Help Needed: Determining Speed of Dart After Being Fired

[physicsgirl101]

again, thanks for everyone's help on the other problems, here is one last question. For part b.) the answer seemed to easy, and I wasn't sure if I was doing it correctly! here's the problem

In a laboratory experiment, you wish to determine the initial speed of a dart just after it leaves a dart gun. The dart, of mass m, is fired with the gun very close to a wooden block of mass, M., whichhangs from a cord of length L, and negligible mass. Assume the size of the clock is negligible compared to L, and the dart is moving horizontally when it hits the left side of the block at its center and becomes embedded init. The block swings up to a maximum angle (theta max) from the vertical. Express your answers to the following in terms of m, Mo, L, (theta max), and g.

b.) The dart and block subsequently swing as a pendulum, Determine the tnesion in the cord when it returns to the lowest point of the swing.

I set up a free body diagram and got T =mg It just seemed to easy- so my final answer was T = (m + Mo)g

thanks for any help!

 

[dextercioby]

The lowest point of the trajectory (which is a circles,assuming the length of the wire is constant) is the equilibrium point.The total force at that point is zero.So the tension in the wire is exactly the gravity force exerted by the Earth on the composed body.

Your result is correct.

Daniel.

 

[apchemstudent]

P.S- I deleted my message since i wanted to include my calcs...

As a restatement... I disagree with Dextercioby since the objects are still moving, not in an equilibrium, which means there's also a centripetal force involved. not just Fg...

Kinetic energy = Potential energy

you can figure out V here and with it you can figure out the centripetal force...

Fnet = FT

Fc = Fnet - Fg

FT = Fg + Fc...

 

[Zaimeen]

that's true, there is a centripetal force acting.

 

[Zaimeen]

P.S- I deleted my message since i wanted to include my calcs...

As a restatement... I disagree with Dextercioby since the objects are still moving, not in an equilibrium, which means there's also a centripetal force involved. not just Fg...

Kinetic energy = Potential energy

you can figure out V here and with it you can figure out the centripetal force...

Fnet = FT

Fc = Fnet - Fg

FT = Fg + Fc...

I think it should be FT = Fg - Fc....think about it!

The total upward force = total downward force

hence, centripetal_force (Fc) + tension (FT) = weight (Fg)

and FT = Fg - Fc

 

[apchemstudent]

I think it should be FT = Fg - Fc....think about it!

The total upward force = total downward force

hence, centripetal_force (Fc) + tension (FT) = weight (Fg)

and FT = Fg - Fc

Actually... i made a mistake...

Fc = Fnet

so

Fnet = FT - Fg

Fc + Fg = FT... , If you think about it, this makes more sense since the tension should be greater than mg, not less in the case Fg - Fc..

 

[Zaimeen]

Actually... i made a mistake...

Fc = Fnet

so

Fnet = FT - Fg

Fc + Fg = FT... , If you think about it, this makes more sense since the tension should be greater than mg, not less in the case Fg - Fc..

I think the FNet = 0, because there is no vertical movement! so, that's why it's more logical to consider upward forces to be equal to downward forces..

since there are two upward forces, namely the centripetal force for the SHM and the tension, these forces equate to the total weight of the swinging object!

 

[dextercioby]

P.S- I deleted my message since i wanted to include my calcs...

As a restatement... I disagree with Dextercioby since the objects are still moving, not in an equilibrium, which means there's also a centripetal force involved. not just Fg...

Kinetic energy = Potential energy

you can figure out V here and with it you can figure out the centripetal force...

Fnet = FT

Fc = Fnet - Fg

FT = Fg + Fc...

I disagree with you,amigo... What are the forces that act on the body??Compute the centripetal force,knowing that
"The dart and block subsequently swing as a pendulum".

I remind you that in the case of pendulum:
\omega=\sqrt{\frac{g}{l}}

The TENSION FORCE IS THE CENTRIPETAL FORCE IN THIS CASE.
So the cp.force=tension=(M_{0}+m)g.

Daniel.

 

[apchemstudent]

I disagree with you,amigo... What are the forces that act on the body??Compute the centripetal force,knowing that
"The dart and block subsequently swing as a pendulum".

I remind you that in the case of pendulum:
\omega=\sqrt{\frac{g}{l}}

The TENSION FORCE IS THE CENTRIPETAL FORCE IN THIS CASE.
So the cp.force=tension=(M_{0}+m)g.

Daniel.

You are wrong, because if the centripetal force was in fact Fg, then the objects would go in a complete circle, not in a pendulum...

You're saying Tension equals centripetal force in this case: that is not true again because that just doesn't make any sense: There are 2 forces acting on the objects. The force of tension from the rope and the force of gravity.

In order to have a Fc = Fg then FT must be twice as strong as Fg at the lowest point. 2Fg does not = Fg...

 

[physicsgirl101]

thank you all for your help- but I'm confused as what the tension should equal, is it the centrifical force, or is it just mg - (m + Mo)g?

 

[apchemstudent]

thank you all for your help- but I'm confused as what the tension should equal, is it the centrifical force, or is it just mg - (m + Mo)g?

The Ft = Fc + Fg

You can determine the Fc from

KE = U(p)

you get the velocity at the bottom of the arc which you can use to determine the centripetal force.

Dextercioby does not make any sense since the Force of tension from the rope is in opposite direction of the Fg at the bottom. If Ft = Fg then there would be no centripetal force, which means the object shouldn't even be moving...

 

[dextercioby]

The second principle of dynamics reads in vector form
(M+m)\vec{a}=(M+m)\vec{g}+\vec{T}(1)
For the lowest point of the trajectory,the acceleration is null/zero and the velocity is maximum
\vec{0}=(M+m)\vec{g}+\vec{T}(2)

Use an axis (call it "y") to project (2) on and find that the 2 forces have the same "support" but different directions.Therefore
T=(M+m)g (3)

,just what i was saying from the very beginning and what u found initially.


Daniel.

 

[dextercioby]

The Ft = Fc + Fg

You can determine the Fc from

KE = U(p)

you get the velocity at the bottom of the arc which you can use to determine the centripetal force.

Dextercioby does not make any sense since the Force of tension from the rope is in opposite direction of the Fg at the bottom. If Ft = Fg then there would be no centripetal force, which means the object shouldn't even be moving...

What is the centripetal force,what is the centrifugal force...??Are they active forces??What forces act on the body...??

Daniel.

 

[apchemstudent]

The second principle of dynamics reads in vector form
(M+m)\vec{a}=(M+m)\vec{g}+\vec{T}(1)
For the lowest point of the trajectory,the acceleration is null/zero and the velocity is maximum
\vec{0}=(M+m)\vec{g}+\vec{T}(2)

Use an axis (call it "y") to project (2) on and find that the 2 forces have the same "support" but different directions.Therefore
T=(M+m)g (3)

,just what i was saying from the very beginning and what u found initially.


Daniel.

Ok, At the bottom you're saying Fc = 0 now since there is no acceleration which implies there's no net force...

As well you said v is at the max... .

(Mo + m) = M

v^2 = 2*g*(L-costheta*L)

Fc = M*v^2/L -----> 2*Mg*L(1-costheta)

You're saying Fc = 0 = 2*Mg * (1-costheta)

The only 0 you get from the above equation if theta was equal to 0 degrees between the vertical and the rope at the highest point swung...

So if you didn't swing it, then Fc = 0, otherwise... NOOOO FC does not equal 0

 

[dextercioby]

You're totally confused,my friend...On the two joint bodies act only two forces all the time,even if it is an oscillator,or a rotator,it doesn't matter.
What it does matter is that at the lowest of the trajectory (which is a circle,assuming the wire is ideal and it doesn't bend either) these two forces are in complete equilibrium.Therefore in modulus they are equal.
The tension force is on the direction of the center of the circle all the time,therefore it is a centripetal force.Gravity on the other hang,at the lowest point of the trajectory can be seen as a centrifugal force,but thatonly in the noninertial reference system (for example,theone attached to the center of the circle).


Daniel.

 

[apchemstudent]

You're totally confused,my friend...On the two joint bodies act only two forces all the time,even if it is an oscillator,or a rotator,it doesn't matter.
What it does matter is that at the lowest of the trajectory (which is a circle,assuming the wire is ideal and it doesn't bend either) these two forces are in complete equilibrium.Therefore in modulus they are equal.
The tension force is on the direction of the center of the circle all the time,therefore it is a centripetal force.Gravity on the other hang,at the lowest point of the trajectory can be seen as a centrifugal force,but thatonly in the noninertial reference system (for example,theone attached to the center of the circle).


Daniel.

Where does it say that the lowest of the trajectory is where the 2 forces, tension and Fg, in complete equilibrium?

As well, you've completely ignored my calculations... You've said nothing that counters my explanation, besides coming up with "What it does matter is that at the lowest of the trajectory (which is a circle,assuming the wire is ideal and it doesn't bend either) these two forces are in complete equilibrium"...

You do know this is not an advanced university or college question?
And besides that "What it does matter is that at the lowest of the trajectory (which is a circle,assuming the wire is ideal and it doesn't bend either) these two forces are in complete equilibrium"

Just makes no sense... if they are in fact in an equilibrium then there is no centripetal force since Ft is supplying the centripetal force and Fg is countering it. Ft is not the centripetal force itself, so don't assume it is...

If there is no centripetal force, then there shouldn't be any change in velocity, and yet the objects are still in motion(change in velocity)...

 

[dextercioby]

Hopefully I'm not mistaking...Weren't u that guy that started the thread:"Redemption:I challange Dextercioby"...? If so,and you still don't believe that my result is correct,and apparently no one interviens in our little "polemics",i advise to start another thread with the title
:"Redemption:I challange Dextercioby,AGAIN!"
and get some other opinions on it.
In case u do that,here's my argument:"The tension force for a circular/mathematical pedulum is a centripetal force,and it the lowest point of the trajectory,the tension is exactly balancing the gravity force and therefore the body is in dynamical equilibrium,meaning the total force acting on it is zero.Since the acceleration is zero,the velocity has a critical point,which in this case is a maximum.So the maximum tension force is equal tot the total gravity,i.e.
|\vec{T}_{max}|=(M+m)g
."
I rest my case...

Daniel.

 

[apchemstudent]

Hopefully I'm not mistaking...Weren't u that guy that started the thread:"Redemption:I challange Dextercioby"...? If so,and you still don't believe that my result is correct,and apparently no one interviens in our little "polemics",i advise to start another thread with the title
:"Redemption:I challange Dextercioby,AGAIN!"
and get some other opinions on it.
In case u do that,here's my argument:"The tension force for a circular/mathematical pedulum is a centripetal force,and it the lowest point of the trajectory,the tension is exactly balancing the gravity force and therefore the body is in dynamical equilibrium,meaning the total force acting on it is zero.Since the acceleration is zero,the velocity has a critical point,which in this case is a maximum.So the maximum tension force is equal tot the total gravity,i.e.
|\vec{T}_{max}|=(M+m)g
."
I rest my case...

Daniel.

i hope some one else will check this out... I've done a problem like this before concerning the ballistic pendulum... And why won't you agree with my calculation? Everything is in the ideal sense...

Gravitational energy is conserved so you can determine the velocity at the bottom of the trajectory... which you can use to determine the Fc and Ft...
It's obvious that Fc is not 0 as I've said...

 

[physicsgirl101]

First... there is obviously major disagreement betweek dextercioby and apchemstudent, so does a third party want to give an opinion?

Second... My professor has never even mentioned the word "centrifugal force"----so unless there's another name for it that my professor has used and I'm just not making the connection, I doubt he would give us a problem involving it... ?

 

[apchemstudent]

First... there is obviously major disagreement betweek dextercioby and apchemstudent, so does a third party want to give an opinion?

Second... My professor has never even mentioned the word "centrifugal force"----so unless there's another name for it that my professor has used and I'm just not making the connection, I doubt he would give us a problem involving it... ?

Yes i agree there should be a third party to check this out... I've already sent this problem out to a different website... I'll keep you posted on what happens...

 

[physicsgirl101]

Thanks, definitely keep me posted, Thanks!

 

[vincentchan]

Hello, I am the third party you guys looking for . I am a big fans of dextercioby. he is very very good at physics. but, sadly, I must say he is WRONG this time, If you review the question carefully, you will see at the lowest point of the trajectory, accelaration is not zero... Its speed doesn't change, but ITS VELOCITY (direction) does... I will vote for apchemstudent this time...

 

[physicsgirl101]

Again, I've never heard the word "centripetal force"----so unless there's another name for it that my professor has used and I'm just not making the connection, I doubt he would give us a problem involving it... ?

 

[dextercioby]

The net acceleration is zero.It is composed however from 2 "pieces"ne is due to gravity:\vec{g} and the other is due to centripetal character of the tension force -\omega^{2} l\vec{r},with
|\vec{r}| =1.

If u see that
\omega=\sqrt{\frac{g}{l}}
,u'll understand why in the lowest point of the trajectory the accleretion is zero and the velocty is maximum.

Daniel.

PS.For the "fan" part...

 

[dextercioby]

Again, I've never heard the word "centripetal force"----so unless there's another name for it that my professor has used and I'm just not making the connection, I doubt he would give us a problem involving it... ?

Voilà,it proves my point.U don't need to know that the tension has a centripetal character.The second law of Newton would do it...

Daniel.

 

[vincentchan]

the gravity can't balance the tension force
If the accelaration is zero, why does the block changes its velocity (direction) after hitting the lowest point>?

 

[vincentchan]

physicsgirl: did u see the formulas f=mv^2/r?

 

[apchemstudent]

the gravity can't balance the tension force
If the accelaration is zero, why does the block changes its velocity (direction) after hitting the lowest point>?

umm... its speed does change due to conservation of energy and the Fc is not constant... but i agree the direction also changes... This is what I'm trying to explain dextercioby... hope you have better luck...

 

[dextercioby]

physicsgirl: did u see the formulas f=mv^2/r?

I'm sure she didn't.She said not to have been taught about circular motion and centripetal and centrifugal forces.

Daniel.

PS.Instead of asking questions,why don't u show us how u would apply the second principle of Mr.Newton...

 

[vincentchan]

centripetal character of the tension force -\omega^{2} l\vec{r},with

dex:
\omega is the angular speed, and it is not equal \sqrt{g/l}, indeed, it is not constant, At the highest point, \omega = 0, on the other hand, \omega = maximun at the lowest point. \sqrt{g/l}is the average \omega under a whole period, so, the tension on the string is greater than mg,,,,,

 

[apchemstudent]

I'm sure she didn't.She said not to have been taught about circular motion and centripetal and centrifugal forces.

Daniel.

PS.Instead of asking questions,why don't u show us how u would apply the second principle of Mr.Newton...

She's just not used to centrifugal forces... Same thing with me, my teacher told me there's no such thing except that people get mixed up with the centrifugal force for centripetal...

 

[physicsgirl101]

physicsgirl: did u see the formulas f=mv^2/r?

i've seen radial acceleration = v^2/r... so yes, i guess, because that's just subbing a = v^2/r into f=ma

right? so yes, we've done some circular motion stuff... but just using that equation

 

[physicsgirl101]

She's just not used to centrifugal forces... Same thing with me, my teacher told me there's no such thing except that people get mixed up with the centrifugal force for centripetal...

i've never heard of centrifugal OR centripetal forces!

 

[vincentchan]

mr dex, b4 your reply, think carefully, the faster the block move, the greater tension on the string...

 

[apchemstudent]

i've never heard of centrifugal OR centripetal forces!

the centripetal force is simply radial acceleration times the mass...

 

[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... 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... 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,
<br /> l(1-cos{\theta_{max}})mg=1/2mv^2<br />
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,
<br /> l(1-cos{\theta_{max}})mg=1/2mv^2<br />
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...

Daniel.