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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) $1884.96J$

Last updated date: 11th Sep 2024
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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.