Question

A particle of mass M just completes the vertical circular motion, derive the expression for the difference in tensions at the highest and lowest points ​

Answer 1

The figure shows the FBD of the particle at the lowest point.

$\setlength{\unitlength}{1mm}\begin {picture}(6,5)\put(0,0){\circle{3}}\put(0,1.5){\vector(0,1){10}}\put(0,-1.5){\vector(0,-1){10}}\put(0,-14){ Mg }\put(0,12){ T_L }\put(1.5,0){\vector(1,0){10}}\put(13,0){ v_L }\put(-10,0){.}\end {picture}$

Since the particle is in circular motion, the weight of the particle and tension in the string provides the centripetal force to the particle, i.e.,

$T_L-Mg=\dfrac {M(v_L)^2}{R}\\\\\\T_L=Mg+\dfrac {M(v_L)^2}{R}$

where R is the length of the string.

The total mechanical energy of the particle at the lowest point, taking this point as the mean position (means h = 0), will be,

$E_L=\dfrac {1}{2}M(v_L)^2+Mg(0)\\\\\\E_L=\dfrac {1}{2}M(v_L)^2$

The figure shows the FBD of the particle at the highest point.

$\setlength{\unitlength}{1mm}\begin {picture}(6,5)\put(0,0){\circle{3}}\put(0,-1.5){\vector(0,-1){10}}\put(-8,-15){ Mg+T_H }\put(-1.5,0){\vector(-1,0){10}}\put(-17,-1){ v_H }\put(-10,10){.}\end {picture}$

As at the lowest point, the tension in the string and the weight of the particle provides the centripetal force for the particle, i.e.,

$Mg+T_H=\dfrac {M(v_H)^2}{R}\\\\\\T_H=\dfrac {M(v_H)^2}{R}-Mg$

Here the total mechanical energy of the particle will be (h = 2R),

$E_H=\dfrac {1}{2}M(v_H)^2+Mg(2R)\\\\\\E_H=\dfrac {1}{2}M(v_H)^2+2MgR$

By the law of conservation of mechanical energy, we have,

$E_H=E_L\\\\\\\dfrac {1}{2}M(v_H)^2+2MgR=\dfrac {1}{2}M(v_L)^2\\\\\\\dfrac {1}{2}M(v_L)^2-\dfrac {1}{2}M(v_H)^2=2MgR\\\\\\\dfrac {(v_L)^2-(v_H)^2}{2}=2gR\\\\\\(v_L)^2-(v_H)^2=4gR\quad\longrightarrow\quad(1)$

Then, the difference in tensions at the highest and the lowest points will be,

$T_L-T_H=\left (Mg+\dfrac {M(v_L)^2}{R}\right)-\left(\dfrac {M(v_H)^2}{R}-Mg\right)\\\\\\T_L-T_H=Mg+\dfrac {M(v_L)^2}{R}-\dfrac {M(v_H)^2}{R}+Mg\\\\\\T_L-T_H=2Mg+\dfrac {M}{R}[(v_L)^2-(v_H)^2]\\\\\\T_L-T_H=2Mg+\dfrac {M}{R}\cdot 4gR\quad\quad[\text{From (1)}]\\\\\\T_L-T_H=2Mg+4Mg\\\\\\\boxed{\boxed{T_L-T_H=6Mg}}$

I.e., the difference in the tensions in the string at the highest and the lowest points is six times the weight of the particle. From the equation we can see that $T_L\ \textgreater\ T_H.$

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