Question

what is dimensional of Rotational kinetic energy please full details​

Answer 1

Answer:

it is equal to dimension of energy because addition and subtraction is only valid under same dimensions ( in this case we add it with transitional kinetic energy)

by the way dimensions of

moment of inertia =ML^2

square of angular velocity= T^-2

on multiplying both dimension and f rotational kinetic energy=M L^2T^-2

Answer 2

Answer:-

Rotational Kinetic Energy:- It is the energy which is possessed by the rotational motion of a body. It is a scalar quantity. If force acting on the body is 0 then

$\sf{E \propto \: \frac{1}{E I} \propto \: \frac{1}{ {MK}^{2} } }$

Explaination:-

Let we know about equations in detail

Consider:-

(a) It is a rigid body and is rotating in it's axis with angular velocity ω.

$\sf{(b) \: The \: body \: is \: considered \: to \: be \: composed} \\ \sf{of \: masses \: m_{1} , m_{2} , m_{3}.....}$

(c) Also consider the attachment as reference.

Then:-

$\sf{The \: linear \: velocity \: of \: the \: particle \: will} \\ \sf{become \: as \: v_{1} = \omega \: r_{1}, \: v_{1} = \omega \: r_{1}......}$

Therefore:-

The total energy of the rotating body will be as

$\sf{KE_{r} = \frac{1}{2} m_1} {v}^{2} _{1} + \frac{1}{2} m_{2} {v}^{2} _{2} + ....... \\ \\ \sf{KE_{r} = { \frac{1}{2} (m_{1} {v}^{2} _{1} + m_{2} {v}^{2} _{2} + ......) { \omega}^{2} }} \\ \\ {\sf \boxed{{KE_{r} = \frac{1}{2}I { \omega}^{2} }}}$

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