Question

What is the discussion and conclusion for experiment MOMENT OF INERTIA OF ROTATING DISC KINETIC OF RIGID BODIES

Answer 1

The kinetic energy associated with the translation motion of the rigid body is,

K 2 -mu 2

But if a body (rigid) is rotating, then it must also have energy. For example in grinding operation the rotation energy of the grinding wheel is used to machine the component.

For a rigid rotating disk the velocity of each particle in different. The angular velocity is same for all the body. Let a small mass m_{​​​​​​j}​​​ have velocity v_{​​​​​j} then the total rotation kinetic energy is

$K=\sum \frac{1}{2}m_{j}v_{j}^{2}$

The velocity of each particle can be explained as the product of the radial distance r_{​​​​​​j} and angular velocity w_{​​​​​​j}​​​​​

$K=\sum \frac{1}{2}m_{j}(r_{j}w_{j})^{2}$​​​​​​

The angular velocity of each particle is same hence,

$K=\sum \frac{1}{2}m_{j}(r_{j})^{2}w^{2}$

$K=\frac{1}{2}w^{2}\sum m_{j}(r_{j})^{2}$

The new quantity which in the summation of small particle mass and the sequare of the radial distance of that small particle is the mass moment of inertia of the rotating body,

$I=\sum m_{j}(r_{j})^{2}$

Hence the expression of kinetic energy is,

$K=\frac{1}{2}Iw^{2}$

It ts the rotational kinetic energy of the body.