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

At any point in a vertical circle, the particle's mechanical energy (KE + PE, measured from a fixed reference) is conserved, allowing KE and PE to be expressed in terms of the angle turned through.

Consider a particle of mass $m$ performing vertical circular motion of radius $r$, with speed $v_L$ at the lowest point $L$ (taken as the reference level, $h=0$). At a point $P$ where the radius makes angle $\theta$ with the downward vertical (measured from $L$), the height risen is

$h = r(1-\cos\theta)$

By conservation of mechanical energy (gravity being the only external force doing work, tension does no work as it's always perpendicular to velocity):

$\frac{1}{2}mv_L^2 = \frac{1}{2}mv_P^2 + mgh$

Kinetic energy at $P$:

$KE_P = \frac{1}{2}mv_P^2 = \frac{1}{2}mv_L^2 - mgr(1-\cos\theta)$

Potential energy at $P$ (relative to the lowest point):

$PE_P = mgh = mgr(1-\cos\theta)$

At specific positions:

• At the lowest point ($\theta=0$): $PE = 0$, $KE = \tfrac12mv_L^2$ (maximum). …