocity... It's velocity is still changing after that so an acceleration must exist... If there was no acceleration, then it's velocity will stay the same magnitude and direction in this case...

 

[vincentchan]

I really don't see the why ppl don't agree with me an apchemstudent. I think we have made a valid point here... unless you think the speed of the block is unrelated to the tension on the string... How could Mr. dex get confuse in a high school's problem...?

 

[dextercioby]

I really don't see the why ppl don't agree with me an apchemstudent. I think we have made a valid point here... unless you think the speed of the block is unrelated to the tension on the string... How could Mr. dex get confuse in a high school's problem...?

Jesus,guys u're really getting funny... :tongue2: I ask u one more time...What is the maximum tension in the wire??

Daniel.

 

[dextercioby]

I'm not confused,i'm clear all along...You're the ones confused...

Daniel.

 

[apchemstudent]

Jesus,guys u're really getting funny... :tongue2: I ask u one more time...What is the maximum tension in the wire??

Daniel.

Do you really want us to say it... or you're just going to disagree with it which means it will be a waste of time typing it out...

 

[dextercioby]

Do you really want us to say it... or you're just going to disagree with it which means it will be a waste of time typing it out...

Yes,i want u to say it...Do u think I'm fooling around??It's a serious matter...

Daniel.

PS.If u think of it as a waste of time,that's your choise...

 

[vincentchan]

I really don't want to do that but it seems i have to do a little bit calculation here to convince Mr dex.

let $$\theta_{max}$$ is the maximun angle between the string and vertical, compare with the lowest point
At the highest point $$E=KE + PE = 0+ l(1-\cos{\theta_{max}})mg$$
At the lowest point $$E=1/2 mv^2$$
by the light of conservation of energy,
$$l(1-cos{\theta_{max}})mg=1/2mv^2$$
$$v=\sqrt{2(1-cos{\theta_{max}})lg}$$
$$F_{cen} = mv^2/l$$
notice the string FELLs the block PULLING DOWNWARD, so do the gravity...
the tension is ADDING $$T_{cen}$$ and mg together, NOT SUBTRUCTING...
$$T = F_{cen} + Mg$$

 

[apchemstudent]

I really don't want to do that but it seems i have to do a little bit calculation here to convince Mr dex.

let $$\theta_{max}$$ is the maximun angle between the string and vertical, compare with the lowest point
At the highest point $$E=KE + PE = 0+ l(1-\cos{\theta_{max}})mg$$
At the lowest point $$E=1/2 mv^2$$
by the light of conservation of energy,
$$l(1-cos{\theta_{max}})mg=1/2mv^2$$
$$v=\sqrt{2(1-cos{\theta_{max}})lg}$$
$$F_{cen} = mv^2/l$$
notice the string FELLs the block PULLING DOWNWARD, so do the gravity...
the tension is ADDING $$T_{cen}$$ and mg together, NOT SUBTRUCTING...
$$T = F_{cen} + Mg$$

I had basically the same setup as vincentchan... I'd just like to add something (i've mentioned before) that proves Fc cannot be 0:

Fc = 2*(Mo+m)*(1-cos(theta))*g

Mo and m cannot be 0, g is a constant:

again... The only 0 this equation has is if theta was 0 degrees which means that the string has not been swung at all, which is untrue...

 

[dextercioby]

Are u telling me that for $\theta_{max} =0$ the linear velocity is zero?

Daniel.

 

[vincentchan]

yes, coz if $$\theta_{max} = 0$$ the ball starts at the lowest point. Therefore, at the lowest point, the speed it zero...

 

[dextercioby]

yes, coz if $$\theta_{max} = 0$$ the ball starts at the lowest point. Therefore, at the lowest point, the speed it zero...

Jesus,do you "listen" to yourself??That is a pendulum,it cannot have zero speed at it's lowest point.It has zero speed where the amplitude is maximum.It has maximum speed at it's lowest point.

Daniel.

 

[vincentchan]

if the maximun amplitude is zero, what is its maximun speed? Mr dex

 

[apchemstudent]

Jesus,do you "listen" to yourself??That is a pendulum,it cannot have zero speed at it's lowest point.It has zero speed where the amplitude is maximum.It has maximum speed at it's lowest point.

Daniel.

you don't get the point... we're trying to say that it WILL have a max of 0 speed if theta was 0. Where theta is the angle between the vertical the highest point swung... Obviously it will have the max speed at the lowest point, which will contribute to force of tension...

 

[dextercioby]

Yep,you're both right. I was trying to make a false point.I'm wrong...This time...Too bad it took me a lot of time to realize...
The correct formula can be found here

Well,i'm human after all.I didn't say i was always right.Nobody is always right,though.I realized i was wrong when i started calculating...
So i'd say it's a black ball for me... :yuck:

Daniel.

 

[Gokul43201]

I get

$$T = g(M+m)(3-2cos \theta _{max})$$

Shocked that dex and cyclo missed this !

Edit : When I looked at this, dexter hadn't posted this last post yet. His link gives the same answer, but in this case, the mass is M+m.

 

[dextercioby]

I get

$$T = g(M+m)(3-2cos \theta _{max})$$

Shocked that dex and cyclo missed this ! :

Well,it happens... :tongue2: Reminds of some chass matches i played years ago...I was blind... :grumpy: Sometimes is just not see obvious things,like this problem...

Daniel.

 

[Pyrrhus]

Well i guess that happens sometimes, i cannot believe it too, when i read the vincent reply i knew something was wrong, nice vincent and apchem noticed, and kept on trying to prove their point

Well if you are going to bring up chess matches, dexter, i used to have an history of one particular one in which i lost the match, and there was a checkmate i didn't see, it was a sad game for me

 

[Gokul43201]

Of course, it happens.

I goofed up in another thread, earlier today, with a wrong substitution, and realized it only much later when I wrote down the formula to substantiate my result.

 

[physicsgirl101]

Alright, now that ya'll have come to a consensus, can you make it understandable to me? (because I'm so confused after hearing everyone's different comments I don't know what is right anymore). Basically, when I've dealt with pendulums, I've only dealt with radial acceleration in the form: a=v^2/r
And I've never heard of centripetal/centrifugal forces...

So can anyone break down the work you did to show it in a simple way that i will understand, please? I'd REALLY appreciate it...

 

[dextercioby]

Okay,the two bodies which are joint together are in a circular movement round the point of suspension "O".
The Second law of Dynamics for this body,projected onto the radial direction reads
$$T-(m+M)g\cos\theta =\frac{(m+M)v^{2}}{l}$$ (1)
,where $\theta$ is the angle at center,"T" is the tension in the in wire and [itex] \frac{v^{2}}{l} [/tex] is the radial acceleration.

Now,write down the law of energy conservation between the highest point (in which the system has only potential energy) and the lowest point in which the system has only kinetic energy:
$$\frac{(m+M)v^{2}}{2}=(m+M)gl(1-\cos\theta_{max})$$ (2)
Note that this formula yields the maximum linear velocity the body (total) can have.Therefore,combining (1) and (2),taking into account that in both formulas,the subscript 'max' for angles and velocity interviens
$$T_{max}=(m+M)g+2(m+M)g(1-\cos\theta_{max})$$(3)

,which can be simplified to
$$T_{max}=(m+M)g(3-2\cos\theta_{max})$$(4)

The reason for why I've set $\theta=0$ is that u must maximize the function T.Which means maximize the velocity (which is done in formula (2)) and maximize the gravity as well,which is done by setting $\cos\theta =1$ or,equivalently,$\theta=0$

Daniel.

 

[Yapper]

a=v^2/r is centripetal acceleration, multiply by mass for force. And I am now also confused, the problem suggest that the solution should use theta max, L, g, m, Mo. Now if centripetal accerleeration is taken into affect it should be FT = (m+Mo)g - Fc. Because centripetal force is in the same direction as force tension.

You can find Fc by using the maximum y height in potential energy formula, and transforming it into kinetic energy, then find the velocity.
With the velocity you can calculate Fc.

I got 2g(L-Lcos(theta))=v^2

(m+Mo)g - [(m+Mo)(2g(L-Lcos(theta)))]/L = Fnet

I think this is right, I coudl be very wrong as I just learned this stuff my self

 

[dextercioby]

The mademoiselle insisted we do it as simple as possible,without using the notion of centripetal force/accleration.I believe that could be done...And my last post simply confirms it...

Daniel.

EDIT:IIRC,in high-school u're being taught centripetal forces/accelerations.It's just that she hasn't been taught these things yet...

 

[Yapper]

Jeez... You guys should use stuff that we learn in High School. As this is the k-12 forums. Applying higher level formulas to HS problems might be easier but confuses us simple minded ones. :)

Edit: Lol You think the second law of dynamics is simple? It looks like you could derive it from HS formulas but I don't think that they give it in that form, thus it might be confusing. The lady new what centripital acceleration was, if you read, she just knew it by a different name.

 

[physicsgirl101]

Okay,the two bodies which are joint together are in a circular movement round the point of suspension "O".
The Second law of Dynamics for this body,projected onto the radial direction reads
$$T-(m+M)g\cos\theta =\frac{(m+M)v^{2}}{l}$$ (1)
,where $\theta$ is the angle at center,"T" is the tension in the in wire and [itex] \frac{v^{2}}{l} [/tex] is the radial acceleration.

Now,write down the law of energy conservation between the highest point (in which the system has only potential energy) and the lowest point in which the system has only kinetic energy:
$$\frac{(m+M)v^{2}}{2}=(m+M)gl(1-\cos\theta_{max})$$ (2)
Note that this formula yields the maximum linear velocity the body (total) can have.Therefore,combining (1) and (2),taking into account that in both formulas,the subscript 'max' for angles and velocity interviens
$$T_{max}=(m+M)g+2(m+M)g(1-\cos\theta_{max})$$(3)

,which can be simplified to
$$T_{max}=(m+M)g(3-2\cos\theta_{max})$$(4)

The reason for why I've set $\theta=0$ is that u must maximize the function T.Which means maximize the velocity (which is done in formula (2)) and maximize the gravity as well,which is done by setting $\cos\theta =1$ or,equivalently,$\theta=0$

Daniel.

Thanks... but it's still a little complicated because I've never used those formulas before... how did you get the law of energy conservation equation?

 

[vincentchan]

do you know the formulas of potential energy and kinetic energy, if no, here is the formulas:
PE= mgh (h is the height between the object and the floor)
KE= 1/2 mv^2
and the total energy is PE+KE
since energy cannot come out from nothing, the total energy for the system at anytime is the same... just compute the total energy at the lowest point and the highest point... set them equal... vala, you have the formulas of conservation of energy...

 

[dextercioby]

Simple.In this case,the law of total energy conservation can be applied and it reads,for two arbitrary states.

$$KE(1)+PE(1)=KE(2)+PE(2)$$
The explicit expression i posted can be deduced from the geometry of the figure.
You chose the lowest point of the circular trajectory as to have 0 potential energy and the upmost to have only potential energy and the kinetic one to be zero.Analyzing the geometry,u can reach the formula almost immediately.

Daniel.

 

[physicsgirl101]

Thanks, I'll let you know if i get the right answer when I do it out----THANKS!

 

[physicsgirl101]

Okay,the two bodies which are joint together are in a circular movement round the point of suspension "O".
The Second law of Dynamics for this body,projected onto the radial direction reads
$$T-(m+M)g\cos\theta =\frac{(m+M)v^{2}}{l}$$ (1)
,where $\theta$ is the angle at center,"T" is the tension in the in wire and [itex] \frac{v^{2}}{l} [/tex] is the radial acceleration.

Now,write down the law of energy conservation between the highest point (in which the system has only potential energy) and the lowest point in which the system has only kinetic energy:
$$\frac{(m+M)v^{2}}{2}=(m+M)gl(1-\cos\theta_{max})$$ (2)
Note that this formula yields the maximum linear velocity the body (total) can have.Therefore,combining (1) and (2),taking into account that in both formulas,the subscript 'max' for angles and velocity interviens
$$T_{max}=(m+M)g+2(m+M)g(1-\cos\theta_{max})$$(3)

,which can be simplified to
$$T_{max}=(m+M)g(3-2\cos\theta_{max})$$(4)

The reason for why I've set $\theta=0$ is that u must maximize the function T.Which means maximize the velocity (which is done in formula (2)) and maximize the gravity as well,which is done by setting $\cos\theta =1$ or,equivalently,$\theta=0$

Daniel.

For your equation for the second law of dynamics... is it basically saying the sum of the forces equals mass x (radial)acceleration which is also equal, in this case, to Tension minus Fg (minus because Fg is negative). .. ? So you set them equal... (?)

 

[physicsgirl101]

YAY! I did it on my own using what you had said about setting energies in different positions equal and I got the same answer as yoU! thanks.

 

[dextercioby]

The second law reads in vector form
$$m\vec{a}=\vec{G}+\vec{N}$$

Project this law on radial direction.U'll find my equation.


Daniel.

 

[physicsgirl101]

The second law reads in vector form
$$m\vec{a}=\vec{G}+\vec{N}$$

Project this law on radial direction.U'll find my equation.


Daniel.

is one of the equations set up T-(M+m)g = ...(that's G + N, right?)
because (M+m)g is negative... ?