Question
Derive an expression for the difference in tensions at the highest and lowest point for a particle performing the vertical circular motion.
✓
Answer
i. Suppose a body of mass 'm' performs V.C.M on a circle of radius $r$ as shown in the figure.
ii. Let,
$T _{ L }=$ tension at the lowest point
$T _{ H }=$ tension at the highest point
$V_L=$ velocity at the lowest point
$v_H=$ velocity at the highest point
iii. At lowest point $L$,
$T _{ L }=\frac{ mv _{ L }^2}{ r }+ mg$
At highest point $H _{\text {, }}$
$T _{ H }=\frac{ mv _{ H }^2}{ r }- mg$
iv. Subtracting (1) by (2),
$ T _{ L }- T _{ H }=\frac{ mv _{ L }^2}{ r }+ mg -\left(\frac{ mv _{ H }^2}{ r }- mg \right)$
$=\frac{ m }{ r }\left( v _{ L }^2- v _{ H }^2\right)+2 mg$
$\therefore T _{ L }- T _{ H }=\frac{ m }{ r }\left( v _{ L }^2- v _{ H }^2\right)+2 mg \ldots \ldots \ldots . .(3) $
$v$. By law of conservation of energy,
$ ( P . E + K . E ) \text { at } L =( P . E + K . E ) \text { at } H$
$\therefore 0+\frac{1}{2} mv _{ L }^2= mg \cdot 2 r +\frac{1}{2} mv _{ H }^2$
$\therefore \frac{1}{2} m \left( v _{ L }^2- v _{ H }^2\right)= mg \cdot 2 r$
$\therefore v _{ L }^2- v _{ H }^2=4 gr \ldots \ldots . .(4) $
vi. From equation (3) and (4),
$ T _{ L }- T _H=\frac{ m }{ r }(4 gr )+2 mg =4 mg +2 mg$
$\therefore T _{ L }- T _{ H }=6 mg $
ii. Let,
$T _{ L }=$ tension at the lowest point
$T _{ H }=$ tension at the highest point
$V_L=$ velocity at the lowest point
$v_H=$ velocity at the highest point
iii. At lowest point $L$,
$T _{ L }=\frac{ mv _{ L }^2}{ r }+ mg$
At highest point $H _{\text {, }}$
$T _{ H }=\frac{ mv _{ H }^2}{ r }- mg$
iv. Subtracting (1) by (2),
$ T _{ L }- T _{ H }=\frac{ mv _{ L }^2}{ r }+ mg -\left(\frac{ mv _{ H }^2}{ r }- mg \right)$
$=\frac{ m }{ r }\left( v _{ L }^2- v _{ H }^2\right)+2 mg$
$\therefore T _{ L }- T _{ H }=\frac{ m }{ r }\left( v _{ L }^2- v _{ H }^2\right)+2 mg \ldots \ldots \ldots . .(3) $
$v$. By law of conservation of energy,
$ ( P . E + K . E ) \text { at } L =( P . E + K . E ) \text { at } H$
$\therefore 0+\frac{1}{2} mv _{ L }^2= mg \cdot 2 r +\frac{1}{2} mv _{ H }^2$
$\therefore \frac{1}{2} m \left( v _{ L }^2- v _{ H }^2\right)= mg \cdot 2 r$
$\therefore v _{ L }^2- v _{ H }^2=4 gr \ldots \ldots . .(4) $
vi. From equation (3) and (4),
$ T _{ L }- T _H=\frac{ m }{ r }(4 gr )+2 mg =4 mg +2 mg$
$\therefore T _{ L }- T _{ H }=6 mg $
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