Question

Obtain an relation for the kinetic energy of the rotation. Hence define the moment of inertia of a body.

Accepted Answer

Let a body moving s with angular velocity ω will also has a translational motion. Let the velocity for translation motion be ‘’v’’.

We know for translational motion the energy poses by body is kinetic and given as

Let a rigid body rotates with constant angular velocity, all the particles perform uniform circular motion with the same angular velocity $ω$ . However, various particles of the body have different linear speeds depending upon their distances from the axis of rotation.
The linear speed of the particle with mass $E_{1} = \frac{1}{2} m_{1} {v_{1}}^{2} = m_{1} {(r_{1} ω)}^{2} . . . . . . . . . . (1)$

Similarly, the kinetic energy of the particle with mass $m_{2}$ is
$E_{1} = \frac{1}{2} m_{2} {v_{2}}^{2} = m_{2} {(r_{2} ω)}^{2} . . . . . . . . . . (1)$ ,and so on.
The rotational kinetic energy (E) of the body is
$E = E_{1} + E_{2} + E_{3 +} . . . . . . . . . . . E_{n}$

= $E_{1} = m_{1} {(r_{1} ω)}^{2} + m_{2} {(r_{2} ω)}^{2} + m_{3} {(r_{3} ω)}^{2} + . . . . .$

= $[m_{1} {r^{2}}_{1} + m_{2} {r^{2}}_{2} + m_{3} {r^{2}}_{3} + . . . .] ω^{2} = {Iω}^{2}$

where ${Σmr}^{2}$ = moment of inertia

Substituting we get, $KE = \frac{1}{2} {Iω}^{2}$

G L Mittal and TARUN MITTAL Solutions for Chapter: Rotational Motion of a Rigid Body : Moment of Inertia, Exercise 1: QUESTIONS

G L Mittal Physics Solutions for Exercise - G L Mittal and TARUN MITTAL Solutions for Chapter: Rotational Motion of a Rigid Body : Moment of Inertia, Exercise 1: QUESTIONS

Questions from G L Mittal and TARUN MITTAL Solutions for Chapter: Rotational Motion of a Rigid Body : Moment of Inertia, Exercise 1: QUESTIONS with Hints & Solutions

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