A particle of mass $m = 0.5 ext{ kg}$ moves along a circle of radius $R = 2.0 ext{ m}$ on a smooth

Question

A particle of mass $m = 0.5	ext{ kg}$ moves along a circle of radius $R = 2.0	ext{ m}$ on a smooth horizontal plane. Its normal acceleration varies with time as $w_n = a t^2$, where $a = 8.0	ext{ m/s}^4$. Find the instantaneous power (in $	ext{W}$) developed by all forces acting on the particle at $t = 2.0	ext{ s}$.

Professor PC Thomas Classes · Answer

The normal acceleration is given by $w_n = \frac{v^2}{R} = a t^2$. Solving for speed gives $v(t) = \sqrt{a R} t$. The tangential acceleration is $w_t = \frac{dv}{dt} = \sqrt{a R}$. Only the tangential force does work, since the normal force is always perpendicular to velocity. The instantaneous power is $P(t) = F_t v = (m w_t) v = m \sqrt{a R} (\sqrt{a R} t) = m a R t$. Substituting $m = 0.5	ext{ kg}$, $a = 8.0	ext{ m/s}^4$, $R = 2.0	ext{ m}$, and $t = 2.0	ext{ s}$, we get $P(2) = 0.5 	imes 8.0 	imes 2.0 	imes 2.0 = 16	ext{ W}$.