# 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 2

^{nd}law: Resultant force**F = ma**
Net force towards the centre of the
circle O is

**T – mg cos**

**θ**

The net force provides the
centripetal acceleration.

Therefore,

**T – mg cos****θ = mv**^{2}/ r
where v = speed of pail; r = radius
of circle

**Tension T =**

**mv**

^{2}/r +**mg cos**

**θ**

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 =**

**mv**

^{2}/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 =**

**mv**

^{2}/r
·
When
the pail is at the highest point (θ = π = 180°), cosθ = –1 and

**Tension T =**

**mv**

^{2}/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 =**

**mv**

^{2}/r –**mg ) > 0**

**mv**

^{2}/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 3

^{rd}law) on the water by the base of the pail. The resultant force towards the centre of the circle is**F = R + mg = mv**

^{2}/ r
The centripetal force is provided by
the reaction R and the weight mg.

**R = mv**

^{2}/r – mg
For the water not to fall, R > 0 (that is, there should be a reaction R on the water).

**( R = mv**

^{2}/r – mg ) > 0**mv**

^{2}/ 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

**mv**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^{2}/r**mv**.)^{2}/r
Remember that the weight acts
vertically downwards (towards the ground). The centripetal force

**mv**acts towards the centre, causing the bucket to move in a circle.^{2}/r
On the other hand, if the pail is
whirled slowly, then

**mg > mv**. The weight is greater than the centripetal force and the water falls out of the bucket. (Here, part of the weight provides the force^{2}/r**mv**. The rest of the weight causes the water to accelerate downward and hence to leave the bucket.)^{2}/r**References:**

1.

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

^{th}edition, pg 63 – 64