Question

A particle of mass m is rotating by means of a string
in a vertical circle. The difference in the tension at the
bottom and top would be- Please tell !

Answer 1

Let the length of the string be $\sf{L,}$ which is also the radius of the vertical circle.

Free body diagram of the particle at the bottom most point is given below.

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put(0,0){\circle*{1}}\put(0,0){\vector(0,-1){10}}\put(0,0){\vector(0,1){10}}\put(-6,-12){ \sf{mg} }\put(-5,10){ \sf{T_B} }\put(0,0){\vector(1,0){10}}\put(10.5,0){ \sf{v_B} }\put(-4,-1){ \sf{m} }\end{picture}$

Here,

• $\sf{mg=}$ Weight of the body.

• $\sf{T_B=}$ Tension in the string at the bottom most point.

• $\sf{v_B=}$ Velocity of the particle at this point.

The centripetal force experienced by the particle at this point is,

$\longrightarrow\sf{T_B-mg=\dfrac{m\left(v_B\right)^2}{L}}$

$\longrightarrow\sf{T_B=mg+\dfrac{m\left(v_B\right)^2}{L}\quad\quad\dots(1)}$

Free body diagram of the particle at the top most point is given below.

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put(0,0){\circle*{1}}\put(0,0){\vector(0,-1){10}}\put(-6,-13.3){ \sf{mg+T_T} }\put(0,0){\vector(-1,0){10}}\put(-14,0){ \sf{v_T} }\put(1.5,-1){ \sf{m} }\end{picture}$

Here,

• $\sf{T_T=}$ Tension in the string at the top most point.

• $\sf{v_T=}$ Velocity of the particle at this point.

The centripetal force experienced by the particle at this point is,

$\longrightarrow\sf{mg+T_T=\dfrac{m\left(v_T\right)^2}{L}}$

$\longrightarrow\sf{T_T=\dfrac{m\left(v_T\right)^2}{L}-mg\quad\quad\dots(2)}$

Well the total mechanical energy is conserved in this vertical circular motion.

$\longrightarrow\sf{K(B)+U(B)=K(T)+U(T)}$

If the bottom most point is taken as the base point,

$\longrightarrow\sf{\dfrac{1}{2}m(v_B)^2+mg(0)=\dfrac{1}{2}m(v_T)^2+mg(2L)}$

since the distance between the top and bottom most points is $\sf{2L.}$

$\longrightarrow\sf{\dfrac{1}{2}m\left[(v_B)^2-(v_T)^2\right]=2mgL}$

$\longrightarrow\sf{\dfrac{m\left[(v_B)^2-(v_T)^2\right]}{L}=4mg\quad\quad\dots(3)}$

To find the difference in the tension in the string each at the top and bottom most points, (2) should be subtracted from (1), because the tension in the bottom most point is the greatest and that at the top most point is the least, due to gravity.

So,

$\longrightarrow\sf{(1)-(2)}$

$\longrightarrow\sf{T_B-T_T=mg+\dfrac{m(v_B)^2}{L}-\dfrac{m(v_T)^2}{L}+mg}$

$\longrightarrow\sf{T_B-T_T=2mg+\dfrac{m\left[(v_B)^2-(v_T)^2\right]}{L}}$

From (3),

$\longrightarrow\sf{T_B-T_T=2mg+4mg}$

$\longrightarrow\underline{\underline{\sf{T_B-T_T}=\bf{6mg}}}$

Therefore, the difference in the tension in the string each at the top and bottom most points is 6 times the weight of the particle.