Vertical Circular Motion

Vertical Circular Motion

Motion of a Particle in a Vertical Circle attached to an In-Extensible light String: 
Suppose a particle of mass m is attached to an inextensible light string of length R. The particle is moving in a vertical circle of radius R about a fixed point O.  It is imparted a velocity u in horizontal direction at lowest point A. Let v be its velocity at point B of the circle as shown in figure. Here
     h = R(1 - cos\(\theta\) )                            . . . . .(i) 
    From conservation of mechanical energy 
        \( \frac{1} {2}m(u^2 - v^2 ) = mgh \)   or     v² = u² - 2gh               . . . . . (ii) 
              

The necessary centripetal force is provided by the resultant of tension 
    T and mg cos\(\theta\)
    T - mg cos\(\theta\) = \( \frac{{mv^2 }} {R} \)              . . . . .(iii) 
Now, following three conditions arise depending on the value of u. 

Case - 1 : (u\( \geqslant \sqrt {5gR} \)  )       
The particle will complete the circle if the string does not slack even at the highest point ( \( \theta = \pi \) ).  Thus, tension in the string should be greater than or equal to zero (T \( \geqslant \) 0) at  \( \theta = \pi \) . In critical case, substituting T = 0 and \( \theta = \pi \)  in equation (iii), we get 
    mg =\( \frac{{mv^2 _{\min } }} {R} \)

or \( v_{\min }^2 = gR \)    
                                 

or    v_min =     (at highest point)   
    Substituting \(\theta\) =\(\pi\)  in equation (i) h = 2R 
    Therefore, from equation (ii)  \( u_{\min }^2 = v_{\min }^2 + 2gh \)    
    or    \( u_{\min }^2 = gR + 2g(2R) = 5gR \)    or  u_min = \( \sqrt {5gR} \) 
    Thus, if u\( \geqslant \sqrt {5gR} \)  the particle will complete the circle. 
    At u =\( \sqrt {5gR} \) velocity at highest point is v =\( \sqrt {gR} \) and tension in the string is zero. 
    Substituting  = 0⁰ and v = u = \( \sqrt {5gR} \) in equation (iii),
we get T = 6 mg or in the critical condition tension in the string at lowest position is 6 mg.  This is shown in figure. 
     If u < \( \sqrt {5gR} \)following two cases are possible. 

Vertical Circular Motion

Case - 2: \( \left( {\sqrt {2gR} < u < \sqrt {5gR} } \right) \) If u <\( \sqrt {5gR} \), the tension in the string will become zero before reaching the highest point. 
From equation (iii), tension in the string becomes zero (T = 0)
    where     cos\(\theta\) = \( \frac{{ - v^2 }} {{Rg}} \) or      cos \(\theta\)= \( \frac{{2gh - u^2 }} {{Rg}} \)  
    Substituting, this value of cos in equation (i), we get 
          \( \frac{{2gh - u^2 }} {{Rg}} = 1 - \frac{h} {R} \)           \( h = h_1 = \frac{1} {3}\left( {\frac{{u^2 + Rg}} {g}} \right) \)  . . .  .(iv)
or we can say that at height h1 tension in the string becomes zero. Further, if u < \( \sqrt {5gR} , \) 
velocity of the particle becomes zero when 
    0 = u² - 2gh        or         h = \( \frac{{u^2 }} {{2g}} = h_2 \)    (say)        . . . .(v)         
 i.e. at height h₂ velocity of particle becomes zero. 
               

Now, the particle will leave the circle if tension in the string becomes zero but velocity is not zero or 
    T = 0 but v  0. This is possible only when   h₁ < h₂ 
    or      \( \frac{{u^2 + Rg}} {{3g}} < \frac{{u^2 }} {{2g}} \)  or   2u² + 2Rg < 3u² 
    or   u² > 2Rg            or  u >\( \sqrt {2Rg} \)
Therefore, if \( \sqrt {2gR} < u < \sqrt {5gR} , \) the particle leaves the circle. 
From equation (v), we can see that h > R if u² > 2gR. Thus, the particle, will leave the circle when h > R or  90⁰ < \(\theta\) < 180⁰.  This situation is shown in the figure. 
      \( \sqrt {2gR} < u < \sqrt {5gR} , \)  or   90⁰ < \(\theta\) < 180⁰ 
 

Vertical Circular Motion

 Case - 3: (0 < u <\( \sqrt {2gR} \)  ) 
The particle will oscillate if velocity of the particle becomes zero but tension in the string is not zero.
                 

 or   v = 0, but T \( \ne \) 0. This is possible when 
          h₂ < h₁ 
    or  \( \frac{{u^2 }} {{2g}} < \frac{{u^2 + Rg}} {{3g}} \)   or   3u² < 2u² + 2Rg    
    or    u² < 2Rg  or u < \( \sqrt {2Rg} \)
Further, if h₁ = h₂, u =\( \sqrt {2Rg} \)  and tension and velocity both becomes zero simultaneously. 
Further, from equation (v), we can say that h < R if u < \( \sqrt {2Rg} \). 
Thus, for 0 < u <\( \sqrt {2Rg} \)  particle oscillates in lower half of the circle (0⁰ < \(\theta\) < 90⁰). This situation is shown in the figure, 
     0 < u <\( \sqrt {2Rg} \)     or   0⁰ < \(\theta\) < 90⁰ .
   

Example : A heavy particle hanging from a fixed point by a light inextensible string of length l is projected horizontally with speed \(\sqrt gl\)  Find the speed of the particle and the inclination of the string to the vertical at the instant of the motion when the tension in the string is equal to the weight of the particle. 

Solution:  
Let tension in the string becomes equal to the weight of the particle when particle reaches the point B and deflection of the string from vertical is\(\theta\) . Resolving mg along the string and perpendicular to the string, we get net radial force on the particle at B i.e.  
         FR = T - mg cos\(\theta\)               . . . . (i)
If v be the speed of the particle at B, then 
          FR = \( \frac{{mv^2 }} {l} \)                . . . . (ii)  
    From (i) and (ii), we get 
    T - mg cos\(\theta\) = \( \frac{{mv^2 }} {l} \)            . . . . (iii)  
    Since at B, T = mg 
      mg(1 - cos\(\theta\)) = \( \frac{{mv^2 }} {l} \) 
     v² = gl(1 - cos\(\theta\))                . . . . .(iv)
Conserving the energy of the particle at point A and B, we have
           \( \frac{1} {2}mv_0^2 = mgl(1 - \cos \theta ) + \frac{1} {2}mv^2 \)
    where v₀ = \( \sqrt {gl} \) and v = \( \sqrt {gl(1 - \cos \theta )} \) 
      gl = 2gl(1 - cos\(\theta\)) + gl(1 - cos\(\theta\))  
     cos\(\theta\) = 2/3                  . . . . .(v)
Putting the value of cos in equation (iv) we get      v  = \( \sqrt {\frac{{gl}} {3}} \).