91Ó°ÊÓ

i) Mass of three particles isM, 2M, 3M whereas $M=0.40\text{ â¶Ä‹k²µ}$

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

iii) Angular speed,$\mathrm{Ó\neg }=5.0\text{ r²¹»å}/\text{s}$$\mathrm{Ó\neg }=5.0\text{ r²¹»å}/\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-

$\begin{array}{c}W=\mathrm{Δ°­·¡}\\ \mathbf{=}\frac{1}{2}{\mathbf{IÓ\neg }}_f^2\mathbf{-}\frac{1}{2}{\mathbf{IÓ\neg }}_i^2\end{array}$

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

According to the work-energy theorem,

$W=\frac{1}{2}I{\mathrm{Ó\neg }}_f^2−\frac{1}{2}I{\mathrm{Ó\neg }}_i^2$

But, moment of inertia of the body is,

$I=∑m{r}^2$

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,

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

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

$I=2\left(0.40\text{ k²µ}\right){\left(0.3\text{″¾}\right)}^2+2\left(0.40\text{ k²µ}\right){\left(0.3\text{″¾}\right)}^2+\left(0.40\text{ k²µ}\right){\left(\sqrt{{(0.5\text{ }\text{m})}^2−{(0.3\text{″¾})}^2}\right)}^2$

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

Therefore,

$W=\frac{1}{2}(0.208\text{ }\text{kg}.{\text{m}}^2){\left(5.0\text{ }\frac{\text{rad}}{\text{s}}\right)}^2−\frac{1}{2}(0.208\text{ k²µ}.{\text{m}}^2){\left(0.0\text{ }\frac{\text{rad}}{\text{s}}\right)}^2$

$W=2.6\text{ }\text{J}$

Hence, the amount of work required to take the body from rest to an angular speed of$5.0\text{ r²¹»å}/\text{s}$is$2.6\text{ J}$.