Thursday, December 31, 2015

Circular motion in a vertical plane

  • Circular motion in a vertical plane

Consider a pail of water whirling at the end of a rope in a vertical circle. The tension T in the rope varies with the position of the pail.
The figure below shows the forces acting on the pail when the rope is at an angle of θ to the vertical.

From Newton’s 2nd law: Resultant force F = ma
Net force towards the centre of the circle O is
T – mg cosθ

The net force provides the centripetal acceleration.
Therefore, T – mg cosθ = mv2 / r
where v = speed of pail; r = radius of circle

Tension T = mv2/r + mg cosθ

Now, let’s consider how the tension changes as the position of the pail changes.

·         When the pail is at the lowest point (θ = 0),
Tension T = mv2/r + mg        (which is the maximum tension)

·         When the rope is horizontal (θ = π/2 (=90°) or θ = 3π/2 (=270°)), cosθ = 0 and thus
Tension T = mv2/r

·         When the pail is at the highest point (θ = π = 180°), cosθ = –1 and
Tension T = mv2/r – mg        (which is the minimum tension)

Now if the pail reaches the highest point (at a distance equal to the radius r form centre O of the circle), the rope must be taut (there should be a tension in the rope, which is not zero).
( T = mv2/r – mg )      > 0
mv2/r > mg
v > √(gr)
Thus, they velocity of the pail at the highest point must be greater than √(gr).

Now consider the forces acting on the water in the pail when it is at the highest point.

For the water not to fall out (that is, the water is in contact with the pail – there is a contact force on the pail), there should be a reaction R (from Newton’s 3rd law) on the water by the base of the pail. The resultant force towards the centre of the circle is
F = R + mg = mv2 / r

The centripetal force is provided by the reaction R and the weight mg.
R = mv2/r – mg

For the water not to fall, R > 0 (that is, there should be a reaction R on the water).
( R = mv2/r – mg )      > 0
mv2 / r > mg

Therefore, water in the pail does not fall out because the centripetal force is greater than the weight of the water. (The weight mg of the water is less than the required force mv2/r towards the centre and so the water stays in. The reaction R of the bucket base on the water provides the rest of the force mv2/r.)

Remember that the weight acts vertically downwards (towards the ground). The centripetal force mv2/r acts towards the centre, causing the bucket to move in a circle.

On the other hand, if the pail is whirled slowly, then mg > mv2/r. The weight is greater than the centripetal force and the water falls out of the bucket. (Here, part of the weight provides the force mv2/r. The rest of the weight causes the water to accelerate downward and hence to leave the bucket.)

1. “Pacific Physics A Level,” by POH LIONG YONG, pg 154 – 155 ‘Motion in a vertical circle’
2. “Advanced Level Physics,” by Nelkon & Parker, 7th edition, pg 63 – 64


  1. i read in a book that the minimum speed of the bucket is when acceleration is equal to g ? why

    1. I have not come across this one. Give me a link or reference of the book. If it's correct I'll add it.


    2. advanced physics by steve adams and jonathan allday
      but i dont understand why its like that if u get it can u pls explain why the minimum speed is when a = g

    3. i don't think i have the book. can you scan the relevant pages and send it to
      I'll try to see then.

  2. when the water is in contact with a pail it exerts a force and in reaction to it the pail exerts a force ryt ?
    so why arent we considering the force exerted by water on pail??
    Please replyy thanks

    1. in the last part we are considering the forces acting on the water and not the pail. SO there is not need to consider that force.

  3. can you please explain me weightlessness

    1. Try to go through solution 989 at

  4. Just like how you put up this article on motion in a verticle circle


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