A pot maker rotates a pot making wheel of radius $3m$ by applying a force of $200N$ tangentially. If wheel completes exactly $\dfrac{3}{2}$ revolutions, work done by him is:
A) $5654.86J$
B) $4321.32J$
C) \[4197.5J\]
D) $1884.96J$

Hint: While calculating work done in a rotational motion, we need to consider all the angular equivalents of linear parameters. Maximum angular calculations are done in radians not in degrees.

Complete step by step answer:
In rotational motion the angular equivalents of linear parameters are,
$
  S \to \theta \\
  v \to \omega \\
  a \to \alpha \\
  F \to \tau $
So, To find work done, we need to find torque$\left( \tau \right)$ first,
$\Rightarrow \tau = r \times F$
$\Rightarrow \tau = 3 \times 200$
$\Rightarrow \tau = 600Nm$
Now to calculate work done by the man,
$\Rightarrow W = \int {\tau d\theta } $
Since torque is constant,
$\Rightarrow W = \tau \Delta \theta $
In this case total angular displacement$\left( \theta \right)$,
$\Rightarrow \Delta \theta = \dfrac{3}{2} \times 2\pi $
$\Rightarrow \Delta \theta = 3\pi $
So, $W = 600 \times 3\pi $
$\Rightarrow W = 1800\pi J$
$\Rightarrow W = 5654.86J$

Therefore, the correct answer is option A.

Note: Work done can also be calculated by work-energy theorem in a rotational motion. It says that work done during a rotational motion is equal to change in kinetic energy. In vector form the dot product of torque vector and radial vector is known as work done.