An object of mass $m$ moves in a circle of radius $r$ - AQA - A-Level Physics - Question 28 - 2019 - Paper 1

Since the object is moving in a circle and completing $n$ revolutions every second, its angular velocity ($heta$) can be calculated as:

$heta = 2\pi n$

The tangential velocity ($v$) can be derived from angular velocity as:

$v = r\theta = r(2\pi n) = 2\pi nr$.

Substituting $v = 2\pi nr$ into the kinetic energy formula gives:

$KE = \frac{1}{2} m (2\pi nr)^2 \newline = \frac{1}{2} m (4\pi^2 n^2 r^2) \newline = 2 m \pi^2 n^2 r^2$

The correct form of the answer can be represented depending on the factors involved. Among the given answer choices, the closest corresponding expression to our derived equation is:

Answer: C) $2 m n^2 r^2$.