Question

Expressionfork.e.ofarotating body

Answer 1

Answer:

Let us consider any one point of a rotating body , let it be at a distance of r1 from the axis

So it's linear velocity be v1 , it's angular Velocity be ω1.

It's Kinetic energy will be K1

$K1 = \dfrac{1}{2}m_{1} {v_{1}}^{2}$

$= > K1 = \dfrac{1}{2}m_{1} {(\omega_{1}r_{1})}^{2}$

$= > K1 = \dfrac{1}{2}m_{1} {\omega_{1}}^{2} {r_{1}}^{2}$

Similarly , all the other points will have similar type of Kinetic Energy .

Total Kinetic Energy will be sum of each particle's Kinetic Energy :

$= > K_{T} = \dfrac{1}{2}m_{1} {\omega_{1}}^{2} {r_{1}}^{2} + \dfrac{1}{2}m_{2} {\omega_{2}}^{2} {r_{2}}^{2} + ......$

Since they are rotating with same angular velocity , then ω1 = ω2 = ω3 = .......

$= > K_{T} = \dfrac{1}{2}m_{1} {\omega}^{2} {r_{1}}^{2} + \dfrac{1}{2}m_{2} {\omega}^{2} {r_{2}}^{2} + ......$

$= > K_{T} = \dfrac{1}{2} \{m_{1} {r_{1}}^{2} + m_{2} {r_{2}}^{2} + .... \} { \omega}^{2}$

$\displaystyle = > K_{T} = \dfrac{1}{2} \{ \sum (m {r}^{2}) \} { \omega}^{2}$

$\displaystyle = > K_{T} = \dfrac{1}{2} I { \omega}^{2}$

So final answer :

$\boxed{ \blue{ \huge{ \bold{K_{T} = \dfrac{1}{2} I { \omega}^{2} }}}}$

Answer 2

Solution :

⏭ Kinetic energy of a rigid body rotating about a given axis :

✏ Consider a rigid body rotating about a line AB with an angular speed ω.

✏ The i$_{th}$ particle is going in a circle of radius r$_i$ with a linear speed v$_i$ = ωr$_i$.

✏ The kinetic energy of the whole body is

$\star \: \sf \: KE = \sum \dfrac{1}{2} m_ i { \omega}^{2} {r _ i}^{2} = \dfrac{1}{2} \sum(m _ i {r _ i}^{2} ) { \omega}^{2} \\ \\ \star \: \boxed{ \red{ \sf{ KE = \dfrac{1}{2} I { \omega}^{2} }}}$

✏ Sometimes it is called rotational kinetic energy.

✏ It is not a new kind of energy as is clear from the derivation.

✏ It is the sum of 1/2mv$^2$ of all the particles.