Tension in a string swinging a rock.

[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?
because 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

 

[Shooting Star]

Hi rl.bhat,

Well, it's the same thing. You are using cosvn of energy, and so am I. Only I did it for the second part, because there was no need to use it for the first part. Your answers are the same as mine.

The only thing that still nags me is question C, as I have written in my earlier post. If the point where one end of the string is fixed is not moved, then you’ll get a different answer for the angle.

 

[rl.bhat]

Tension

I have rewritten the part A because I can demonstrate the problem to the students by using a pendulum.
I have treated the part 3 indipendent of A and B. beacuse from vertical circular motion we cannot shift to conical pendulum. I can demonstrate this also to the students. In all the three cases the point of suspension is the same.

 

[Shooting Star]

Well, it's nagging me because of that reason. Actually, the question C is not properly formulated. If you treat C independently from A and B, and just keep the KE of C as of B, you’ll get one answer. The PE + KE is more in this case than from case B.

But if you can somehow apply some force to the rock perp to its velocity at the point where it’s at the bottom of the vertical swing, it won’t change the total energy but maybe change the direction so that now it swings horizontally like a conical pendulum. In this case the angle is different. I have no idea whether such a thing can be done or not.