Question

Q] Derive an expression for the difference in tensions at highest and lowest point for a
particle performing vertical circular motion​

Answer 1

✠ Question Given :

• Derive an expression for the difference in tensions at highest and lowest point for a particle performing vertical circular motion.

✠ Required Solution :

• At Diagram :

• ⇒ T - mgCosθ = mv²/ R ( centrifugal force)

• ⇒ T = mv²/ R - mgCosθ

✯ At Highest Point :

• ( θ = 180° , Cos180° = -1 )

• ⇒ T min = mv²/R - mg

• For string not to be sla.ck

• ⇒ T min > 0 ( looping to loop)

• ⇒ mv²/R - mg > 0

• ⇒ V = √ rg

✯ At Lowest Point :

• ⇒ ( Cos180°= +1 , θ = 0 )

• ⇒ Tmax = mv²/R + mg ____ (eq1)

• Apply conservation of energy

• ⇒ 1/2 mv² ± 1/2 mv² + mg (2r)

• ⇒ 1/2 mv² > 1/2 mv² + mg (r2)

• ⇒V ²/2 > 5gr /2

✠ V = √ 5gr ( min velocity required)

• Putting equation (2) in (1)

• ⇒ Tmax = m (5gr) /r + mg

• ⇒ Tmax = 6 mg

✯ At Middle Point :

• By law of conservation of energy

• ⇒ Tm = TL

• ⇒ 1/2 mv² + mg(r) = 1/2 mv²

• ⇒ 1/2 mv² > m(√5gr) ² - mgr

• ⇒ 1/2 mv² > 5mgr/ 2 - mgr

• ⇒ 1/2mv² > 3mgr/2

• ⇒ Vm = √ 3 great

• ⇒ T > m ( 3gr) /r + mg cosθ

• ⇒ T > 3 mg

✠ Note :

• ⇒ Required Diagram is in attachment :D

_____________________________

Answer 2

Explanation:

I attached one picture moderator please refer it