Question
A particle of mass m, just completes the vertical circular motion. Derive the expression for the difference in tensions at the highest and the lowest points
✓
Answer
Consider a body of mass ' $M$ ' moving in a vertical circle of radius ' $R$ '. At any position $P$, the forces are acting body are weight Mg vertically downward and tension ' $T$ ' towards centre. Centripetal force $F _{ C }= T - Mg \cos \theta$
$T=FC+Mg \cos \theta$
The body is moving such that it can move in a vertical circle.
At position \(B : \theta=180^{\circ}, \cos \theta=-1 . T_B=F_C-M g\)
\(\therefore T_B=\frac{M V_B^2}{R}-M g\)
\(\therefore T_B=\frac{M g R}{R}-M g\)
\(\therefore T_B=0\)
At position A : \(\theta=0^{\circ}, \cos \theta=1, T_B=F_C-M g\)
\(T_A=F_C+M g\)
\(\therefore T_A=\frac{M V_A^2}{R}+M g\)
\(\therefore T_A=\frac{5 M g R}{R}+M g\)
\(\therefore T_A=6 M g\)
Difference in tension \(T_B-T_A=6 Mg\)
$T=FC+Mg \cos \theta$
The body is moving such that it can move in a vertical circle.
At position \(B : \theta=180^{\circ}, \cos \theta=-1 . T_B=F_C-M g\)
\(\therefore T_B=\frac{M V_B^2}{R}-M g\)
\(\therefore T_B=\frac{M g R}{R}-M g\)
\(\therefore T_B=0\)
At position A : \(\theta=0^{\circ}, \cos \theta=1, T_B=F_C-M g\)
\(T_A=F_C+M g\)
\(\therefore T_A=\frac{M V_A^2}{R}+M g\)
\(\therefore T_A=\frac{5 M g R}{R}+M g\)
\(\therefore T_A=6 M g\)
Difference in tension \(T_B-T_A=6 Mg\)
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