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,,,,,