Pendulum conservation of energy

  • Thread starter alingy1
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  • #1
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Please look at pictures.

There is something really weird going on.
They say the bob has to make a vertical circle. Then, in the solutions, they say that, for that to happen, speed at top must be greater than zero.

BUT, since the only force acting is mg, that means that mg=mv^2/r. <=> v=sqroot(rg)

This is my solution:
vbob+bullet=sqroot(rg)=4.43m/s.
m1 vbullet=(m1+m2) vbob+bullet.

vbullet=137 m/s.

Who is right?
 

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Answers and Replies

  • #2
TSny
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You are correct about the block and bullet needing a speed of √(gR) at the top.

But did you take into account that gravity will slow down the block and bullet as they travel from the bottom to the top of the circle?
 
  • #3
Simon Bridge
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They say the bob has to make a vertical circle. Then, in the solutions, they say that, for that to happen, speed at top must be greater than zero.
If the speed at the top were exactly zero, then why would the bob continue in a circle at all? Why not just fall down?

BUT, since the only force acting is mg, that means that mg=mv^2/r. <=> v=sqroot(rg)
This equation is saying that the centripetal force is completely provided by the gravitational force ... this can only happen when the gravitational force is pointing centripetally: towards the center of the circle.
Is that always the case?
 
  • #4
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Good remark. I see the flaw in my logic.
Here is the correction:
Energy at top=1/2 mv^2 + mgh=6.08 J + 24.304 J=30.384 J
Energy at the bottom= 30.384 J=1/2 mv^2 initial of block + bullet.
vinitial of block + bullet =9.9 m/s.
m1 vbullet=(m1+m2) vbob+bullet.

vbullet= 306.9 m/s^2.

Seems right?
 
  • #5
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By the way, I think I understand what is going on. The word "around" can either mean at the opposite side of the initial position OR actually moving a full 360° in a circle. This is what happens when language mixes in with physics ;)
 
  • #6
TSny
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Good remark. I see the flaw in my logic.
Here is the correction:
Energy at top=1/2 mv^2 + mgh=6.08 J + 24.304 J=30.384 J
Energy at the bottom= 30.384 J=1/2 mv^2 initial of block + bullet.
vinitial of block + bullet =9.9 m/s.
m1 vbullet=(m1+m2) vbob+bullet.

vbullet= 306.9 m/s^2.

Seems right?

Looks good. Often there's an advantage to not plugging numbers in until the last. If you set up the energy conservation is symbols you have

Ebottom = Etop

(1/2)(m1+m2)Vb2 = (1/2)(m1+m2)Vt2 + (m1+m2)g(2R)

Vb2 = Vt2 +4gR

Vb2 = gR + 4gR = 5gR

Vb = √(5gR)

Vbullet = 31Vbottom = 31√(5gR)
 
  • #7
TSny
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By the way, I think I understand what is going on. The word "around" can either mean at the opposite side of the initial position OR actually moving a full 360° in a circle. This is what happens when language mixes in with physics ;)

Yes. But if the block and bullet do not have a speed of at least √(gR) at the top, then the tension in the string will go slack before reaching maximum height. The block will then not stay on a circular trajectory, it will go into parabolic projectile motion. The problem states that the block goes all the way around a vertical circle.
 
  • #8
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TSny, if the ball has just a bit more energy than the potential energy change in the height difference, the ball will make it to the top (at the opposite side "around in the second sense given to the word in the dictionary") BUT, it will at that moment stop going in a vertical circle and go into a parabolic motion.

Am I making sense?
 
  • #9
TSny
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To stay on the circle all the way to the top you would need to have a minimum speed of √(gR) at the top of the circle or a minimum speed of √(5gR) at the bottom. Any speed less than this and the systems will not stay on the circle all the way to the top.

If the kinetic energy at the bottom is just barely greater than the PE change corresponding to a height of 2R, then you would have to throw the block straight upward to reach a height of 2R. That would not be staying on the circle.
 

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