Tension in a string swinging a rock.

[pinkyjoshi65]

A rock of mass 200g is attached to a 0.75m long string and swung in a vertical plane.
A) what is the speed that the rock must travel so that its weight at the top of the swing is 0?
B) what is the tension in teh string at the bottom of the swing?
C) if the rock is now moved to a horizontal swinging position, what is the angle of the string to the horizontal?

I got the answer for part a) and part c). I am not sure about Part b).
Here is what i did..:
A) At A: Net force= -T-mg
= -(T+mg)
But weight at A should be zero.
Hence, Net force= -T = -mg= -0.2*9.8= 1.96 N (Down)
F_c= mv^2/r= 0.2*v^2/0.75
v^2= 1.96*0.75/0.2
v = 2.71m/s
B)
At B: Net force= T-mg
F_C= mv^2/r
= 0.2*2.71*2.71/0.75
= 1.958 N
Net Force= T-mg
T= Net Force+mg
T= 1.958+0.2*9.8
T= 1.958+1.96
T= 3.918N

 

[nrqed]

A rock of mass 200g is attached to a 0.75m long string and swung in a vertical plane.
A) what is the speed that the rock must travel so that its weight at the top of the swing is 0?
B) what is the tension in teh string at the bottom of the swing?
C) if the rock is now moved to a horizontal swinging position, what is the angle of the string to the horizontal?

I got the answer for part a) and part c). I am not sure about Part b).
Here is what i did..:
A) At A: Net force= -T-mg
= -(T+mg)
But weight at A should be zero.
Hence, Net force= -T = -mg= -0.2*9.8= 1.96 N (Down)
F_c= mv^2/r= 0.2*v^2/0.75
v^2= 1.96*0.75/0.2
v = 2.71m/s
B)
At B: Net force= T-mg
F_C= mv^2/r
= 0.2*2.71*2.71/0.75
= 1.958 N
Net Force= T-mg
T= Net Force+mg
T= 1.958+0.2*9.8
T= 1.958+1.96
T= 3.918N

The first question does not make sense. The weight is mg and that's fixed. Do they mean that the tension must be zero?

 

[pinkyjoshi65]

no they mean that the weight is zero. I know that dosent make sense, but i guess the answer i got is the right one. But I seem to have trouble with part b..Firse i used the conservation of energy formula, but i found out the there is not change in h.

 

[saket]

I guess, you have already discussed this problem in some other thread.
Anyways..

But weight at A should be zero.
Hence, Net force= -T = -mg= -0.2*9.8= 1.96 N (Down)

weight at A should be zero => T = 0. So, net force = -m*g. [I guess, you mis-typed. If not, please take note of this.]

 

[saket]

what does your F_c stand for... centripetal force or centrifugal force??

i am asking because, in either one of the part, you have taken wrong sign for it.

 

[pinkyjoshi65]

I did type net force=-mg..:S

 

[pinkyjoshi65]

centripetal force

 

[nrqed]

no they mean that the weight is zero. I know that dosent make sense, but i guess the answer i got is the right one. But I seem to have trouble with part b..Firse i used the conservation of energy formula, but i found out the there is not change in h.

I know that you get the right answer but the reasoning is illogical.

How do you get from net force = - (T+mg) to syaing that -T = -mg ?

No prof would mark this as correct even if the final answer is right.

The only thing that makes any sense is to say that the tension is zero there fore the net force is -mg. You set this to m a_y with a_y given by -mv^2/R. This tehn gives the same answer but is now logically consistent.

 

[saket]

Furthermore, as now the string (read, rock) is at the bottom-most point... how come you are using same speed... 2.71 m/s?? This was speed of the rock when it was at A, isn't it?

 

[pinkyjoshi65]

Ok..Net force= -T-mg or i can write this as -(T+mg) ok..?
Then since the weight is Zero, the total net force will be -(T+0)= -T
T=mg Hence -T=-mg..this is how i got it..Any help for part b)..is it right?

 

[saket]

Okay, if centripetal..

F_c= mv^2/r= 0.2*v^2/0.75

F_c would be in downward direction at A.. thus.. F_c = -mv^2/r= -0.2*v^2/0.75
[Note the -ve sign]

 

[pinkyjoshi65]

umm..yes thts what i thought saket. But when I used the conservation of energy to find out the velocity of the rock at the botton, and checked it with one of my teachers in school, she said that this was wrong..:S:S. She asked me to use..Net force= T+mg..But how can mg b positive..it should be negative..:S

 

[saket]

Ok..Net force= -T-mg or i can write this as -(T+mg) ok..?
Then since the weight is Zero, the total net force will be -(T+0)= -T
T=mg Hence -T=-mg..this is how i got it..Any help for part b)..is it right?

No, that's a wrong perception! This is what I was trying to explain.

Okay, what is weight? Mass*gravitational acceleration?? Ya, fine.. but that is when you are weighing yourself on a weighing machine.. then your acceleration is zero!

If, you are accelerating upwards with an acceleration "a", your weight will be m*(g + a).
If, you are accelerating downwards with an acceleration "a", your weight will be m*(g - a), where a is not greater than g. In this only, if the lift is accelerating downwards with acceleration "g", your weight is zero! (since, a = g) ... This is what they call, a situation of "weightlessness"! haven't you heard about it?

 

[nrqed]

Ok..Net force= -T-mg or i can write this as -(T+mg) ok..?
Then since the weight is Zero, the total net force will be -(T+0)= -T
T=mg Hence -T=-mg..this is how i got it..Any help for part b)..is it right?

I will look at part b in a second.

But realize that what you wrote above is wrong. You first say mg=0 and in the next line you say T=mg...therefore T should be zero according to what you just wrote!

 

[Shooting Star]

B)
At B: Net force= T-mg
F_C= mv^2/r
= 0.2*2.71*2.71/0.75
= 1.958 N

The v at the top is different from the v at the bottom. Find v at bottom by using conservation of total energy. Then find the tension in the usual way.

C) Again, the total energy is conserved.

It's not so tough!

 

[pinkyjoshi65]

nrged..Ok so i got the net force as -T.
Now I have to find T
the formula for finding T is T=mg
So-T=-mg...Isin't that right..?..:S

 

[saket]

So, in this problem, weight becomes zero when T = 0 and not when m*g = 0!

 

[saket]

pleeeeeeeeeezzeee... "nrqed" and "pinkyjoshi65" .. read my posts! you two are discussing something which is fundamentally wrong.

 

[saket]

Okay, a more well-known example is that of an artificial satellite orbitting around Earth. An astronaut in it feels "weightlessness", because he is falling towards Earth with an acceleration "g" at any point of the path! (i.e. a = g)

 

[nrqed]

No, that's a wrong perception! This is what I was trying to explain.

Okay, what is weight? Mass*gravitational acceleration?? Ya, fine.. but that is when you are weighing yourself on a weighing machine.. then your acceleration is zero!

If, you are accelerating upwards with an acceleration "a", your weight will be m*(g + a).
If, you are accelerating downwards with an acceleration "a", your weight will be m*(g - a), where a is not greater than g. In this only, if the lift is accelerating downwards with acceleration "g", your weight is zero! (since, a = g) ... This is what they call, a situation of "weightlessness"! haven't you heard about it?

I won't get into a debate here but I find that pedagogically speaking, it's confusing to talk about "gravitational acceleration" as if there are different types of accelerations. I find that's it's better to distinguish "apparent wight" from true weight which is simply the force of gravity on an object and is always equal to mg. And apparent weight involves the notion of non-inertial frames which is a subtle point.

This problem can be done without ever talking about non-inertial frames, apparent weight or "gravitational acceleration". It's simply a simple application of Newton's second law!

 

[pinkyjoshi65]

I am sorry saket, yes i get what you are saying. mg cannot be zero. Thanks Nrqed and Saket for correcting me.

 

[saket]

umm..yes thts what i thought saket. But when I used the conservation of energy to find out the velocity of the rock at the botton, and checked it with one of my teachers in school, she said that this was wrong..:S:S. She asked me to use..Net force= T+mg..But how can mg b positive..it should be negative..:S

I am not sure what your teacher meant! [Or, maybe you took her wrong.. don't take it personally. It happens with students. It used to happen/happens with me as well.]
But, yes, if I were to solve this problem, I would certainly use energy conservation to get speed at B.

 

[nrqed]

A rock of mass 200g is attached to a 0.75m long string and swung in a vertical plane.
A) what is the speed that the rock must travel so that its weight at the top of the swing is 0?
B) what is the tension in teh string at the bottom of the swing?
C) if the rock is now moved to a horizontal swinging position, what is the angle of the string to the horizontal?

I got the answer for part a) and part c). I am not sure about Part b).
Here is what i did..:
A) At A: Net force= -T-mg
= -(T+mg)
But weight at A should be zero.
Hence, Net force= -T = -mg= -0.2*9.8= 1.96 N (Down)
F_c= mv^2/r= 0.2*v^2/0.75
v^2= 1.96*0.75/0.2
v = 2.71m/s
B)
At B: Net force= T-mg
F_C= mv^2/r
= 0.2*2.71*2.71/0.75
= 1.958 N
Net Force= T-mg
T= Net Force+mg
T= 1.958+0.2*9.8
T= 1.958+1.96
T= 3.918N

EDIT


The question b is ambiguous. . Assuming that the speed is kept constant during the entire motion then your answer b is correct. using conservation of energy won't work because there is some other force doing work on the mass in this case.

However if there is only the tension in teh rope an dthe motion is purely around a fixed center, then the speed cannot be constant and the problem must be done using conservation of energy

 

[pinkyjoshi65]

I am not taking physics now. I am just refreshing the concepts on my own since I am starting university in January. Sometimes I go to my old school and ask a few questions to a physics teacher.

 

[pinkyjoshi65]

Thank You, Sorry for not getting what you were trying to say..!..:)

 

[saket]

... "apparent wight" from true weight ...
This problem can be done without ever talking about non-inertial frames, apparent weight or "gravitational acceleration". It's simply a simple application of Newton's second law!

Bingo! (& sorry) This term I was forgetting. Yes, true weight is m*g. But we are talking of apparent weight to be equal to zero.

And, can't help "nrqed" ... the question itself has put the term 'weight' .. so we have to talk about apparent weight, atleast.

 

[saket]

EDIT


The question b is ambiguous. . Assuming that the speed is kept constant during the entire motion then your answer b is correct. using conservation of energy won't work because there is some other force doing work on the mass in this case.

However if there is only the tension in teh rope an dthe motion is purely around a fixed center, then the speed cannot be constant and the problem must be done using conservation of energy

Yes, I have assumed centre is fixed.
@ "pinkyjoshi65"
Why don't u give a try using energy conservation?

 

[pinkyjoshi65]

I have and i got the wrong answer.

 

[saket]

I have and i got the wrong answer.

Show your attempt.

 

[pinkyjoshi65]

At B: Net force= T-mg
Change in height= 0.75m
Total energy at A= E_k + E_p= E_p
= mgh= 0.2*9.8*1.5
= 2.94 J
E= 0.5mv^2
2.94= 0.5*0.2*v^2
v^2= 2.94/0.1= 29.4
v=5.42m/s
F_C= mv^2/r
= 0.2*5.42*5.42/0.75
= 7.83 N
Net Force= T-mg
7.83= T-0.2*9.8
7.83= T-1.96
T= 9.79 N

 

[learningphysics]

At B: Net force= T-mg
Change in height= 0.75m
Total energy at A= E_k + E_p= E_p
= mgh= 0.2*9.8*1.5
= 2.94 J

You left out the kinetic energy at the top... you calculated v = 2.71m/s at the top... not 0.

so energy at A = mgh + (1/2)mv^2 = 2.94 + (1/2)(0.2)(2.71)^2 etc...

 

[nrqed]

Bingo! (& sorry) This term I was forgetting. Yes, true weight is m*g. But we are talking of apparent weight to be equal to zero.

And, can't help "nrqed" ... the question itself has put the term 'weight' .. so we have to talk about apparent weight, atleast.

Ok, I see what you are saying. Then one would start the solution by imposing that the acceleration is equal to g and downward 9which then implies that the tension is zero).
we are on the same wavelength then.

best regards,

Patrick

 

[rl.bhat]

Tension

pinkyjoshi65, I went through your merathone Q & A.
Your first part of the question was
A) what is the speed that the rock must travel so that its weight at the top of the swing is 0?
If you rewright that in this way:
A) what should be the minimum speed that the rock must be imparted at the bottom of a vertical plane, so that its weight at the top of the swing is 0?
Becuase the rock cannot move to the top as it is.
In this case part B and C also make better sence.

 

[Shooting Star]

A) what is the speed that the rock must travel so that its weight at the top of the swing is 0?
If you rewright that in this way:
A) what should be the minimum speed that the rock must be imparted at the bottom of a vertical plane, so that its weight at the top of the swing is 0?

There is no need to reformulate the question A. There is a speed at the top such that the tension is instantaneously zero. That is the speed asked for and it has been found correctly.

As nrqed pointed out right at the beginning, it should read “tension”, not “weight”. But it is also true that if the tension is zero, the ball is in free fall at that point, and in the ball frame, there is weightlessness. But I do not think that the question was asked with that in mind.

Q B utilizes conservation of energy, so the answer should be simple to find out.

It’s the 3rd Q which is not very clear. The rock had only horizontal velo at the bottom. Now, what is meant by moving it to a horizontal swinging position? Does it mean that the rock continues to move in a circle at the same height which it was in at the bottom of the swing with the same speed? Or does it mean that the original centre remains fixed and the ball has the same total energy? The latter implies that the height of the rock will increase and the PE will increase a bit, so KE will decrease.

After 33+1 posts, will the OP try to solve for the correct answers?

 

[rl.bhat]

Tension

My approach of the problem is this.
Let the rock receive an energy = 1/2*mVo^2 at the bottom of the vertical circle. When it reaches the top of the vertical circle, applying the conservation energy, we get 1/2*mVo^2 = 2mgR + 1/2mV^2 where V is the velocity at the top. At this point if the T = 0, mV^2/R = mg.That gives V^2 = Rg, or V = 2.711/s (Ans. for A)
Substituing this value in the above equation, we get
1/2*mVo^2 = 2mgR + 1/2*m*gR
or Vo^2 = 5gR and tension at the bottom = mVo^2/R + mg =m*5Rg/R + mg =0.2*5*g +0.2g = 1.2g = 11.76N (Ans.for B)
In part C rock moves in the form of conical pendulum with the velocity Vo. If the string makes an angle theta with horizontal, the radius of the horizontal circle will be Rcos(theta).
And tan(theta) = mg/[mVo^2/Rcos(theta)] = mg/[5mgR/Rcos(theta)]
=cos(theta)/5 or 5sin(theta) =cos^2(theta) =1-sin^2(theta)
or sin^2(theta) + 5sin(theta) -1 =0 solving this we get theta = 11.1degree