Imagine that a body of mass (m) attached to one end of a massless and inextensible string executes circular motion in a vertical plane with the other end of the string fixed. The length of the string becomes the radius (r) of the circular path (See figure).
Let us discuss the motion of the body by taking the free body diagram (FBD) at a position where the position vector (\(\vec r\)) makes an angle θ with the vertically downward direction and the instantaneous velocity is as shown in Figure.
There are two forces acting on the mass.
1. Gravitational force which acts downward
2. Tension along the string.
Applying Newton’s second law on the mass, in the tangential direction,
mg sinθ = m\(a_t\)
mg sinθ = -m (\(\frac{dv}{dt}\)) .............. (i)
Where, \(a_t\) = \(-\frac{dv}{dt}\) is tangential retardation
In the radial direction,
T - mg cosθ = m\(a_r\)
T - mg cosθ =\(\frac{mv^2}{r}\) ..........(ii)
Where, \(a_r\) = \(\frac{v^2}{r}\) is the centripetal acceleration.
The circle can be divided into four sections A, B, C, D for better understanding of the motion.
The four important facts to be understood from the two equations are as follows:
(i) The mass is having tangential acceleration (g sin θ) for all values of θ (except θ = 0°), it is clear that this vertical circular motion is not a uniform circular motion
(ii) From the equations (ii) and (i) it is understood that as the magnitude of velocity is not a constant in the course of motion, the tension in the string is also not constant.
Hence velocity cannot vanish, even when the tension vanishes.
These points are to be kept in mind while solving problems related to motion in vertical circle.
To start with let us consider only two positions, say the lowest point 1 and the highest point 2 as shown in Figure for further analysis. Let the velocity of the body at the lowest point 1 be \(\vec v_1\), at the highest point 2 be \(\vec v_2\) and \(\vec v\) at any other point. The direction of velocity is tangential to the circular path at all points. Let \(\vec T_1\) be the tension in the string at the lowest point and \(\vec T_2\) be , the tension at the highest point and \(\vec T\) be the tension at any other point. Tension at each point acts towards the center. The tensions and velocities at these two points can be found by applying the law of conservation of energy.
For the lowest point (1)
When the body is at the lowest point 1, the gravitational force \(m\vec g\) which acts on the body (vertically downwards) and another one is the tension \(\vec T_1\) , acting vertically upwards, i.e. towards the center. From the equation (ii), we get
For the highest point (2)
At the highest point 2, both the gravitational force mg on the body and the tension T2 act downwards,
i.e. towards the center again.
From equations (iv) and (ii), it is understood that T1 > T2. The difference in tension T1 – T2 is obtained by subtracting equation (iv) from equation (ii).
The term [\(v^2_1 - v^2_2\)] can be found easily by applying law of conservation of energy at point 1 and also at point 2.
Note: The tension will not do any work on the mass as the tension and the direction of motion is always perpendicular.
The gravitational force is doing work on the mass, as it is a conservative force the total energy of the mass is conserved throughout the motion.
Total energy at point 1 (E1) is same as the total energy at a point 2 (E2)
E1 = E2
Potential energy at point 1, U1 = 0 (by taking reference as point 1)
Similarly, Potential energy at point 2, U2 = mg (2r) (h is 2r from point 1)
From the law of conservation of energy given in equation (vi), we get
Substituting equation (vii) in equation (iv) we get,
Therefore, the difference in tension is
T1 – T2 = 6 mg …(viii)
Minimum speed at the highest point (2) The body must have a minimum speed at point 2 otherwise, the string will slack before reaching point 2 and the body will not loop the circle. To find this minimum speed let us take the tension T2 = 0 in equation (iv).
The body must have a speed at point 2, \(v_2 \geq \sqrt{gr}\) to stay in the circular path.
Maximum speed at the lowest point 1 To have this minimum speed (\(v_2 = \sqrt{gr}\)) at point 2, the body must have minimum speed also at point 1.
By making use of equation (vii) we can find the minimum speed at point 1.
Substituting equation (ix) in (vii),
The body must have a speed at point 1,
\(v_1 \geq \sqrt{5gr}\) to stay in the circular path. From equations (ix) and (x), it is clear that the minimum speed at the lowest point 1 should be v 5 times more than the minimum speed at the highest point 2, so that the body loops without leaving the circle.
