Question

The expression for rotational kinetic energy is​

Answer 1

${ \huge \bf { \mid{ \overline{ \underline{Question}}} \mid}}$

The expression for rotational kinetic energy is?

$\huge{\bold{\underline{\underline{Answer}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Let I be the Moment of Inertia.
• Let ω be the Angular velocity.

$\huge{\bold{\underline{Explanation:-}}}$

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Let us Consider a rigid body rotating with a constant angular velocity "ω" about an axis.

As the body rotates, all the particles will be in uniform circular motion.

Kinetic energy of 1st Particle,

$\large{\tt K.E_1 = \dfrac{1}{2}mv^2}$

But we know,

⇒ v = rω,

Substituting it,

$\large{\tt \leadsto K.E_1 = \dfrac{1}{2} \times m_1 \times (r_1 \omega)^2}$

(As mass , & Radius will vary for each particle)

$\large{ \leadsto{\underline{\tt K.E_1 = \dfrac{1}{2} \times m_1 \times r_1^2 \: \omega^2}}}$

Similarly Kinetic energy of 2nd Particle.

$\large{\tt K.E_2 = \dfrac{1}{2}mv^2}$

But we know,

⇒ v = rω,

Substituting it,

$\large{\tt \leadsto K.E_2 = \dfrac{1}{2} \times m_2 \times (r_2 \omega)^2}$

(As mass & Radius will vary for each particle)

$\large{ \leadsto {\underline{\tt K.E_2 = \dfrac{1}{2} \times m_2 \times r_2^2 \: \omega^2}}}$

Note:-

• Angular velocity will be same

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Kinetic energy of the body will be..

$\large{\boxed{\tt K.E_R = K.E_1 + K.E_2 + ..... + K.E_n}}$

Substituting the values,

$\large{\tt \leadsto K.E_R = \dfrac{1}{2}. m_1 . r_1^2 \: \omega^2 + \dfrac{1}{2}. m_2. r_2^2 \: \omega^2 + ... +\dfrac{1}{2} m_n r_n^2 \: \omega^2}$

$\large{\tt \leadsto K.E_R = \dfrac{1}{2} . \omega^2 \bigg[m_1 . r_1^2 + m_2. r_2^2 + ... + m_n r_n^2 \bigg] }$

$\large{\tt \leadsto K.E_R = \dfrac{1}{2} . \omega^2 \bigg[\displaystyle \tt \sum^N_{i=1}m_ir_i^2 \bigg]}$

As we know,

$\large{\leadsto \displaystyle \tt \sum^N_{i=1}m_ir_i^2 = I (Moment \: of \: Inertia)}$

The equation becomes,

$\huge{\boxed{\boxed{\tt K.E_R = \dfrac{1}{2}I \omega^2}}}$

Hence derived!

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