Question

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

Answer 1

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,
$$\begin{gathered} S\to \theta \\ v\to \omega \\ a\to \alpha \\ F\to \tau \end{gathered}$$
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,
$\Rightarrow W=\tau \mathrm{\Delta }\theta$
In this case total angular displacement$\left(\theta \right)$,
$\Rightarrow \mathrm{\Delta }\theta ={\displaystyle \frac{3}{2}}\times 2\pi$
$\Rightarrow \mathrm{\Delta }\theta =3\pi$
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.