Obtain an expression for the critical vertical of a body revolving in a vertical circle.

Explanation

Imagine that a body of mass (m) attached to one end of a massless and inextensible string executes circular motion ir a vertical plane with the other end of the string fixed. The length of the string becomes the radius (F) 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 (7) makes an angle 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 sin theta =ma_t`
`mg sin theta=-m ((dv)/(dt))`
where, `a+(t)=-(dv)/(dt)` is tangential retardation.
In the radial direction,
`T-mg cos theta = ma_r`
`T-mg cos theta = (mv^(2))/(r)`
where, `a_(r)=v^(2)/r` is the centripetal acceleration.
The circle can be divided into four sections A, B, C and D for better understanding of the motion. The four important facts to be understood from the two equations are as follows:
(1) The mass is having tangential acceleration (g sine) for all values of 0 (except 0 = 0), it 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 constant in the course of motion, the tension in the string is also not constant.
(iii) The equation (ii), `T=mg cos theta=(mv^(2))/(r)` highlight that in sections A and D of the circles, `("for "-pi/2 lt theta lt pi/2, cos theta " is positive")`, the term `mg cos theta` is always greatest than zero. Hence the tension cannot vanish even when the velocity vanishes.
(iv) The equation (ii), `(mv^(2))/r=T-mg cos theta` further highlights that in sections B and C of the circles `("for "pi/2 lt theta lt (3pi)/(2), cos theta" is negative")`, the second term is always greater than zero. Hence velocity cannot vanish, even when the tension vanisher These points are to be kept in mind while solving problems related to motion in vanishes.
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 `vecv_(1)` at the highest point be and v at any other point. The direction of velocity is tangential to the circular path at all points. Let `vecT_(1)` The the tension in the string at the lowest point and `vecT_(2)` be the tension at the highest point and `vecT` be the tensionat any other point. Tension at cach point acts towards the center. The tensions and velocities at these two points can be found by applying the law of conservation of Motion in vertical dels hown for energy.
For the lowest point (1) When the body is at the lowest point (1), the gravitational force `m vecg` me which acts on the body (vertically downwards) and another one is the tension `vecT_1` noting vertically upwards, i.e., towards the center. From the equation (b), we get
`T_(1)-mg=(mv_(1)^(2))/(r)` .........(1)
`T_(1)=(mv_(1)^(2))/(r)+mg` ....(2)
For the highest point (2)
At the highest point 2, both the gravitational force `m vecg` on the body and the tension downwards, i.e., towards the center again. `T_(2)+mg =(mv_(2)^(2))/(r_(2))` ........(3)
`T_(2)=(mv^(2))/(r_(2))-mg` ..........(4)
From equations (4) and (2), it is understood that `T_(1) gt T_(2)` The difference in tension `T_(1)-T_(2)` is obtained by subtracting equation (4) from equation (2)
`T_(1)-T_(2)=(mv_(1)^(2))/(r)+mg-((mv_(2)^(2))/(r)-mg) =(mv_(1)^(2))/(r)+mg-(mv_(2)^(2))/(r)+mg`
`T_(1)-T_(2)=m/r [v_(1)^(2)-v_(2)^(2)]+2mg ........(5)`
The term `[v_(1)^(2)-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 pointl `1 (E_1)` is same as the total energy at a point `2 (E_2)`
`E_1= E_2`
Potential energy at point 1, `U_(1)=0` (by taking reference as point 1)
Kinetic energy at point 1, `KE_1 = mv_(1)^(2)`
Total Energy at point `1, E_(1)=U_(1)+KE_(1)=0+1/2 mv_(1)^(2)=1/2 mv_(1)^(2)`
Similarly, Potential energy at point 2, `U_2= mg (2r) `(h is 2r from point 1)
Kinetic Energy at point `2, KE=1/2 mv_(2)^(2)`
Total energy at point `2, E_(2)=U_(2)+KE_(2)=2mgr+1/2 mv_(2)^(2)`
From the law of conservation of energy given in equation (6), we get `1/2 mv_(1)^(2)=2mgr+1/2 mv_(2)^(2)`
After rearranging `1/2 m(v_(1)^(2)-v_(2)^(2))=2mgr `
`v_(1)^(2)-v_(2)^(2)=4gr ......(7)`
Substituting equation in in equations
`T_(1)-T_(2)=m/r [4gr] +2mg`
Therefore, the difference in tension is
`T_(1)-T_(2)=6mg ........(8)`
Minimum speed at the highest point (2)
The body must have a minimum speed upon the wind reaching point and the body will not loop het take the tension `T_2=0` in equation) (4)
`0=(mv_(2)^(2))/(r)-mg`
`(mv_(2)^(2))/(r)=mg`
`v_(2)^(2)=mg`
`v_(2)^(2)=rg`
`v_(2)=sqrt(gr)` ......(9)
The body must have a speed at point, `2, v_(2) sqrtgr` to stay in the circular path.
Maximum speed at the lowest point 1.
To have this minimum speed `(v_(2)=sqrt(gr))` pin the band we have also of point By making use of equation (7) we can find them the minimum speed at point 1.
`v_(1)^(2)-v_(2)^(2)=4gr`
Substituting equation (9) in (7),
`v_(1)^(2)-gr=4gr`
`v_(1)^(2)=5gr`
`v_(1)=sqrt5(gr) ........(10)`
The body must have a speed at point to say the work
From equations (9) and (10), 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.