B Explain FBD in vertical circle

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1. May 25, 2017

Deepak verma

Hi ,

just curious about the F.B.D of a pebble moving in a vertical circle, which component that balances the weight of the pebble at the horizontal position , as tension is providing the required centripetal force , weight is acting downwards , which component balances it ?

2. May 25, 2017

Staff: Mentor

Why do you think the weight needs to be balanced?

3. May 25, 2017

Deepak verma

it has to be , otherwise system won't be working as it does . There is no counter force to balance it. a system is in equilibrium has all ∑force = 0 .

4. May 25, 2017

Staff: Mentor

The system is not in equilibrium.

5. May 25, 2017

Deepak verma

not in equilibrium but forces should be balanced as it is a case of uniform circular motion no tangential acceleration only centripetal .
[Mentor's note: This post has been edited to remove some unnecessary personal argumentation]

Last edited by a moderator: May 25, 2017
6. May 25, 2017

Staff: Mentor

Even in uniform circular motion the forces are not balanced; the object is accelerating in some direction and this requires some non-zero net force.

7. May 25, 2017

Staff: Mentor

But motion in a vertical circle (such as a rock tied to a rope, spun in a circle) is not uniform circular motion. There will be a tangential acceleration.

As Nugatory already pointed out, for accelerated motion (which this is an example of) the forces aren't "balanced" but comply with Newton's 2nd law.

8. May 25, 2017

Deepak verma

@Nugatory , I understand there should be some net force but that force is centripetal force acting towards center not the weight , weight has to be balanced as it has no tangential acceleration.

My main concern is at a particular point that is when string is horizontal weight simply acts downwards and as no counter component is there it should not complete the circle because acceleration g is acting downwards.

9. May 25, 2017

Staff: Mentor

What is the tension in the string at that moment? What is the acceleration of the weight at that moment?

The situation might be a bit easier to visualize if you imagine the object to be moving along a circular track instead of the end of a stretched string.

10. May 25, 2017

Deepak verma

Tension would be equal to centripetal force and acceleration is towards the center v sqaure divided by radius of circle . See everything is defined except weight component.

11. May 25, 2017

Staff: Mentor

And the weight is defined too; in fact, it's the only force in the problem which is constant in both magnitude and direction.

You're mistaken about the direction of the acceleration; there are only two points in the circle where the tangential component of the acceleration is zero, and they aren't where the string is horizontal.

12. May 25, 2017

nasu

What point of the trajectory are you referring to? Top, bottom, on the side?

13. May 25, 2017

Deepak verma

Side.

14. May 25, 2017

Deepak verma

If it is a uniform circular motion , then there won't be any tangential acceleration throughout the motion , i'm talking theoretically not practically , I know there is no motion as uniform circular motion but still my doubt persist. Weight at other position counter balance tension or provide tension and centripetal force not at horizontal position that is my concern

15. May 25, 2017

nasu

On the horizontal position of the string the weight is tangential so there is tangential acceleration. The tension is radial so there is centripetal acceleration.
This motion is not uniform, not even theoretically. The speed changes as the body moves along the circular trajectory. It has a maximum speed at the bottom point and a minimum speed at the top point. There is no point on the trajectory where the forces are balanced.
If the forces are balanced (net force equal to zero) the only motion possible is in a straight line, with constant speed.

16. May 25, 2017

Staff: Mentor

You've specified motion in a vertical circle, so it isn't uniform circular motion.

Conservation of energy requires that the sum of the kinetic energy and the potential energy of the weight be constant; but the potential energy is greater at the top of the circle than at the bottom, so the kinetic energy and hence the speed must therefore be less at the top than at the bottom. More informally, the speed must change because gravity is slowing the weight on the upswing and speeding it up on the downswing. The speed can only change if there is a tangential component to the acceleration; it does change, so therefore there must be a tangential component to the acceleration.

Last edited: May 26, 2017
17. May 25, 2017

Deepak verma

@thank you nasu , i got the point . thank you everyone .

18. May 26, 2017

Deepak verma

And one more thing , so I can conclude , there is no vertical uniform circular motion theoretically as well. But horizontally it would be possible right?

19. May 26, 2017

Staff: Mentor

Sure. And using the same example -- the rock on a rope, but now spun horizontally -- the weight of the rock would be balanced by the tension in the rope. The rope would not be horizontal, but at an angle.

20. May 26, 2017

Deepak verma

yeah in that case :

Tcosθ = mg and , Tsinθ= mv(square)/r