Critical Velocity & Circular motion in Vertical plane
We have earlier discussed on circular motion in horizontal plane and in that context discussed in length about Centripetal force and Centrifugal force.Today we will take a few related items to discuss, which will not only strengthen your understanding on these topics but also take it to a higher level. Discussion pointers would be (1) circular motion in vertical plane and (2) Concept of Critical Velocity. So let’s start!
Circular motion in vertical plane has some basic difference with the same in horizontal plane. In the last post on Centripetal Force, we have showed that while rotating in a circle in Horizontal plane, the gravitational force has no role (no work done) on the rotating object. But during the circular motion in vertical plane, the role of gravitation is omnipresent.
Deriving Equation of Critical Velocity
Let’s take a rock of mass m tied to a light and inextensible string and start to rotate it along a vertical circular path of radius R.
What can be observed from its motion? Yes, it’s not uniform.
While descending from the highest point (A) of the circular track to the lowest point B, its velocity gradually increases. Also the tension of the string increases gradually as the rock comes from A to B. If you rotate the rock with the string tied with your finger, then you can feel this difference.
Let’s consider that the velocity at A and B are V1 and V2 respectively. Here we have to consider the Tension in the string as well.
Now please note that both at A and B, the net force required for the circular motion to continue is Centripetal Force.
As we know, Centripetal Force is a real force which has to be provided by some agent/s or it’s to be generated by the mutual interaction of multiple agents in a system.
Now we will analyze with Free Body diagrams to find out who supplies this centripetal force at point A and B.
At the highest point (A) if we consider the forces acting on the rock then these are,
(i) The weight (mg) of the rock acting vertically downwards
(ii) The tension exerted by the string (T1) on the rock (direction is away from the rock along the string)
So these 2 forces are mutually acting to supply the required Centripetal force at point A.
Now considering the position of A, all these 2 vectors are pointing vertically downwards i.e. towards the centre.
So we can write this equation for the point A
T1 + mg = (mV1^2)/R ……………………..(1)
Now consider point B. The force components acting on the rock here are as listed below:
(i)The weight (mg) of the rock is acting vertically downwards.
(ii) The tension exerted by the string (T2) on the rock (As said earlier, the string tension at B is more than that at point A. We mark this say, as T2)
Again if we consider the position of B, weight acts vertically downwards. Direction of T2 is again away from the rock and along the string, so at B it’s vertically upwards. As these 2 force components mutually act to supply the centripetal force at point B also, so we can write this equation for the point B
T2 – mg = (mV2^2)/R …………………….. (2)
Now let’s move towards deriving the expression of Critical Velocity.
At the highest point, let’s consider that the tension of the string is zero (T1 = 0) and the weight of the rock is sufficient enough to supply the centripetal force required to sustain the circular motion.
In that case, we can write,
mg = (mV1^2)/R
or, V1 = (gr)^(1/2) …………………….. (3) [Equation of critical velocity]
Please note that, this is the minimum velocity at point A which can keep the rock in its circular path. If its velocity falls below this value, then the string will slack and the rock will drop down finally.
This minimum velocity of a vertically rotating object at the highest point of rotation is called Critical Velocity.
Minimum Velocity at the Lowest Point to attain the Critical velocity at the highest point
We can easily calculate the expression of minimum value of V2 at point B, which is a must to maintain the Critical Velocity at point A.
Total Energy of the rock at point B = its Kinetic energy = ½ . mV2^2 ………….. (4)
Total energy of the rock at point A = its Kinetic Energy + its Potential Energy
= ½ .mV1^2 + mg. (2R)
= ½ .m.gR + 2mgR [ substituting V1 with critical velocity i.e. (gR)^(1/2)]
Equating equation 4 and 5 following the Law of conservation of Energy, we get,
V2^2 = 5gR
or, V2 = (5gR)^(1/2) …………..(6)
So we have studied the concept of Critical velocity as well as derived its expression. Also we have derived the equation of the minimum velocity required at the lowest point of the circle, to attain just the critical velocity at the highest point of circular track.
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Related Study: Centripetal force