The rigid body shown in Fig.$\mathbf{\text{10-57}}$consists of three particles connected by massless rods. It is to be rotated about an axis perpendicular to its plane through point P. If$\mathbf{M}\mathbf{\text{=0.40kg}}$ , $\mathbf{a}\mathbf{\text{=30cm}}$ , and $\mathbf{b}\mathbf{\text{=50cm}}$, how much work is required to take the body from rest to an angular speed of $\mathbf{\text{5.0\hspace{0.17em}rad/s}}$?

i) Mass of three particles isM, 2M, 3M whereas $M=0.40\text{\hspace{0.17em}kg}$

ii) Distance, $a=30\text{cm}=0.3\text{m}$and $b=50\text{cm}=0.5\text{m}$

iii) Angular speed,$\omega =5.0\text{\hspace{0.17em}rad}/\text{s}$$\omega =5.0\text{\hspace{0.17em}rad}/\text{s}$

When a rigid body rotates about an axis perpendicular to its plane through point p, the work required to take from rest to an angular speed of$\mathbf{\text{5.0rad/s}}$can be found by using the work energy theorem. Therefore, by using the formula of work-energy theorem, find the amount of work required to take the body from rest to an angular speed of$\mathbf{\text{5.0rad/s}}$

The work-energy theorem is given as-

$\begin{array}{c}W=\mathrm{\Delta KE}\\ \mathbf{=}\frac{1}{2}{\mathbf{I\omega }}_f^2\mathbf{-}\frac{1}{2}{\mathbf{I\omega }}_i^2\end{array}$

where, W is work done, KE is kinetic energy, Iis moment of inertia and $\mathbf{\omega }$ is angular velocity.

According to the work-energy theorem,

$W=\frac{1}{2}I{\omega }_f^2-\frac{1}{2}I{\omega }_i^2$

But, moment of inertia of the body is,

$I=\sum m{r}^2$

For lower masses 2M, its distances from the axis of rotation is,

$r_1=r_2=a$

Similarly, for particle on the top, its distance from the axis of rotation is,

$r_3=\sqrt{b^2-a^2}$

Therefore,

$I=m{a}^2+m{a}^2+m{\left(\sqrt{b^2-a^2}\right)}^2$

$I=2M{a}^2+2M{a}^2+M{\left(\sqrt{b^2-a^2}\right)}^2$

$I=2\left(0.40\text{\hspace{0.17em}kg}\right){\left(0.3\text{\hspace{0.17em}m}\right)}^2+2\left(0.40\text{\hspace{0.17em}kg}\right){\left(0.3\text{\hspace{0.17em}m}\right)}^2+\left(0.40\text{\hspace{0.17em}kg}\right){\left(\sqrt{{(0.5\text{m})}^2-{(0.3\text{\hspace{0.17em}m})}^2}\right)}^2$

$I=0.208\text{kg}.{\text{m}}^2$

Therefore,

$W=\frac{1}{2}(0.208\text{kg}.{\text{m}}^2){\left(5.0\frac{\text{rad}}{\text{s}}\right)}^2-\frac{1}{2}(0.208\text{\hspace{0.17em}kg}.{\text{m}}^2){\left(0.0\frac{\text{rad}}{\text{s}}\right)}^2$

$W=2.6\text{J}$

Hence, the amount of work required to take the body from rest to an angular speed of$5.0\text{\hspace{0.17em}rad}/\text{s}$is$2.6\text{\hspace{0.17em}J}$.