Question

Obtain expressions of energy at different positions in the vertical circular motion.

Answer 1

energy at any point is (Kp) = 1/2 ( mv^2)

Answer 2

"Energy of a body at the lowest point L when looping a loop:

Energy of the body has two components a) kinetic and b) potential.

$E_{ K }=\frac { 1 }{ 2 } mv^{ 2 }=\frac { 1 }{ 2 } m(\sqrt { 5gr } )^{ 2 }=\frac { 5 }{ 2 } mgr$

Therefore, $R_p = mgh = mg(0) = 0$

$E_{ T }=E_{ K }+E_{ P }=\frac { 5 }{ 2 } mgr+0=\frac { 5 }{ 2 } mgr$

Energy of a body at the highest point L when looping a loop:

Energy of the body has two components a) kinetic and b) potential.

$E_{ K }=\frac { 1 }{ 2 } mv^{ 2 }=\frac { 1 }{ 2 } m(\sqrt { gr } )^{ 2 }=1/2mgr$

Therefore, $R_p = mgh = mg(2r) = 2mgr$

$E_{ T }=E_{ K }+E_{ P }=\frac { 1 }{ 2 } mgr+2mgr=\frac { 5 }{ 2 } mgr$

Energy of a body when string is horizontal and when looping a loop:

Energy of the body has two components a) kinetic and b) potential.

$E_{ K }=\frac { 1 }{ 2 } mv^{ 2 }=\frac { 1 }{ 2 } m(\sqrt { 3gr } )^{ 2 }=\frac { 3 }{ 2 } mgr$

Therefore, $R_p = mgh = mg(r) = mgr$

$E_{ T }=E_{ K }+E_{ P }=\frac { 3 }{ 2 } mgr+mgr=\frac { 5 }{ 2 } mgr$"